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Let L:R2→R2L: R^{2} \rightarrow R^{2} be defined by \[ L(\boldsymbol{x})=L\binom{x_...
May 3, 2024
Let L:R2→R2L: R^{2} \rightarrow R^{2} be defined by L(x)=L(x1x2)=(x1+x2−x1) L(\boldsymbol{x})=L\binom{x_{1}}{x_{2}}=\binom{x_{1}+x_{2}}{-x_{1}} We want to show that LL is a linear transformation by showing that L(ax+by)=aL(x)+bL(y) L(a \boldsymbol{x}+b \boldsymbol{y})=a L(\boldsymbol{x})+b L(\boldsymbol{y}) for all scalars a,ba, b. We'll first compute the left side of this equation in two steps (remembering to use WebWork's angular bracket notation (e.g., ⟨x1,x2⟩)\langle x 1, x 2\rangle) for vectors. Enter subscripted variables x1x_{1} as x1x 1.): ax+by=a⟨x1,x2⟩+b⟨y1,y2⟩= a \boldsymbol{x}+b \boldsymbol{y}=a\langle x 1, x 2\rangle+b\langle y 1, y 2\rangle= □\square Now take LL of the preceding vector to get: L(ax+by)= L(a \boldsymbol{x}+b \boldsymbol{y})= □\square OK, now let's compute the right side of the equation. First, aL(x)=a L(\boldsymbol{x})= □\square and bL(y)=b L(\boldsymbol{y})= □\square So aL(x)+bL(y)=a L(\boldsymbol{x})+b L(\boldsymbol{y})= □\square If your answers are correct, the two sides should be equal.
Solution by Steps
step 1
To show that LL is a linear transformation, we need to verify that L(ax+by)=aL(x)+bL(y)L(a \boldsymbol{x} + b \boldsymbol{y}) = a L(\boldsymbol{x}) + b L(\boldsymbol{y})
step 2
First, we compute ax+by=a⟨x1,x2⟩+b⟨y1,y2⟩a \boldsymbol{x} + b \boldsymbol{y} = a\langle x_1, x_2\rangle + b\langle y_1, y_2\rangle
step 3
Distribute the scalars aa and bb to get ⟨ax1,ax2⟩+⟨by1,by2⟩\langle a x_1, a x_2\rangle + \langle b y_1, b y_2\rangle
step 4
Add the vectors component-wise to get ⟨ax1+by1,ax2+by2⟩\langle a x_1 + b y_1, a x_2 + b y_2\rangle
step 5
Now apply LL to the vector ⟨ax1+by1,ax2+by2⟩\langle a x_1 + b y_1, a x_2 + b y_2\rangle to get L(⟨ax1+by1,ax2+by2⟩)L(\langle a x_1 + b y_1, a x_2 + b y_2\rangle)
step 6
Using the definition of LL, we get L(⟨ax1+by1,ax2+by2⟩)=⟨(ax1+by1)+(ax2+by2),−(ax1+by1)⟩L(\langle a x_1 + b y_1, a x_2 + b y_2\rangle) = \langle (a x_1 + b y_1) + (a x_2 + b y_2), -(a x_1 + b y_1) \rangle
step 7
Next, we compute aL(x)=aL(⟨x1,x2⟩)=a⟨x1+x2,−x1⟩a L(\boldsymbol{x}) = a L(\langle x_1, x_2\rangle) = a \langle x_1 + x_2, -x_1 \rangle
step 8
Distribute the scalar aa to get a⟨x1+x2,−x1⟩=⟨a(x1+x2),−ax1⟩a \langle x_1 + x_2, -x_1 \rangle = \langle a(x_1 + x_2), -a x_1 \rangle
step 9
Similarly, compute bL(y)=bL(⟨y1,y2⟩)=b⟨y1+y2,−y1⟩b L(\boldsymbol{y}) = b L(\langle y_1, y_2\rangle) = b \langle y_1 + y_2, -y_1 \rangle
step 10
Distribute the scalar bb to get b⟨y1+y2,−y1⟩=⟨b(y1+y2),−by1⟩b \langle y_1 + y_2, -y_1 \rangle = \langle b(y_1 + y_2), -b y_1 \rangle
step 11
Now add aL(x)a L(\boldsymbol{x}) and bL(y)b L(\boldsymbol{y}) component-wise to get ⟨a(x1+x2)+b(y1+y2),−ax1−by1⟩\langle a(x_1 + x_2) + b(y_1 + y_2), -a x_1 - b y_1 \rangle
step 12
Finally, compare the results from step 6 and step 11 to verify that L(ax+by)=aL(x)+bL(y)L(a \boldsymbol{x} + b \boldsymbol{y}) = a L(\boldsymbol{x}) + b L(\boldsymbol{y})
Answer
L(ax+by)=aL(x)+bL(y)L(a \boldsymbol{x} + b \boldsymbol{y}) = a L(\boldsymbol{x}) + b L(\boldsymbol{y}) is verified, hence LL is a linear transformation.
Key Concept
Linear Transformation
Explanation
A function L:Rn→RmL: \mathbb{R}^n \rightarrow \mathbb{R}^m is a linear transformation if it satisfies the properties of additivity and homogeneity of degree 1, which are shown by the equation L(ax+by)=aL(x)+bL(y)L(a \boldsymbol{x} + b \boldsymbol{y}) = a L(\boldsymbol{x}) + b L(\boldsymbol{y}) for all vectors x,y\boldsymbol{x}, \boldsymbol{y} in Rn\mathbb{R}^n and scalars a,ba, b.
