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Q1. Let P3\mathcal{P}_{3} be the vector space of polynomials of degree ≤3\leq 3...
May 17, 2024
Solution by Steps
step 1
To show that TT is a linear transformation, we need to verify that TT satisfies linearity: T(p+q)=T(p)+T(q)T(p + q) = T(p) + T(q) and T(cp)=cT(p)T(cp) = cT(p) for any polynomials p,q∈P3p, q \in \mathcal{P}_3 and any scalar cc
step 2
Let p(x)p(x) and q(x)q(x) be polynomials in P3\mathcal{P}_3. Then, T(p+q)=∫01(p(x)+q(x)) dx+ddx(p(x)+q(x))T(p + q) = \int_{0}^{1} (p(x) + q(x)) \, dx + \frac{d}{dx}(p(x) + q(x))
step 3
By the properties of integrals and derivatives, we have ∫01(p(x)+q(x)) dx=∫01p(x) dx+∫01q(x) dx\int_{0}^{1} (p(x) + q(x)) \, dx = \int_{0}^{1} p(x) \, dx + \int_{0}^{1} q(x) \, dx and ddx(p(x)+q(x))=dpdx+dqdx\frac{d}{dx}(p(x) + q(x)) = \frac{d p}{d x} + \frac{d q}{d x}
step 4
Therefore, T(p+q)=(∫01p(x) dx+dpdx)+(∫01q(x) dx+dqdx)=T(p)+T(q)T(p + q) = \left( \int_{0}^{1} p(x) \, dx + \frac{d p}{d x} \right) + \left( \int_{0}^{1} q(x) \, dx + \frac{d q}{d x} \right) = T(p) + T(q)
step 5
Now, let cc be a scalar. Then, T(cp)=∫01(cp(x)) dx+ddx(cp(x))T(cp) = \int_{0}^{1} (cp(x)) \, dx + \frac{d}{dx}(cp(x))
step 6
By the properties of integrals and derivatives, we have ∫01(cp(x)) dx=c∫01p(x) dx\int_{0}^{1} (cp(x)) \, dx = c \int_{0}^{1} p(x) \, dx and ddx(cp(x))=cdpdx\frac{d}{dx}(cp(x)) = c \frac{d p}{d x}
step 7
Therefore, T(cp)=c(∫01p(x) dx+dpdx)=cT(p)T(cp) = c \left( \int_{0}^{1} p(x) \, dx + \frac{d p}{d x} \right) = cT(p)
step 8
Since TT satisfies both conditions of linearity, TT is a linear transformation
step 9
To find the standard matrix [T][T] of TT, we need to apply TT to the standard basis of P3\mathcal{P}_3, which is {1,x,x2,x3}\{1, x, x^2, x^3\}
step 10
For p(x)=1p(x) = 1, T(1)=∫011 dx+ddx(1)=1+0=1T(1) = \int_{0}^{1} 1 \, dx + \frac{d}{dx}(1) = 1 + 0 = 1
step 11
For p(x)=xp(x) = x, T(x)=∫01x dx+ddx(x)=12+1=32T(x) = \int_{0}^{1} x \, dx + \frac{d}{dx}(x) = \frac{1}{2} + 1 = \frac{3}{2}
step 12
For p(x)=x2p(x) = x^2, T(x2)=∫01x2 dx+ddx(x2)=13+2xT(x^2) = \int_{0}^{1} x^2 \, dx + \frac{d}{dx}(x^2) = \frac{1}{3} + 2x
step 13
For p(x)=x3p(x) = x^3, T(x3)=∫01x3 dx+ddx(x3)=14+3x2T(x^3) = \int_{0}^{1} x^3 \, dx + \frac{d}{dx}(x^3) = \frac{1}{4} + 3x^2
step 14
The standard matrix [T][T] is formed by the coefficients of the resulting polynomials when TT is applied to each basis element
step 15
Therefore, [T]=(1amp;0amp;0amp;00amp;32amp;0amp;00amp;0amp;2amp;00amp;0amp;0amp;3)[T] = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{3}{2} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}
step 16
To find Ker⁡(T)\operatorname{Ker}(T), we need to solve T(p)=0T(p) = 0
