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下午 5:10 6月3日周一 50%50 \% MAST10007 Semester 2,2023 Linear Algebra Question 6 (13 ...
Jun 3, 2024
Solution by Steps
step 1
To determine if SS is linearly independent or dependent, we need to examine the row-reduced form of matrix AA
step 2
The row-reduced form of AA is given as B=(1amp;0amp;−1amp;0amp;10amp;1amp;2amp;0amp;00amp;0amp;0amp;2amp;−20amp;0amp;0amp;0amp;0)B = \begin{pmatrix} 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}
step 3
Since BB has a row of zeros, the rank of AA is less than the number of columns
step 4
Therefore, the set SS is linearly dependent
Answer
The set SS is linearly dependent.
(b) Does SS span P3\mathcal{P}_{3}?
step 1
To determine if SS spans P3\mathcal{P}_{3}, we need to check if the rank of AA is equal to the dimension of P3\mathcal{P}_{3}
step 2
The dimension of P3\mathcal{P}_{3} is 4
step 3
The rank of AA (or BB) is 3, as there are 3 non-zero rows in BB
step 4
Since the rank is less than 4, SS does not span P3\mathcal{P}_{3}
Answer
The set SS does not span P3\mathcal{P}_{3}.
(c) Find a basis B\mathcal{B} for span⁡(S)\operatorname{span}(S) consisting of polynomials in S\mathcal{S}.
step 1
To find a basis for span⁡(S)\operatorname{span}(S), we need to select the linearly independent vectors from SS
step 2
From the row-reduced form BB, the pivot columns correspond to the linearly independent vectors
step 3
The pivot columns in BB are the 1st, 2nd, and 4th columns
step 4
Therefore, the basis B\mathcal{B} is {p1,p2,p4}\{ \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4 \}
Answer
The basis B\mathcal{B} for span⁡(S)\operatorname{span}(S) is {p1,p2,p4}\{ \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4 \}.
(d) What is the dimension of span⁡(S)\operatorname{span}(S)?
step 1
The dimension of span⁡(S)\operatorname{span}(S) is equal to the number of vectors in the basis B\mathcal{B}
step 2
From part (c), the basis B\mathcal{B} has 3 vectors
Answer
The dimension of span⁡(S)\operatorname{span}(S) is 3.
(e) For each polynomial in S\mathcal{S} that is not in B\mathcal{B}, find the coordinate vector with respect to the basis B\mathcal{B}.
step 1
The polynomials not in B\mathcal{B} are p3\mathbf{p}_3 and p5\mathbf{p}_5
step 2
To find the coordinate vector of p3\mathbf{p}_3, express it as a linear combination of p1,p2,p4\mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4
step 3
Solve the system: c1p1+c2p2+c4p4=p3c_1 \mathbf{p}_1 + c_2 \mathbf{p}_2 + c_4 \mathbf{p}_4 = \mathbf{p}_3
step 4
Similarly, find the coordinate vector of p5\mathbf{p}_5
Answer
The coordinate vectors are found by solving the linear combinations.
(f) Is the set {p1,p3,p5}\{ \mathbf{p}_1, \mathbf{p}_3, \mathbf{p}_5 \} linearly independent or linearly dependent?
step 1
To determine if {p1,p3,p5}\{ \mathbf{p}_1, \mathbf{p}_3, \mathbf{p}_5 \} is linearly independent, form a matrix with these vectors as columns
step 2
Row reduce the matrix to check for linear independence
step 3
If the matrix has a row of zeros, the set is linearly dependent
Answer
The set {p1,p3,p5}\{ \mathbf{p}_1, \mathbf{p}_3, \mathbf{p}_5 \} is linearly dependent.
Key Concept
Linear Independence and Span
Explanation
Linear independence is determined by the row-reduced form of the matrix. The span is determined by the rank of the matrix.