(1 point) Let L:R2→R2L: R^{2} \rightarrow R^{2} be defined by L(x)=L(x1x2)=(x1x2−x1) L(\boldsymbol{x})=L\binom{x_{1}}{x_{2}}=\binom{x_{1} x_{2}}{-x_{1}} Show that LL is not a linear transformation by finding vectors x\boldsymbol{x}, and, y\boldsymbol{y} such that L(x+y)≠L(x)+L(y)L(\boldsymbol{x}+\boldsymbol{y}) \neq L(\boldsymbol{x})+L(\boldsymbol{y}) x=⟨x1,x2⟩=\boldsymbol{x}=\left\langle x_{1}, x_{2}\right\rangle= □\square y=⟨y1,y2⟩=\boldsymbol{y}=\left\langle y_{1}, y_{2}\right\rangle= □\square Prove your answer by calculating (for your choice of x,y\boldsymbol{x}, \boldsymbol{y} ): L(x+y)=L(\boldsymbol{x}+\boldsymbol{y})= □\square and L(x)+L(y)= L(\boldsymbol{x})+L(\boldsymbol{y})= □\square
Solution by Steps
step 1
To show that LL is not a linear transformation, we need to find vectors x\boldsymbol{x} and y\boldsymbol{y} such that L(x+y)≠L(x)+L(y)L(\boldsymbol{x}+\boldsymbol{y}) \neq L(\boldsymbol{x})+L(\boldsymbol{y})
step 2
Let's choose x=⟨1,0⟩\boldsymbol{x} = \langle 1, 0 \rangle and y=⟨0,1⟩\boldsymbol{y} = \langle 0, 1 \rangle
step 3
Calculate L(x+y)L(\boldsymbol{x}+\boldsymbol{y}): Since x+y=⟨1+0,0+1⟩=⟨1,1⟩\boldsymbol{x}+\boldsymbol{y} = \langle 1+0, 0+1 \rangle = \langle 1, 1 \rangle, we apply LL to get L(⟨1,1⟩)=(1⋅1−1)=(1−1)L(\langle 1, 1 \rangle) = \binom{1 \cdot 1}{-1} = \binom{1}{-1}
step 4
Calculate L(x)L(\boldsymbol{x}): Applying LL to x=⟨1,0⟩\boldsymbol{x} = \langle 1, 0 \rangle gives L(⟨1,0⟩)=(1⋅0−1)=(0−1)L(\langle 1, 0 \rangle) = \binom{1 \cdot 0}{-1} = \binom{0}{-1}
step 5
Calculate L(y)L(\boldsymbol{y}): Applying LL to y=⟨0,1⟩\boldsymbol{y} = \langle 0, 1 \rangle gives L(⟨0,1⟩)=(0⋅1−0)=(00)L(\langle 0, 1 \rangle) = \binom{0 \cdot 1}{-0} = \binom{0}{0}
step 6
Calculate L(x)+L(y)L(\boldsymbol{x})+L(\boldsymbol{y}): Adding the results from steps 4 and 5 gives (0−1)+(00)=(0−1)\binom{0}{-1} + \binom{0}{0} = \binom{0}{-1}
step 7
Compare L(x+y)L(\boldsymbol{x}+\boldsymbol{y}) and L(x)+L(y)L(\boldsymbol{x})+L(\boldsymbol{y}): We have L(x+y)=(1−1)L(\boldsymbol{x}+\boldsymbol{y}) = \binom{1}{-1} and L(x)+L(y)=(0−1)L(\boldsymbol{x})+L(\boldsymbol{y}) = \binom{0}{-1}. Since (1−1)≠(0−1)\binom{1}{-1} \neq \binom{0}{-1}, LL is not a linear transformation
Answer
L(x+y)=(1−1)L(\boldsymbol{x}+\boldsymbol{y}) = \binom{1}{-1} and L(x)+L(y)=(0−1)L(\boldsymbol{x})+L(\boldsymbol{y}) = \binom{0}{-1}, thus L(x+y)≠L(x)+L(y)L(\boldsymbol{x}+\boldsymbol{y}) \neq L(\boldsymbol{x})+L(\boldsymbol{y}), proving LL is not a linear transformation.
Key Concept
Linear Transformation Properties
Explanation
A linear transformation LL must satisfy L(x+y)=L(x)+L(y)L(\boldsymbol{x}+\boldsymbol{y}) = L(\boldsymbol{x})+L(\boldsymbol{y}) for all vectors x\boldsymbol{x} and y\boldsymbol{y}. If this condition fails for any vectors, then LL is not a linear transformation. In this case, the chosen vectors x\boldsymbol{x} and y\boldsymbol{y} demonstrate that LL does not preserve vector addition, hence it is not linear.
Let L:R2→R2L: R^{2} \rightarrow R^{2} be a linear transformation. If L((1,2)T)=(2,3)TL\left((1,2)^{T}\right)=(2,3)^{T}, then L((−3,−6)T)=L\left((-3,-6)^{T}\right)=
Solution by Steps
step 1
Given that L:R2→R2L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} is a linear transformation and L((1,2)T)=(2,3)TL\left((1,2)^{T}\right)=(2,3)^{T}, we want to find L((−3,−6)T)L\left((-3,-6)^{T}\right)
step 2
By the properties of linear transformations, we know that L(av)=aL(v)L(a\mathbf{v})=aL(\mathbf{v}) for any vector v\mathbf{v} in R2\mathbb{R}^{2} and scalar aa
step 3
Notice that (−3,−6)T=−3(1,2)T(-3,-6)^{T} = -3(1,2)^{T}. Applying the linear transformation property from step 2, we get L((−3,−6)T)=L(−3(1,2)T)=−3L((1,2)T)L\left((-3,-6)^{T}\right) = L\left(-3(1,2)^{T}\right) = -3L\left((1,2)^{T}\right)
step 4
Substituting the known value of L((1,2)T)L\left((1,2)^{T}\right), we have L((−3,−6)T)=−3(2,3)TL\left((-3,-6)^{T}\right) = -3(2,3)^{T}
step 5
Calculating the scalar multiplication, we get L((−3,−6)T)=(−6,−9)TL\left((-3,-6)^{T}\right) = (-6,-9)^{T}
Answer
L((−3,−6)T)=(−6,−9)TL\left((-3,-6)^{T}\right) = (-6,-9)^{T}
Key Concept
Linear Transformation and Scalar Multiplication
Explanation
The solution relies on the property of linear transformations that states L(av)=aL(v)L(a\mathbf{v})=aL(\mathbf{v}), where aa is a scalar and v\mathbf{v} is a vector. This property allows us to find the image of any scalar multiple of a vector under the transformation.
(1 point) If T:R3→R3T: R^{3} \rightarrow R^{3} is a linear transformation such that T([100])=[−3−43],T[010])=[−411],T[001])=[42−1], \left.\left.T\left(\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]\right)=\left[\begin{array}{c} -3 \\ -4 \\ 3 \end{array}\right], T\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right)=\left[\begin{array}{c} -4 \\ 1 \\ 1 \end{array}\right], T\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]\right)=\left[\begin{array}{c} 4 \\ 2 \\ -1 \end{array}\right], then T([5−1−1])=[□□□]T\left(\left[\begin{array}{c}5 \\ -1 \\ -1\end{array}\right]\right)=\left[\begin{array}{c}\square \\ \square \\ \square\end{array}\right]
Solution by Steps
step 1
To find the image of a vector under a linear transformation, we use the transformation matrix and multiply it by the vector
step 2
The transformation matrix is formed by the images of the basis vectors. In this case, the matrix is A=[−3amp;−4amp;4−4amp;1amp;23amp;1amp;−1]A = \begin{bmatrix} -3 & -4 & 4 \\ -4 & 1 & 2 \\ 3 & 1 & -1 \end{bmatrix}
step 3
The vector we want to transform is v⃗=[5−1−1]\vec{v} = \begin{bmatrix} 5 \\ -1 \\ -1 \end{bmatrix}
step 4
Multiply the transformation matrix AA by the vector v⃗\vec{v} to get the image under the transformation TT
step 5
The multiplication is Av⃗=[−3amp;−4amp;4−4amp;1amp;23amp;1amp;−1][5−1−1]=[−3(5)+−4(−1)+4(−1)−4(5)+1(−1)+2(−1)3(5)+1(−1)+−1(−1)]A\vec{v} = \begin{bmatrix} -3 & -4 & 4 \\ -4 & 1 & 2 \\ 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} 5 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3(5) + -4(-1) + 4(-1) \\ -4(5) + 1(-1) + 2(-1) \\ 3(5) + 1(-1) + -1(-1) \end{bmatrix}
step 6
Simplify the result to find the image of v⃗\vec{v} under TT
step 7
The simplified result is [−15+4−4−20−1−215−1+1]=[−15−2315]\begin{bmatrix} -15 + 4 - 4 \\ -20 - 1 - 2 \\ 15 - 1 + 1 \end{bmatrix} = \begin{bmatrix} -15 \\ -23 \\ 15 \end{bmatrix}
Answer
[−15−2315]\begin{bmatrix} -15 \\ -23 \\ 15 \end{bmatrix}
Key Concept
Linear Transformation of Vectors
Explanation
To find the image of a vector under a linear transformation, we multiply the transformation matrix by the vector. The transformation matrix is constructed using the images of the standard basis vectors.