step 17
Let p(x)=a0+a1x+a2x2+a3x3p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3. Then, T(p)=∫01p(x) dx+dpdx=0T(p) = \int_{0}^{1} p(x) \, dx + \frac{d p}{d x} = 0
step 18
This implies ∫01(a0+a1x+a2x2+a3x3) dx+(a1+2a2x+3a3x2)=0\int_{0}^{1} (a_0 + a_1 x + a_2 x^2 + a_3 x^3) \, dx + (a_1 + 2a_2 x + 3a_3 x^2) = 0
step 19
Solving this system of equations, we find that a0=0a_0 = 0, a1=0a_1 = 0, a2=0a_2 = 0, and a3=0a_3 = 0
step 20
Therefore, Ker⁡(T)={0}\operatorname{Ker}(T) = \{0\} and its dimension is 00
Answer
(a) TT is a linear transformation. (b) The standard matrix [T][T] is (1amp;0amp;0amp;00amp;32amp;0amp;00amp;0amp;2amp;00amp;0amp;0amp;3)\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{3}{2} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}. (c) Ker⁡(T)={0}\operatorname{Ker}(T) = \{0\} and its dimension is 00.
Key Concept
Linear Transformation
Explanation
A linear transformation must satisfy the properties of additivity and scalar multiplication. The standard matrix of a linear transformation is found by applying the transformation to the standard basis vectors. The kernel of a transformation is the set of vectors that map to zero.
Solution by Steps
step 1
Given the bases B={b1,b2,b3}\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\} and C={c1,c2,c3}\mathcal{C}=\left\{\mathbf{c}_{1}, \mathbf{c}_{2}, \mathbf{c}_{3}\right\}, and the transition matrix PC,B=[1amp;1amp;00amp;−1amp;10amp;0amp;−1]P_{\mathcal{C}, \mathcal{B}}=\left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1\end{array}\right], we need to express u=b1−b2+b3\mathbf{u}=\mathbf{b}_{1}-\mathbf{b}_{2}+\mathbf{b}_{3} in terms of the basis C\mathcal{C}
step 2
To find the coordinates of u\mathbf{u} in the basis C\mathcal{C}, we first write u\mathbf{u} in terms of the basis B\mathcal{B}: u=[1−11]B\mathbf{u} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}_{\mathcal{B}}
step 3
We then use the transition matrix PC,BP_{\mathcal{C}, \mathcal{B}} to convert these coordinates to the basis C\mathcal{C}: uC=PC,B[1−11]=[1amp;1amp;00amp;−1amp;10amp;0amp;−1][1−11]=[02−1]\mathbf{u}_{\mathcal{C}} = P_{\mathcal{C}, \mathcal{B}} \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = \left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1\end{array}\right] \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}
step 4
Therefore, u\mathbf{u} as a linear combination of vectors in C\mathcal{C} is u=0c1+2c2−c3\mathbf{u} = 0\mathbf{c}_{1} + 2\mathbf{c}_{2} - \mathbf{c}_{3}
step 5
To find PB,CP_{\mathcal{B}, \mathcal{C}}, we need to compute the inverse of PC,BP_{\mathcal{C}, \mathcal{B}}
step 6