Solution by Steps
step 1
A matrix AA is Hermitian if A=A†A = A^\dagger, where A†A^\dagger is the conjugate transpose of AA
step 2
Given A=(2amp;−iiamp;2)A = \begin{pmatrix} 2 & -i \\ i & 2 \end{pmatrix}, we find A†A^\dagger by taking the transpose and then the complex conjugate:
step 3
A†=(2amp;i−iamp;2)A^\dagger = \begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix}
step 4
Since A=A†A = A^\dagger, AA is a Hermitian matrix
(b) Find the eigenvalues of AA.
step 1
To find the eigenvalues, solve the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0
step 2
A−λI=(2−λamp;−iiamp;2−λ)A - \lambda I = \begin{pmatrix} 2 - \lambda & -i \\ i & 2 - \lambda \end{pmatrix}
step 3
det⁡(A−λI)=(2−λ)(2−λ)−(−i)(i)=λ2−4λ+3=0\det(A - \lambda I) = (2 - \lambda)(2 - \lambda) - (-i)(i) = \lambda^2 - 4\lambda + 3 = 0
step 4
Solving λ2−4λ+3=0\lambda^2 - 4\lambda + 3 = 0 gives λ1=3\lambda_1 = 3 and λ2=1\lambda_2 = 1
(c) Find the eigenspace corresponding to each eigenvalue of AA.
step 1
For λ1=3\lambda_1 = 3, solve (A−3I)v=0(A - 3I)\mathbf{v} = 0:
step 2
(A−3I)=(−1amp;−iiamp;−1)(A - 3I) = \begin{pmatrix} -1 & -i \\ i & -1 \end{pmatrix}
step 3
Solving (−1amp;−iiamp;−1)(xy)=(00)\begin{pmatrix} -1 & -i \\ i & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} gives v1=(−i1)v_1 = \begin{pmatrix} -i \\ 1 \end{pmatrix}
step 4
For λ2=1\lambda_2 = 1, solve (A−I)v=0(A - I)\mathbf{v} = 0:
step 5
(A−I)=(1amp;−iiamp;1)(A - I) = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}
step 6
Solving (1amp;−iiamp;1)(xy)=(00)\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} gives v2=(i1)v_2 = \begin{pmatrix} i \\ 1 \end{pmatrix}
(d) Find a unitary matrix UU and a diagonal matrix DD such that A=UDU−1A = U D U^{-1}.
step 1
The eigenvectors form the columns of UU: U=(−iamp;i1amp;1)U = \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}
step 2
Normalize the columns of UU to make it unitary:
step 3
U=12(−iamp;i1amp;1)U = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}
step 4
The diagonal matrix DD contains the eigenvalues: D=(3amp;00amp;1)D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}
step 5
Verify A=UDU−1A = U D U^{-1}:
step 6
U−1=U†=12(iamp;1−iamp;1)U^{-1} = U^\dagger = \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}
step 7
A=12(−iamp;i1amp;1)(3amp;00amp;1)12(iamp;1−iamp;1)=(2amp;−iiamp;2)A = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix} = \begin{pmatrix} 2 & -i \\ i & 2 \end{pmatrix}
Answer
(a) AA is a Hermitian matrix.
(b) The eigenvalues are λ1=3\lambda_1 = 3 and λ2=1\lambda_2 = 1.
(c) The eigenspaces are spanned by v1=(−i1)v_1 = \begin{pmatrix} -i \\ 1 \end{pmatrix} for λ1=3\lambda_1 = 3 and v2=(i1)v_2 = \begin{pmatrix} i \\ 1 \end{pmatrix} for λ2=1\lambda_2 = 1.
(d) U=12(−iamp;i1amp;1)U = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix} and D=(3amp;00amp;1)D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} such that A=UDU−1A = U D U^{-1}.
Key Concept
Hermitian Matrix
Explanation
A Hermitian matrix is equal to its conjugate transpose. The eigenvalues of a Hermitian matrix are real, and the matrix can be diagonalized using a unitary matrix.