Which of the following transformations are linear? A. {y1=0y2=x1x2\left\{\begin{array}{l}y_{1}=0 \\ y_{2}=x_{1} x_{2}\end{array}\right. B. {y1=−5x1y2=3x1y3=6x1\left\{\begin{array}{l}y_{1}=-5 x_{1} \\ y_{2}=3 x_{1} \\ y_{3}=6 x_{1}\end{array}\right. c. {y1=9x1+x2y2=−x1\left\{\begin{array}{l}y_{1}=9 x_{1}+x_{2} \\ y_{2}=-x_{1}\end{array}\right. D. {y1=x1+7y2=x2\left\{\begin{array}{l}y_{1}=x_{1}+7 \\ y_{2}=x_{2}\end{array}\right. E. {y1=0y2=4x2\left\{\begin{array}{l}y_{1}=0 \\ y_{2}=4 x_{2}\end{array}\right. F. {y1=x22y2=x3y3=x1\left\{\begin{array}{l}y_{1}=x_{2}^{2} \\ y_{2}=x_{3} \\ y_{3}=x_{1}\end{array}\right.
Solution by Steps
step 2
For transformation A, y1=0y_1 = 0 and y2=x1x2y_2 = x_1 x_2, we see that y2y_2 is not linear because it involves the product of x1x_1 and x2x_2
step 3
For transformation B, y1=−5x1y_1 = -5x_1, y2=3x1y_2 = 3x_1, and y3=6x1y_3 = 6x_1 are all linear because they satisfy both additivity and homogeneity of degree 1
step 4
For transformation C, y1=9x1+x2y_1 = 9x_1 + x_2 and y2=−x1y_2 = -x_1 are linear because they satisfy both properties of a linear transformation
step 5
For transformation D, y1=x1+7y_1 = x_1 + 7 is not linear because of the constant term 7, which violates homogeneity of degree 1
step 6
For transformation E, y1=0y_1 = 0 and y2=4x2y_2 = 4x_2 are linear because they satisfy both properties of a linear transformation
step 7
For transformation F, y1=x22y_1 = x_2^2 is not linear because it involves the square of x2x_2, which violates both additivity and homogeneity of degree 1
B, C, E
Key Concept
Linear Transformation
Explanation
A transformation is linear if it satisfies additivity and homogeneity of degree 1. Nonlinear terms, such as products or powers of variables, or constants added to variables, violate these properties.
(1 point) Which of the following transformations are linear? Select all of the linear transformations. There may be more than one correct answer. Be sure you can justify your answers. A. T(A)=ATT(A)=A^{T} from R3×5R^{3 \times 5} to R5×3R^{5 \times 3} B. T(A)=det⁡(A)T(A)=\operatorname{det}(A) from R5×5R^{5 \times 5} to RR C. T(A)=A[2amp;−69amp;4]−[2amp;69amp;1]AT(A)=A\left[\begin{array}{cc}2 & -6 \\ 9 & 4\end{array}\right]-\left[\begin{array}{ll}2 & 6 \\ 9 & 1\end{array}\right] A from R2×2R^{2 \times 2} to R2×2R^{2 \times 2} D. T(A)=3AT(A)=3 A from R6×5R^{6 \times 5} to R6×5R^{6 \times 5} E. T(A)=ASA−1T(A)=A S A^{-1} from R2×2R^{2 \times 2} to R2×2R^{2 \times 2}, where S=[−4amp;36amp;−9]S=\left[\begin{array}{cc}-4 & 3 \\ 6 & -9\end{array}\right] F. T(A)=[4amp;7−2amp;9]AT(A)=\left[\begin{array}{cc}4 & 7 \\ -2 & 9\end{array}\right] A from R2×3R^{2 \times 3} to R2×3R^{2 \times 3}
Solution by Steps
step 1
To determine if a transformation is linear, it must satisfy two properties: additivity (T(A+B)=T(A)+T(B)T(A+B) = T(A) + T(B)) and homogeneity (T(cA)=cT(A)T(cA) = cT(A)), where AA and BB are matrices and cc is a scalar
step 2
For transformation A, T(A)=ATT(A)=A^{T}, it satisfies both additivity and homogeneity because the transpose of a sum is the sum of the transposes, and the transpose of a scalar multiple is the scalar multiple of the transpose
step 3
For transformation B, T(A)=det⁡(A)T(A)=\operatorname{det}(A), it does not satisfy additivity because the determinant of a sum is not generally the sum of the determinants
step 4
For transformation C, T(A)=A[2amp;−69amp;4]−[2amp;69amp;1]AT(A)=A\left[\begin{array}{cc}2 & -6 \\ 9 & 4\end{array}\right]-\left[\begin{array}{ll}2 & 6 \\ 9 & 1\end{array}\right] A, it does not satisfy additivity because the subtraction of the constant matrix breaks the additivity property
step 5
For transformation D, T(A)=3AT(A)=3A, it satisfies both additivity and homogeneity because scalar multiplication distributes over matrix addition and scalar multiplication is associative
step 6
For transformation E, T(A)=ASA−1T(A)=ASA^{-1}, it satisfies both additivity and homogeneity because matrix multiplication is distributive over addition and associative with scalar multiplication
step 7
For transformation F, T(A)=[4amp;7−2amp;9]AT(A)=\left[\begin{array}{cc}4 & 7 \\ -2 & 9\end{array}\right] A, it satisfies both additivity and homogeneity because matrix multiplication is distributive over addition and associative with scalar multiplication
Answer
The linear transformations are A, D, E, and F.
Key Concept
Linear Transformation Properties
Explanation
A transformation is linear if it satisfies both additivity and homogeneity. Transformations A, D, E, and F satisfy these properties, while B and C do not.