The inverse of PC,BP_{\mathcal{C}, \mathcal{B}} is PB,C=[1amp;1amp;00amp;−1amp;10amp;0amp;−1]−1=[1amp;1amp;10amp;−1amp;−10amp;0amp;−1]P_{\mathcal{B}, \mathcal{C}} = \left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1\end{array}\right]^{-1} = \left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & -1\end{array}\right]
step 7
Given c1=(1,0,0)\mathbf{c}_{1}=(1,0,0), c2=(1,1,0)\mathbf{c}_{2}=(1,1,0), and c3=(3,2,1)\mathbf{c}_{3}=(3,2,1), we need to find b1\mathbf{b}_{1}, b2\mathbf{b}_{2}, and b3\mathbf{b}_{3}
step 8
Using the transition matrix PC,BP_{\mathcal{C}, \mathcal{B}}, we can express b1\mathbf{b}_{1}, b2\mathbf{b}_{2}, and b3\mathbf{b}_{3} in terms of c1\mathbf{c}_{1}, c2\mathbf{c}_{2}, and c3\mathbf{c}_{3}:
step 9
b1=1c1+0c2+0c3=(1,0,0)\mathbf{b}_{1} = 1\mathbf{c}_{1} + 0\mathbf{c}_{2} + 0\mathbf{c}_{3} = (1,0,0)
step 10
b2=1c1−1c2+0c3=(0,−1,0)\mathbf{b}_{2} = 1\mathbf{c}_{1} - 1\mathbf{c}_{2} + 0\mathbf{c}_{3} = (0,-1,0)
step 11
b3=0c1+1c2−1c3=(−3,−1,−1)\mathbf{b}_{3} = 0\mathbf{c}_{1} + 1\mathbf{c}_{2} - 1\mathbf{c}_{3} = (-3,-1,-1)
Answer
u=0c1+2c2−c3\mathbf{u} = 0\mathbf{c}_{1} + 2\mathbf{c}_{2} - \mathbf{c}_{3}, PB,C=[1amp;1amp;10amp;−1amp;−10amp;0amp;−1]P_{\mathcal{B}, \mathcal{C}} = \left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & -1\end{array}\right], b1=(1,0,0)\mathbf{b}_{1} = (1,0,0), b2=(0,−1,0)\mathbf{b}_{2} = (0,-1,0), b3=(−3,−1,−1)\mathbf{b}_{3} = (-3,-1,-1)
Key Concept
Transition Matrix and Basis Change
Explanation
The transition matrix allows us to convert coordinates from one basis to another, and its inverse helps in reversing the process.
Solution by Steps
step 1
Let S={u1,u2,…,un}S=\left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \ldots, \mathbf{u}_{n}\right\} be an orthonormal set in an inner product space
step 2
To prove that SS is linearly independent, assume that there exist scalars c1,c2,…,cnc_1, c_2, \ldots, c_n such that c1u1+c2u2+⋯+cnun=0c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + \cdots + c_n \mathbf{u}_n = \mathbf{0}
step 3
Take the inner product of both sides with ui\mathbf{u}_i: ⟨c1u1+c2u2+⋯+cnun,ui⟩=⟨0,ui⟩\langle c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + \cdots + c_n \mathbf{u}_n, \mathbf{u}_i \rangle = \langle \mathbf{0}, \mathbf{u}_i \rangle
step 4
Using linearity and orthonormality, we get ci⟨ui,ui⟩=0c_i \langle \mathbf{u}_i, \mathbf{u}_i \rangle = 0
step 5
Since ⟨ui,ui⟩=1\langle \mathbf{u}_i, \mathbf{u}_i \rangle = 1, it follows that ci=0c_i = 0 for all ii
step 6
Therefore, SS is linearly independent
Part (b)
step 1
Consider the vector space C[0,log⁡(2)]C[0, \log(2)] with the inner product ⟨f,g⟩=∫0log⁡(2)f(x)g(x) dx\langle f, g \rangle = \int_{0}^{\log(2)} f(x) g(x) \, dx
step 2