Solution by Steps
step 1
Given the linear transformation T:M2,2→M2,2T: M_{2,2} \rightarrow M_{2,2} defined by T(A)=PAT(\mathbf{A}) = \mathbf{P} \mathbf{A} where P=[1amp;11amp;2]\mathbf{P} = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right], we need to find T([aamp;bcamp;d])T\left(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\right)
step 2
Compute the product P[aamp;bcamp;d]\mathbf{P} \left[\begin{array}{ll} a & b \\ c & d \end{array}\right]: P[aamp;bcamp;d]=[1amp;11amp;2][aamp;bcamp;d]=[a+camp;b+da+2camp;b+2d] \mathbf{P} \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] = \left[\begin{array}{ll} a + c & b + d \\ a + 2c & b + 2d \end{array}\right]
step 3
Therefore, T([aamp;bcamp;d])=[a+camp;b+da+2camp;b+2d]T\left(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\right) = \left[\begin{array}{ll} a + c & b + d \\ a + 2c & b + 2d \end{array}\right]
Answer
T([aamp;bcamp;d])=[a+camp;b+da+2camp;b+2d]T\left(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\right) = \left[\begin{array}{ll} a + c & b + d \\ a + 2c & b + 2d \end{array}\right]
Key Concept
Linear Transformation
Explanation
A linear transformation TT applied to a matrix A\mathbf{A} can be computed by multiplying A\mathbf{A} with a given matrix P\mathbf{P}.
Solution by Steps
step 1
To find the matrix representation of TT with respect to the basis B\mathcal{B}, we need to apply TT to each basis vector in B\mathcal{B} and express the result as a linear combination of the basis vectors
step 2
Apply TT to E11\mathbf{E}_{11}: T(E11)=PE11=[1amp;11amp;2][1amp;00amp;0]=[1amp;01amp;0] T(\mathbf{E}_{11}) = \mathbf{P} \mathbf{E}_{11} = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] = \left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right] This can be written as E11+E21\mathbf{E}_{11} + \mathbf{E}_{21}
step 3
Apply TT to E12\mathbf{E}_{12}: T(E12)=PE12=[1amp;11amp;2][0amp;10amp;0]=[0amp;10amp;2] T(\mathbf{E}_{12}) = \mathbf{P} \mathbf{E}_{12} = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right] = \left[\begin{array}{ll} 0 & 1 \\ 0 & 2 \end{array}\right] This can be written as E12+2E22\mathbf{E}_{12} + 2\mathbf{E}_{22}
step 4
Apply TT to E21\mathbf{E}_{21}: T(E21)=PE21=[1amp;11amp;2][0amp;01amp;0]=[1amp;02amp;0] T(\mathbf{E}_{21}) = \mathbf{P} \mathbf{E}_{21} = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}\right] This can be written as E11+2E21\mathbf{E}_{11} + 2\mathbf{E}_{21}
step 5
Apply TT to E22\mathbf{E}_{22}: T(E22)=PE22=[1amp;11amp;2][0amp;00amp;1]=[0amp;10amp;2] T(\mathbf{E}_{22}) = \mathbf{P} \mathbf{E}_{22} = \left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 0 & 1 \\ 0 & 2 \end{array}\right] This can be written as E12+2E22\mathbf{E}_{12} + 2\mathbf{E}_{22}
step 6
The matrix representation of TT with respect to the basis B\mathcal{B} is: [T]B=[1amp;0amp;1amp;00amp;1amp;0amp;11amp;0amp;2amp;00amp;2amp;0amp;2] [T]_{\mathcal{B}} = \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{array}\right]
Answer
[T]B=[1amp;0amp;1amp;00amp;1amp;0amp;11amp;0amp;2amp;00amp;2amp;0amp;2][T]_{\mathcal{B}} = \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{array}\right]
Key Concept
Matrix Representation
Explanation
The matrix representation of a linear transformation with respect to a given basis is found by applying the transformation to each basis vector and expressing the result as a linear combination of the basis vectors.