(1 point) If T:P1→P1T: P_{1} \rightarrow P_{1} is a linear transformation such that T(1+2x)=−3−2xT(1+2 x)=-3-2 x and T(3+5x)=4−3xT(3+5 x)=4-3 x, then T(2−2x)= T(2-2 x)=
Solution by Steps
step 1
Given the linear transformation T:P1→P1T: P_{1} \rightarrow P_{1}, we know that T(1+2x)=−3−2xT(1+2x)=-3-2x and T(3+5x)=4−3xT(3+5x)=4-3x
step 2
To find T(2−2x)T(2-2x), we can express 2−2x2-2x as a linear combination of 1+2x1+2x and 3+5x3+5x
step 3
Notice that 2−2x=12(1+2x)+12(3+5x)−2x2-2x = \frac{1}{2}(1+2x) + \frac{1}{2}(3+5x) - 2x
step 4
Apply the linearity of TT to the expression: T(2−2x)=T(12(1+2x)+12(3+5x)−2x)T(2-2x) = T\left(\frac{1}{2}(1+2x) + \frac{1}{2}(3+5x) - 2x\right)
step 5
Using the linearity property, distribute TT over the sum and scalar multiples: T(2−2x)=12T(1+2x)+12T(3+5x)−T(2x)T(2-2x) = \frac{1}{2}T(1+2x) + \frac{1}{2}T(3+5x) - T(2x)
step 6
Since T(1+2x)=−3−2xT(1+2x)=-3-2x and T(3+5x)=4−3xT(3+5x)=4-3x, substitute these into the equation: T(2−2x)=12(−3−2x)+12(4−3x)−T(2x)T(2-2x) = \frac{1}{2}(-3-2x) + \frac{1}{2}(4-3x) - T(2x)
step 7
Simplify the expression: T(2−2x)=−32−x+2−32x−T(2x)T(2-2x) = -\frac{3}{2}-x + 2-\frac{3}{2}x - T(2x)
step 8
Since TT is a linear transformation, T(cx)=cT(x)T(cx) = cT(x) for any scalar cc. Therefore, T(2x)=2T(x)T(2x) = 2T(x)
step 9
We need to find T(x)T(x). Notice that x=12(1+2x)−12x = \frac{1}{2}(1+2x) - \frac{1}{2}
step 10
Apply TT to xx: T(x)=T(12(1+2x)−12)T(x) = T\left(\frac{1}{2}(1+2x) - \frac{1}{2}\right)
step 11
Using linearity: T(x)=12T(1+2x)−12T(1)T(x) = \frac{1}{2}T(1+2x) - \frac{1}{2}T(1)
step 12
Since T(1+2x)=−3−2xT(1+2x)=-3-2x and T(1)T(1) can be found by setting x=0x=0 in T(1+2x)T(1+2x), which gives T(1)=−3T(1)=-3
step 13
Substitute these into the equation: T(x)=12(−3−2x)−12(−3)T(x) = \frac{1}{2}(-3-2x) - \frac{1}{2}(-3)
step 14
Simplify to find T(x)T(x): T(x)=−32−x+32=−xT(x) = -\frac{3}{2}-x + \frac{3}{2} = -x
step 15
Now substitute T(x)T(x) into the expression for T(2x)T(2x): T(2x)=2T(x)=2(−x)=−2xT(2x) = 2T(x) = 2(-x) = -2x
step 16
Finally, substitute T(2x)T(2x) into the expression for T(2−2x)T(2-2x): T(2−2x)=−32−x+2−32x−(−2x)T(2-2x) = -\frac{3}{2}-x + 2-\frac{3}{2}x - (-2x)
step 17
Simplify the expression to get the final result: T(2−2x)=−32−x+2−32x+2x=12−xT(2-2x) = -\frac{3}{2}-x + 2-\frac{3}{2}x + 2x = \frac{1}{2} - x
Answer
T(2−2x)=12−xT(2-2x) = \frac{1}{2} - x
Key Concept
Linear Transformation and Linearity
Explanation
The key concept is understanding that a linear transformation preserves addition and scalar multiplication. This means that if TT is a linear transformation, then for any vectors vv and ww and any scalar cc, T(v+w)=T(v)+T(w)T(v+w)=T(v)+T(w) and T(cv)=cT(v)T(cv)=cT(v). By expressing the vector 2−2x2-2x as a linear combination of vectors that we know the transformation of, we can use these properties to find T(2−2x)T(2-2x).
(1 point) Which of the following transformations are linear? Select all of the linear transformations. There may be more than one correct answer. Be sure you can justify your answers. A. T(f(t))=f(5)T(f(t))=f(5) from P7P_{7} to RR B. T(f(t))=f′(t)+9f(t)+2T(f(t))=f^{\prime}(t)+9 f(t)+2 from C∞C^{\infty} to C∞C^{\infty} C. T(f(t))=t4f′(t)T(f(t))=t^{4} f^{\prime}(t) from P2P_{2} to P5P_{5} D. T(f(t))=f(t)f′(t)T(f(t))=f(t) f^{\prime}(t) from P3P_{3} to P5P_{5} E. T(f(t))=f′(t)+6f(t)T(f(t))=f^{\prime}(t)+6 f(t) from C∞C^{\infty} to C∞C^{\infty} F. T(f(t))=f(−t)T(f(t))=f(-t) from P8P_{8} to P8P_{8}
Solution by Steps
step 1
To determine if a transformation is linear, it must satisfy two properties: additivity (T(u+v)=T(u)+T(v)T(u + v) = T(u) + T(v)) and homogeneity of degree 1 (T(cu)=cT(u)T(cu) = cT(u)), where uu and vv are functions in the domain of TT, and cc is a scalar
step 2
For option A, T(f(t))=f(5)T(f(t)) = f(5), we check if T(f(t)+g(t))=T(f(t))+T(g(t))T(f(t) + g(t)) = T(f(t)) + T(g(t)) and T(cf(t))=cT(f(t))T(cf(t)) = cT(f(t)). Since T(f(t)+g(t))=f(5)+g(5)=T(f(t))+T(g(t))T(f(t) + g(t)) = f(5) + g(5) = T(f(t)) + T(g(t)) and T(cf(t))=cf(5)=cT(f(t))T(cf(t)) = cf(5) = cT(f(t)), A is a linear transformation
step 3
For option B, T(f(t))=f′(t)+9f(t)+2T(f(t)) = f'(t) + 9f(t) + 2, we check the same properties. The constant term 2 violates additivity, so B is not a linear transformation
step 4
For option C, T(f(t))=t4f′(t)T(f(t)) = t^4 f'(t), we check the properties. Since T(cf(t))=t4(cf(t))′=ct4f′(t)=cT(f(t))T(cf(t)) = t^4 (cf(t))' = ct^4 f'(t) = cT(f(t)) and T(f(t)+g(t))=t4(f(t)+g(t))′=t4f′(t)+t4g′(t)=T(f(t))+T(g(t))T(f(t) + g(t)) = t^4 (f(t) + g(t))' = t^4 f'(t) + t^4 g'(t) = T(f(t)) + T(g(t)), C is a linear transformation
step 5
For option D, T(f(t))=f(t)f′(t)T(f(t)) = f(t)f'(t), we check the properties. The product f(t)f′(t)f(t)f'(t) does not satisfy additivity, so D is not a linear transformation
step 6
For option E, T(f(t))=f′(t)+6f(t)T(f(t)) = f'(t) + 6f(t), we check the properties. Since T(cf(t))=(cf(t))′+6cf(t)=cf′(t)+6cf(t)=cT(f(t))T(cf(t)) = (cf(t))' + 6cf(t) = cf'(t) + 6cf(t) = cT(f(t)) and T(f(t)+g(t))=(f(t)+g(t))′+6(f(t)+g(t))=f′(t)+g′(t)+6f(t)+6g(t)=T(f(t))+T(g(t))T(f(t) + g(t)) = (f(t) + g(t))' + 6(f(t) + g(t)) = f'(t) + g'(t) + 6f(t) + 6g(t) = T(f(t)) + T(g(t)), E is a linear transformation
step 7
For option F, T(f(t))=f(−t)T(f(t)) = f(-t), we check the properties. Since T(cf(t))=f(−ct)=cf(−t)=cT(f(t))T(cf(t)) = f(-ct) = cf(-t) = cT(f(t)) and T(f(t)+g(t))=f(−t)+g(−t)=T(f(t))+T(g(t))T(f(t) + g(t)) = f(-t) + g(-t) = T(f(t)) + T(g(t)), F is a linear transformation
Answer
The linear transformations are A, C, E, and F.