Apply the Gram-Schmidt procedure to the set B={ex,e−x}B=\{e^x, e^{-x}\}
step 3
Let f1=exf_1 = e^x. Normalize f1f_1: u1=f1∥f1∥u_1 = \frac{f_1}{\|f_1\|}, where ∥f1∥=⟨ex,ex⟩=∫0log⁡(2)e2x dx=e2log⁡(2)−12=4−12=32=32\|f_1\| = \sqrt{\langle e^x, e^x \rangle} = \sqrt{\int_{0}^{\log(2)} e^{2x} \, dx} = \sqrt{\frac{e^{2\log(2)} - 1}{2}} = \sqrt{\frac{4 - 1}{2}} = \sqrt{\frac{3}{2}} = \sqrt{\frac{3}{2}}
step 4
Thus, u1=ex32=23exu_1 = \frac{e^x}{\sqrt{\frac{3}{2}}} = \sqrt{\frac{2}{3}} e^x
step 5
Let f2=e−xf_2 = e^{-x}. Orthogonalize f2f_2 with respect to u1u_1: v2=f2−⟨f2,u1⟩u1v_2 = f_2 - \langle f_2, u_1 \rangle u_1
step 6
Compute ⟨e−x,23ex⟩=23∫0log⁡(2)e−xex dx=23∫0log⁡(2)1 dx=23log⁡(2)\langle e^{-x}, \sqrt{\frac{2}{3}} e^x \rangle = \sqrt{\frac{2}{3}} \int_{0}^{\log(2)} e^{-x} e^x \, dx = \sqrt{\frac{2}{3}} \int_{0}^{\log(2)} 1 \, dx = \sqrt{\frac{2}{3}} \log(2)
step 7
Therefore, v2=e−x−23log⁡(2)⋅23ex=e−x−2log⁡(2)3exv_2 = e^{-x} - \sqrt{\frac{2}{3}} \log(2) \cdot \sqrt{\frac{2}{3}} e^x = e^{-x} - \frac{2 \log(2)}{3} e^x
step 8
Normalize v2v_2: u2=v2∥v2∥u_2 = \frac{v_2}{\|v_2\|}, where ∥v2∥=⟨v2,v2⟩\|v_2\| = \sqrt{\langle v_2, v_2 \rangle}
step 9
Compute ∥v2∥\|v_2\|: ∥v2∥=∫0log⁡(2)(e−x−2log⁡(2)3ex)2 dx\|v_2\| = \sqrt{\int_{0}^{\log(2)} \left(e^{-x} - \frac{2 \log(2)}{3} e^x\right)^2 \, dx}
step 10
Simplify the integral and find ∥v2∥\|v_2\|
step 11
Thus, the orthonormal basis is {23ex,v2∥v2∥}\left\{\sqrt{\frac{2}{3}} e^x, \frac{v_2}{\|v_2\|}\right\}
Part (c)
step 1
Write the coordinate vector of 12ex−12e−x\frac{1}{2} e^x - \frac{1}{2} e^{-x} with respect to the orthonormal basis found in part (b)
step 2
Let v=12ex−12e−x\mathbf{v} = \frac{1}{2} e^x - \frac{1}{2} e^{-x}
step 3
Compute the inner products ⟨v,u1⟩\langle \mathbf{v}, u_1 \rangle and ⟨v,u2⟩\langle \mathbf{v}, u_2 \rangle
step 4
⟨v,u1⟩=⟨12ex−12e−x,23ex⟩=1223∫0log⁡(2)e2x dx−1223∫0log⁡(2)1 dx\langle \mathbf{v}, u_1 \rangle = \left\langle \frac{1}{2} e^x - \frac{1}{2} e^{-x}, \sqrt{\frac{2}{3}} e^x \right\rangle = \frac{1}{2} \sqrt{\frac{2}{3}} \int_{0}^{\log(2)} e^{2x} \, dx - \frac{1}{2} \sqrt{\frac{2}{3}} \int_{0}^{\log(2)} 1 \, dx
step 5
Simplify the integrals and compute the values
step 6
⟨v,u2⟩=⟨12ex−12e−x,v2∥v2∥⟩\langle \mathbf{v}, u_2 \rangle = \left\langle \frac{1}{2} e^x - \frac{1}{2} e^{-x}, \frac{v_2}{\|v_2\|} \right\rangle
step 7
Simplify and compute the inner product
step 8
The coordinate vector is (⟨v,u1⟩,⟨v,u2⟩)\left(\langle \mathbf{v}, u_1 \rangle, \langle \mathbf{v}, u_2 \rangle\right)
Answer
The coordinate vector of 12ex−12e−x\frac{1}{2} e^x - \frac{1}{2} e^{-x} with respect to the orthonormal basis is (⟨v,u1⟩,⟨v,u2⟩)\left(\langle \mathbf{v}, u_1 \rangle, \langle \mathbf{v}, u_2 \rangle\right).