Key Concept
Linear Transformation Properties
Explanation
A transformation is linear if it satisfies additivity and homogeneity of degree 1. Options A, C, E, and F satisfy these properties, while options B and D do not due to the presence of a constant term and product of functions, respectively.
(1 point) Let T:P3→P3T: P_{3} \rightarrow P_{3} be the linear transformation satisfying T(1)=3x2+9,T(x)=−4x+7,T(x2)=−5x2−x−1. T(1)=3 x^{2}+9, \quad T(x)=-4 x+7, \quad T\left(x^{2}\right)=-5 x^{2}-x-1 . Find the image of an arbitrary quadratic polynomial ax2+bx+ca x^{2}+b x+c. T(ax2+bx+c)= T\left(a x^{2}+b x+c\right)=
Solution by Steps
step 1
To find the image of ax2+bx+cax^2+bx+c under the transformation TT, we express it as a linear combination of the basis elements 11, xx, and x2x^2
step 2
We apply the transformation TT to each term separately: T(ax2)=aT(x2)T(ax^2) = aT(x^2), T(bx)=bT(x)T(bx) = bT(x), and T(c)=cT(1)T(c) = cT(1)
step 3
Substitute the given transformations: T(x2)=−5x2−x−1T(x^2)=-5x^2-x-1, T(x)=−4x+7T(x)=-4x+7, and T(1)=3x2+9T(1)=3x^2+9
step 4
Calculate the image of each term: T(ax2)=a(−5x2−x−1)T(ax^2)=a(-5x^2-x-1), T(bx)=b(−4x+7)T(bx)=b(-4x+7), and T(c)=c(3x2+9)T(c)=c(3x^2+9)
step 5
Combine the images of the terms to get the image of ax2+bx+cax^2+bx+c: T(ax2+bx+c)=a(−5x2−x−1)+b(−4x+7)+c(3x2+9)T(ax^2+bx+c) = a(-5x^2-x-1) + b(-4x+7) + c(3x^2+9)
step 6
Simplify the expression to find the final image: T(ax2+bx+c)=(−5a)x2+(−a−4b)x+(3c+7b−a)T(ax^2+bx+c) = (-5a)x^2 + (-a - 4b)x + (3c + 7b - a)
Answer
T(ax2+bx+c)=(−5a)x2+(−a−4b)x+(3c+7b−a)T(ax^2+bx+c) = (-5a)x^2 + (-a - 4b)x + (3c + 7b - a)
Key Concept
Linear Transformation of a Polynomial
Explanation
The image of a polynomial under a linear transformation is found by applying the transformation to each term of the polynomial and then combining the results. Each term of the polynomial is transformed separately, and the linearity of the transformation allows us to add the results to get the final image.
(1 point) Let VV be a vector space, and T:V→VT: V \rightarrow V a linear transformation such that T(5v⃗1+3v⃗2)=−3v⃗1+5v⃗2T\left(5 \vec{v}_{1}+3 \vec{v}_{2}\right)=-3 \vec{v}_{1}+5 \vec{v}_{2} and T(3v⃗1+2v⃗2)=−2v⃗1+5v⃗2T\left(3 \vec{v}_{1}+2 \vec{v}_{2}\right)=-2 \vec{v}_{1}+5 \vec{v}_{2}. Then T(v⃗1)=□v⃗1+□v⃗2T(v⃗2)=□v⃗1+□v⃗2 \begin{array}{l} T\left(\vec{v}_{1}\right)=\square \vec{v}_{1}+\square \vec{v}_{2} \\ T\left(\vec{v}_{2}\right)=\square \vec{v}_{1}+\square \vec{v}_{2} \end{array} T(4v⃗1−4v⃗2)=□v⃗1+□v⃗2 T\left(4 \vec{v}_{1}-4 \vec{v}_{2}\right)=\square \vec{v}_{1}+\square \vec{v}_{2}
Solution by Steps
step 1
To find T(v⃗1)T(\vec{v}_{1}) and T(v⃗2)T(\vec{v}_{2}), we need to solve the system of linear equations given by the transformation properties
step 2
We have two equations: T(5v⃗1+3v⃗2)=−3v⃗1+5v⃗2T(5\vec{v}_{1}+3\vec{v}_{2})=-3\vec{v}_{1}+5\vec{v}_{2} and T(3v⃗1+2v⃗2)=−2v⃗1+5v⃗2T(3\vec{v}_{1}+2\vec{v}_{2})=-2\vec{v}_{1}+5\vec{v}_{2}
step 3
Since TT is a linear transformation, we can express T(av⃗1+bv⃗2)T(a\vec{v}_{1}+b\vec{v}_{2}) as aT(v⃗1)+bT(v⃗2)aT(\vec{v}_{1})+bT(\vec{v}_{2})
step 4
Applying linearity to the first equation: 5T(v⃗1)+3T(v⃗2)=−3v⃗1+5v⃗25T(\vec{v}_{1})+3T(\vec{v}_{2})=-3\vec{v}_{1}+5\vec{v}_{2}
step 5
Applying linearity to the second equation: 3T(v⃗1)+2T(v⃗2)=−2v⃗1+5v⃗23T(\vec{v}_{1})+2T(\vec{v}_{2})=-2\vec{v}_{1}+5\vec{v}_{2}
step 6
We can now set up a system of equations based on the coefficients of v⃗1\vec{v}_{1} and v⃗2\vec{v}_{2} from steps 4 and 5
step 7
Let T(v⃗1)=av⃗1+bv⃗2T(\vec{v}_{1}) = a\vec{v}_{1} + b\vec{v}_{2} and T(v⃗2)=cv⃗1+dv⃗2T(\vec{v}_{2}) = c\vec{v}_{1} + d\vec{v}_{2}
step 8
Substituting T(v⃗1)T(\vec{v}_{1}) and T(v⃗2)T(\vec{v}_{2}) into the system gives us two new equations: 5a+3c=−35a+3c=-3 and 5b+3d=55b+3d=5, and 3a+2c=−23a+2c=-2 and 3b+2d=53b+2d=5
step 9
Solving this system of equations will give us the values of aa, bb, cc, and dd
step 10
After solving, we find that a=1a=1, b=0b=0, c=0c=0, and d=1d=1
step 11
Therefore, T(v⃗1)=1v⃗1+0v⃗2T(\vec{v}_{1}) = 1\vec{v}_{1} + 0\vec{v}_{2} and T(v⃗2)=0v⃗1+1v⃗2T(\vec{v}_{2}) = 0\vec{v}_{1} + 1\vec{v}_{2}
step 12
To find T(4v⃗1−4v⃗2)T(4\vec{v}_{1}-4\vec{v}_{2}), we use the linearity of TT again
step 13
We calculate T(4v⃗1−4v⃗2)T(4\vec{v}_{1}-4\vec{v}_{2}) as 4T(v⃗1)−4T(v⃗2)4T(\vec{v}_{1})-4T(\vec{v}_{2})
step 14
Substituting the values from step 11, we get T(4v⃗1−4v⃗2)=4(1v⃗1+0v⃗2)−4(0v⃗1+1v⃗2)T(4\vec{v}_{1}-4\vec{v}_{2}) = 4(1\vec{v}_{1} + 0\vec{v}_{2}) - 4(0\vec{v}_{1} + 1\vec{v}_{2})
step 15
Simplifying, we find T(4v⃗1−4v⃗2)=4v⃗1−4v⃗2T(4\vec{v}_{1}-4\vec{v}_{2}) = 4\vec{v}_{1} - 4\vec{v}_{2}
Answer
T(v⃗1)=1v⃗1+0v⃗2T(\vec{v}_{1}) = 1\vec{v}_{1} + 0\vec{v}_{2}, T(v⃗2)=0v⃗1+1v⃗2T(\vec{v}_{2}) = 0\vec{v}_{1} + 1\vec{v}_{2}, and T(4v⃗1−4v⃗2)=4v⃗1−4v⃗2T(4\vec{v}_{1}-4\vec{v}_{2}) = 4\vec{v}_{1} - 4\vec{v}_{2}.
Key Concept
Linearity of Transformations
Explanation
The key concept is using the properties of a linear transformation to solve for the images of basis vectors and then applying linearity to find the image of any vector expressed in terms of those basis vectors.
(1 point) Let VV be a vector space, v,u∈Vv, u \in V, and let T1:V→VT_{1}: V \rightarrow V and T2:V→VT_{2}: V \rightarrow V be linear transformations such that T1(v)=6v−2u,T1(u)=−6v+4u,T2(v)=4v−7u,T2(u)=7v+5u. \begin{array}{c} T_{1}(v)=6 v-2 u, \quad T_{1}(u)=-6 v+4 u, \\ T_{2}(v)=4 v-7 u, \quad T_{2}(u)=7 v+5 u . \end{array} Find the images of vv and uu under the composite of T1T_{1} and T2T_{2}. (T2T1)(v)=(T2T1)(u)= \begin{array}{l} \left(T_{2} T_{1}\right)(v)= \\ \left(T_{2} T_{1}\right)(u)= \end{array}
Solution by Steps
step 1
To find the images of vv and uu under the composite of T1T_1 and T2T_2, we first need to find the matrix representations of T1T_1 and T2T_2
step 2
The matrix representation of T1T_1 is [6amp;−2−6amp;4]\begin{bmatrix} 6 & -2 \\ -6 & 4 \end{bmatrix} and the matrix representation of T2T_2 is [4amp;−77amp;5]\begin{bmatrix} 4 & -7 \\ 7 & 5 \end{bmatrix}
step 3
To find the composite transformation T2T1T_2 T_1, we multiply the matrices of T1T_1 and T2T_2
step 4
The product of the matrices is [6amp;−2−6amp;4][4amp;−77amp;5]=[(6)(4)+(−2)(7)amp;(6)(−7)+(−2)(5)(−6)(4)+(4)(7)amp;(−6)(−7)+(4)(5)]\begin{bmatrix} 6 & -2 \\ -6 & 4 \end{bmatrix} \begin{bmatrix} 4 & -7 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} (6)(4) + (-2)(7) & (6)(-7) + (-2)(5) \\ (-6)(4) + (4)(7) & (-6)(-7) + (4)(5) \end{bmatrix}
step 5
Simplifying the matrix multiplication, we get [24−14amp;−42−10−24+28amp;42+20]=[10amp;−524amp;62]\begin{bmatrix} 24 - 14 & -42 - 10 \\ -24 + 28 & 42 + 20 \end{bmatrix} = \begin{bmatrix} 10 & -52 \\ 4 & 62 \end{bmatrix}
step 6
The matrix [10amp;−524amp;62]\begin{bmatrix} 10 & -52 \\ 4 & 62 \end{bmatrix} represents the composite transformation T2T1T_2 T_1
step 7
To find (T2T1)(v)\left(T_{2} T_{1}\right)(v), we multiply the composite matrix by the vector representation of vv, which is [10]\begin{bmatrix} 1 \\ 0 \end{bmatrix}, since vv is the first basis vector
step 8
Multiplying the composite matrix by the vector representation of vv, we get [10amp;−524amp;62][10]=[104]\begin{bmatrix} 10 & -52 \\ 4 & 62 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 4 \end{bmatrix}
step 9
To find (T2T1)(u)\left(T_{2} T_{1}\right)(u), we multiply the composite matrix by the vector representation of uu, which is [01]\begin{bmatrix} 0 \\ 1 \end{bmatrix}, since uu is the second basis vector
step 10
Multiplying the composite matrix by the vector representation of uu, we get [10amp;−524amp;62][01]=[−5262]\begin{bmatrix} 10 & -52 \\ 4 & 62 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -52 \\ 62 \end{bmatrix}
Answer
(T2T1)(v)=[104]\left(T_{2} T_{1}\right)(v) = \begin{bmatrix} 10 \\ 4 \end{bmatrix} and (T2T1)(u)=[−5262]\left(T_{2} T_{1}\right)(u) = \begin{bmatrix} -52 \\ 62 \end{bmatrix}.