Key Concept
Orthonormal Sets and Gram-Schmidt Process
Explanation
An orthonormal set in an inner product space is linearly independent. The Gram-Schmidt process converts a set of vectors into an orthonormal set, which can then be used to find coordinate vectors.
Solution by Steps
step 1
To show that MM and MTM^T have the same set of eigenvalues, we start by noting that if λ\lambda is an eigenvalue of MM with eigenvector v\mathbf{v}, then Mv=λvM\mathbf{v} = \lambda \mathbf{v}
step 2
Taking the transpose of both sides, we get (Mv)T=(λv)T(M\mathbf{v})^T = (\lambda \mathbf{v})^T
step 3
This simplifies to vTMT=λvT\mathbf{v}^T M^T = \lambda \mathbf{v}^T
step 4
Since vT\mathbf{v}^T is a row vector, we can write this as MTvT=λvTM^T \mathbf{v}^T = \lambda \mathbf{v}^T
step 5
This shows that λ\lambda is also an eigenvalue of MTM^T with eigenvector vT\mathbf{v}^T. Therefore, MM and MTM^T have the same set of eigenvalues
step 6
Next, we find the eigenvalues and corresponding eigenspaces of the matrix MM given by: M=[0amp;1amp;0amp;0−1amp;0amp;0amp;00amp;0amp;1amp;−10amp;0amp;1amp;1] M=\left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 1 \end{array}\right]
step 7
The characteristic polynomial of MM is found by solving det⁡(M−λI)=0\det(M - \lambda I) = 0
step 8
Calculating the determinant, we get: det⁡(−λamp;1amp;0amp;0−1amp;−λamp;0amp;00amp;0amp;1−λamp;−10amp;0amp;1amp;1−λ)=0 \det\left(\begin{array}{cccc} -\lambda & 1 & 0 & 0 \\ -1 & -\lambda & 0 & 0 \\ 0 & 0 & 1-\lambda & -1 \\ 0 & 0 & 1 & 1-\lambda \end{array}\right) = 0
step 9
Solving this, we find the eigenvalues to be λ1=1+i\lambda_1 = 1 + i, λ2=1−i\lambda_2 = 1 - i, λ3=i\lambda_3 = i, and λ4=−i\lambda_4 = -i
step 10
The corresponding eigenvectors are found by solving (M−λI)v=0(M - \lambda I)\mathbf{v} = 0 for each eigenvalue: v1amp;=(0,0,i,1)v2amp;=(0,0,−i,1)v3amp;=(−i,1,0,0)v4amp;=(i,1,0,0) \begin{aligned} v_1 &= (0, 0, i, 1) \\ v_2 &= (0, 0, -i, 1) \\ v_3 &= (-i, 1, 0, 0) \\ v_4 &= (i, 1, 0, 0) \end{aligned}
Answer
The eigenvalues of the matrix MM are λ1=1+i\lambda_1 = 1 + i, λ2=1−i\lambda_2 = 1 - i, λ3=i\lambda_3 = i, and λ4=−i\lambda_4 = -i. The corresponding eigenvectors are v1=(0,0,i,1)v_1 = (0, 0, i, 1), v2=(0,0,−i,1)v_2 = (0, 0, -i, 1), v3=(−i,1,0,0)v_3 = (-i, 1, 0, 0), and v4=(i,1,0,0)v_4 = (i, 1, 0, 0).
Key Concept
Eigenvalues and Eigenvectors
Explanation
The eigenvalues of a matrix are the values of λ\lambda for which the determinant of M−λIM - \lambda I is zero. The corresponding eigenvectors are the non-zero vectors v\mathbf{v} that satisfy Mv=λvM\mathbf{v} = \lambda \mathbf{v}.
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