Key Concept
Composite Linear Transformations
Explanation
To find the image of vectors under the composite of two linear transformations, we multiply the matrix representations of the transformations in the order they are applied and then apply the resulting matrix to the vectors.
(1 point) Find the characteristic polynomial of the matrix amp;A=[1amp;−49amp;−4]p(x)=□. \begin{array}{ll} & A=\left[\begin{array}{ll} 1 & -4 \\ 9 & -4 \end{array}\right] \\ p(x)=\square . \end{array}
Generated Graph
Solution by Steps
step 1
To find the characteristic polynomial of the matrix AA, we need to calculate the determinant of A−λIA - \lambda I, where II is the identity matrix and λ\lambda is a scalar
step 2
The matrix A−λIA - \lambda I is given by (1−λamp;−49amp;−4−λ)\begin{pmatrix} 1-\lambda & -4 \\ 9 & -4-\lambda \end{pmatrix}
step 3
The characteristic polynomial is found by calculating the determinant of A−λIA - \lambda I, which is (1−λ)(−4−λ)−(−4)(9)(1-\lambda)(-4-\lambda) - (-4)(9)
step 4
Expanding the determinant, we get λ2+3λ+32\lambda^2 + 3\lambda + 32
Answer
p(λ)=λ2+3λ+32p(\lambda) = \lambda^2 + 3\lambda + 32
Key Concept
Characteristic Polynomial
Explanation
The characteristic polynomial of a matrix is found by calculating the determinant of the matrix subtracted by λ\lambda times the identity matrix. It is a polynomial in λ\lambda that provides important information about the matrix, including its eigenvalues.
(1 point) Find the eigenvalues of the matrix C=[−10amp;−16amp;−120amp;0amp;06amp;8amp;8]. C=\left[\begin{array}{ccc} -10 & -16 & -12 \\ 0 & 0 & 0 \\ 6 & 8 & 8 \end{array}\right] . The eigenvalues are □\square (Enter your answers as a comma separated list. The list you enter should have repeated items if there are eigenvalues with multiplicity greater than one.)
Solution by Steps
step 1
To find the eigenvalues of the matrix CC, we calculate the determinant of C−λIC - \lambda I, where II is the identity matrix and λ\lambda is a scalar
step 2
The characteristic polynomial is obtained by calculating the determinant of C−λIC - \lambda I
step 3
Solving the characteristic polynomial will give us the eigenvalues of the matrix CC
step 4
According to the asksia-ll calculation, the eigenvalues are λ1=−4\lambda_1 = -4, λ2=2\lambda_2 = 2, and λ3=0\lambda_3 = 0
Answer
-4, 2, 0
Key Concept
Eigenvalues of a Matrix
Explanation
The eigenvalues of a matrix are the scalars λ\lambda for which there exists a non-zero vector vv such that Cv=λvCv = \lambda v. They are found by solving the characteristic polynomial of the matrix.
(1 point) Find eigenvalues and eigenvectors for the matrix [32amp;12−60amp;−22]\left[\begin{array}{cc}32 & 12 \\ -60 & -22\end{array}\right]. The smaller eigenvalue □\square has an eigenvector □\square has an eigenvector □\square The larger eigenvalue □\square □\square
Solution by Steps
step 1
To find the eigenvalues and eigenvectors of the matrix (32amp;12−60amp;−22)\begin{pmatrix} 32 & 12 \\ -60 & -22 \end{pmatrix}, we solve the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0
step 2
The characteristic equation for this matrix is (32−λ)(−22−λ)−(12)(−60)=0(32 - \lambda)(-22 - \lambda) - (12)(-60) = 0
step 3
Solving this equation gives us the eigenvalues λ1=8\lambda_1 = 8 and λ2=2\lambda_2 = 2
step 4
To find the eigenvectors, we substitute each eigenvalue into the equation (A−λI)v=0(A - \lambda I)v = 0 and solve for vv
step 5
For λ1=8\lambda_1 = 8, the eigenvector is v1=(−1,2)v_1 = (-1, 2)
step 6
For λ2=2\lambda_2 = 2, the eigenvector is v2=(−2,5)v_2 = (-2, 5)
Answer
The smaller eigenvalue is 22 with an eigenvector (−2,5)(-2, 5). The larger eigenvalue is 88 with an eigenvector (−1,2)(-1, 2).
Key Concept
Eigenvalues and Eigenvectors
Explanation
Eigenvalues are scalars associated with a linear system of equations (or a matrix) that, when multiplied by a non-zero vector (the eigenvector), result in the vector being scaled by that scalar, without changing its direction.
(1 point) Find the eigenvalues and eigenvectors of the matrix [−6amp;−4amp;6−6amp;−4amp;6−8amp;−4amp;8].  \left[\begin{array}{lll} -6 &amp; -4 &amp; 6 \\ -6 &amp; -4 &amp; 6 \\ -8 &amp; -4 &amp; 8 \end{array}\right] \text {. } From smallest to largest, the eigenvalues are \lambda_{1}<\lambda_{2}<\lambda_{3} where Note: you may want to use a graphing calculator to estimate the roots of the polynomial which defines the eigenvalues.
Solution by Steps
step 1
To find the eigenvalues and eigenvectors of the matrix, we first set up the characteristic polynomial by subtracting λ\lambda times the identity matrix from the given matrix and setting the determinant to zero
step 2
The characteristic polynomial is found to be λ3+6λ2−4λ=0\lambda^3 + 6\lambda^2 - 4\lambda = 0
step 3
Solving the characteristic polynomial, we find the eigenvalues: λ1=−4\lambda_1 = -4, λ2=2\lambda_2 = 2, and λ3=0\lambda_3 = 0
step 4
For each eigenvalue, we solve the system (A−λI)v=0(A - \lambda I)v = 0 to find the corresponding eigenvectors
step 5
The eigenvectors corresponding to the eigenvalues are v1=(1,1,1)v_1 = (1, 1, 1) for λ1=−4\lambda_1 = -4, v2=(1,1,2)v_2 = (1, 1, 2) for λ2=2\lambda_2 = 2, and v3=(1,0,1)v_3 = (1, 0, 1) for λ3=0\lambda_3 = 0
Answer
The eigenvalues in ascending order are λ1=−4\lambda_1 = -4, λ2=0\lambda_2 = 0, λ3=2\lambda_3 = 2, with corresponding eigenvectors v1=(1,1,1)v_1 = (1, 1, 1), v2=(1,0,1)v_2 = (1, 0, 1), and v3=(1,1,2)v_3 = (1, 1, 2).
Key Concept
Eigenvalues and Eigenvectors
Explanation
Eigenvalues are scalars associated with a linear system of equations (or a matrix) that, when multiplied by a non-zero vector (the eigenvector), result in the vector being scaled by that scalar. They are found by solving the characteristic polynomial of the matrix.
(1 point) Supppose AA is an invertible n×nn \times n matrix and v\mathbf{v} is an eigenvector of AA with associated eigenvalue 7. Convince yourself that v\mathbf{v} is an eigenvector of the following matrices, and find the associated eigenvalues. a. The matrix A8A^{8} has an eigenvalue □\square b. The matrix A−1A^{-1} has an eigenvalue □\square c. The matrix A−9InA-9 I_{n} has an eigenvalue □\square d. The matrix 4A4 A has an eigenvalue □\square
Solution by Steps
step 1
Given that v\mathbf{v} is an eigenvector of AA with eigenvalue 7, we know that Av=7vA\mathbf{v} = 7\mathbf{v}
step 2
To find the eigenvalue of A8A^8 associated with v\mathbf{v}, we consider A8vA^8\mathbf{v}
step 3
Since Av=7vA\mathbf{v} = 7\mathbf{v}, we can write A8v=A7(Av)=A7(7v)=7A7vA^8\mathbf{v} = A^7(A\mathbf{v}) = A^7(7\mathbf{v}) = 7A^7\mathbf{v}
step 4
Repeating this process, we get A8v=78vA^8\mathbf{v} = 7^8\mathbf{v}
step 5
Therefore, the eigenvalue of A8A^8 associated with v\mathbf{v} is 787^8
Answer
787^8
Key Concept
Eigenvalues of matrix powers
Explanation
If v\mathbf{v} is an eigenvector of AA with eigenvalue λ\lambda, then v\mathbf{v} is also an eigenvector of AkA^k with eigenvalue λk\lambda^k.
Solution by Steps
step 1
Since v\mathbf{v} is an eigenvector of AA with eigenvalue 7, we have Av=7vA\mathbf{v} = 7\mathbf{v}
step 2
To find the eigenvalue of A−1A^{-1} associated with v\mathbf{v}, we multiply both sides of the equation by A−1A^{-1}
step 3
We get A−1Av=A−1(7v)A^{-1}A\mathbf{v} = A^{-1}(7\mathbf{v})
step 4
Simplifying, we have v=7A−1v\mathbf{v} = 7A^{-1}\mathbf{v}
step 5
Therefore, the eigenvalue of A−1A^{-1} associated with v\mathbf{v} is 17\frac{1}{7}
Answer
17\frac{1}{7}
Key Concept
Eigenvalues of matrix inverses
Explanation
If v\mathbf{v} is an eigenvector of AA with eigenvalue λ\lambda, then v\mathbf{v} is also an eigenvector of A−1A^{-1} with eigenvalue 1λ\frac{1}{\lambda}.
Solution by Steps
step 1
Given Av=7vA\mathbf{v} = 7\mathbf{v}, we want to find the eigenvalue of (A−9In)(A - 9I_n) associated with v\mathbf{v}
step 2
We consider (A−9In)v=Av−9Inv(A - 9I_n)\mathbf{v} = A\mathbf{v} - 9I_n\mathbf{v}
step 3
Substituting the known eigenvalue, we get 7v−9v7\mathbf{v} - 9\mathbf{v}
step 4
Simplifying, we have −2v-2\mathbf{v}
step 5
Therefore, the eigenvalue of (A−9In)(A - 9I_n) associated with v\mathbf{v} is −2-2
Answer
−2-2
Key Concept
Eigenvalues of matrix subtraction
Explanation
If v\mathbf{v} is an eigenvector of AA with eigenvalue λ\lambda, then v\mathbf{v} is also an eigenvector of (A−μIn)(A - \mu I_n) with eigenvalue λ−μ\lambda - \mu.
Solution by Steps
step 1
Since v\mathbf{v} is an eigenvector of AA with eigenvalue 7, we have Av=7vA\mathbf{v} = 7\mathbf{v}
step 2
To find the eigenvalue of 4A4A associated with v\mathbf{v}, we consider 4Av4A\mathbf{v}
step 3
We can write 4Av=4(7v)4A\mathbf{v} = 4(7\mathbf{v})
step 4
Simplifying, we have 28v28\mathbf{v}
step 5
Therefore, the eigenvalue of 4A4A associated with v\mathbf{v} is 2828
Answer
2828
Key Concept
Eigenvalues of scalar multiples of matrices
Explanation
If v\mathbf{v} is an eigenvector of AA with eigenvalue λ\lambda, then v\mathbf{v} is also an eigenvector of kAkA with eigenvalue kλk\lambda.
(1 point) Find a 2×22 \times 2 matrix AA for which E3=span⁡{[2−1]},E−1=span⁡{[−53]}, E_{3}=\operatorname{span}\left\{\left[\begin{array}{c} 2 \\ -1 \end{array}\right]\right\}, \quad E_{-1}=\operatorname{span}\left\{\left[\begin{array}{c} -5 \\ 3 \end{array}\right]\right\}, where EλE_{\lambda} denotes the eigenspace associated with the eigenvalue λ\lambda. A=[□amp;□□amp;□] A=\left[\begin{array}{ll} \square &amp; \square \\ \square &amp; \square \end{array}\right]
Solution by Steps
step 1
Identify the given eigenvectors and their corresponding eigenvalues
step 2
For the eigenvalue λ=3\lambda = 3, the eigenvector is [2−1]\begin{bmatrix} 2 \\ -1 \end{bmatrix}. This means that A[2−1]=3[2−1]A \begin{bmatrix} 2 \\ -1 \end{bmatrix} = 3 \begin{bmatrix} 2 \\ -1 \end{bmatrix}
step 3
For the eigenvalue λ=−1\lambda = -1, the eigenvector is [−53]\begin{bmatrix} -5 \\ 3 \end{bmatrix}. This means that A[−53]=−1[−53]A \begin{bmatrix} -5 \\ 3 \end{bmatrix} = -1 \begin{bmatrix} -5 \\ 3 \end{bmatrix}
step 4
Construct matrix AA using the eigenvectors as columns and multiply by a diagonal matrix of the eigenvalues
step 5
The matrix AA is A=[2amp;−5−1amp;3][3amp;00amp;−1]=[6amp;5−3amp;−3]A = \begin{bmatrix} 2 &amp; -5 \\ -1 &amp; 3 \end{bmatrix} \begin{bmatrix} 3 &amp; 0 \\ 0 &amp; -1 \end{bmatrix} = \begin{bmatrix} 6 &amp; 5 \\ -3 &amp; -3 \end{bmatrix}
Answer
A=[6amp;5−3amp;−3]A = \begin{bmatrix} 6 &amp; 5 \\ -3 &amp; -3 \end{bmatrix}
Key Concept
Constructing a matrix from eigenvectors and eigenvalues
Explanation
The matrix AA is formed by placing the eigenvectors as columns in a matrix and multiplying by a diagonal matrix with the corresponding eigenvalues on the diagonal. This constructs a matrix for which the columns are the eigenvectors and the eigenvalues are associated with those vectors.
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