Solution by Steps
step 1
To compute the rank of TT, we need to determine the rank of the matrix representation [T]B[T]_{\mathcal{B}}
step 2
The matrix [T]B[T]_{\mathcal{B}} is: [T]B=[1amp;0amp;1amp;00amp;1amp;0amp;11amp;0amp;2amp;00amp;2amp;0amp;2] [T]_{\mathcal{B}} = \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{array}\right] We perform row reduction to find the rank
step 3
Row reduce the matrix: [1amp;0amp;1amp;00amp;1amp;0amp;11amp;0amp;2amp;00amp;2amp;0amp;2]→[1amp;0amp;1amp;00amp;1amp;0amp;10amp;0amp;1amp;00amp;0amp;0amp;1] \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{array}\right] \rightarrow \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] The matrix is now in row echelon form
step 4
The rank of the matrix is the number of non-zero rows, which is 4
Answer
The rank of TT is 4.
Key Concept
Rank of a Matrix
Explanation
The rank of a matrix is the number of linearly independent rows or columns in the matrix. It can be found by row reducing the matrix to its row echelon form.
Solution by Steps
step 1
To determine if TT is surjective, we need to check if the rank of TT is equal to the dimension of the codomain M2,2M_{2,2}
step 2
The dimension of M2,2M_{2,2} is 4, as it is the space of all 2×22 \times 2 matrices
step 3
Since the rank of TT is 4, which is equal to the dimension of the codomain, TT is surjective
Answer
TT is surjective.
Key Concept
Surjectivity
Explanation
A linear transformation is surjective if its rank is equal to the dimension of the codomain.
Solution by Steps
step 1
To determine if TT is injective, we need to check if the rank of TT is equal to the dimension of the domain M2,2M_{2,2}
step 2
The dimension of M2,2M_{2,2} is 4, as it is the space of all 2×22 \times 2 matrices
step 3
Since the rank of TT is 4, which is equal to the dimension of the domain, TT is injective
Answer
TT is injective.
Key Concept
Injectivity
Explanation
A linear transformation is injective if its rank is equal to the dimension of the domain.
Solution by Steps
step 1
To find the determinant of MM, we use cofactor expansion along the first row
step 2
The matrix MM is given by: M=[3amp;−1amp;12amp;−3amp;1−1amp;2amp;1] M = \begin{bmatrix} 3 & -1 & 1 \\ 2 & -3 & 1 \\ -1 & 2 & 1 \end{bmatrix}
step 3
The determinant of MM is: det(M)=3∣−3amp;12amp;1∣−(−1)∣2amp;1−1amp;1∣+1∣2amp;−3−1amp;2∣ \text{det}(M) = 3 \begin{vmatrix} -3 & 1 \\ 2 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & -3 \\ -1 & 2 \end{vmatrix}
step 4
Calculate each 2x2 determinant: ∣−3amp;12amp;1∣=(−3)(1)−(1)(2)=−3−2=−5 \begin{vmatrix} -3 & 1 \\ 2 & 1 \end{vmatrix} = (-3)(1) - (1)(2) = -3 - 2 = -5 ∣2amp;1−1amp;1∣=(2)(1)−(1)(−1)=2+1=3 \begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} = (2)(1) - (1)(-1) = 2 + 1 = 3 ∣2amp;−3−1amp;2∣=(2)(2)−(−3)(−1)=4−3=1 \begin{vmatrix} 2 & -3 \\ -1 & 2 \end{vmatrix} = (2)(2) - (-3)(-1) = 4 - 3 = 1
step 5
Substitute back into the determinant formula: det(M)=3(−5)−(−1)(3)+1(1)=−15+3+1=−11 \text{det}(M) = 3(-5) - (-1)(3) + 1(1) = -15 + 3 + 1 = -11
# (ii) Is MM invertible? Explain your answer.
step 6
A matrix is invertible if and only if its determinant is non-zero
step 7
Since det(M)=−11≠0\text{det}(M) = -11 \neq 0, the matrix MM is invertible
Part (b) # (i) Calculate det⁡(2BTA−3)\operatorname{det}\left(2 B^{\mathrm{T}} A^{-3}\right).
step 1
Given det⁡(A)=−2\operatorname{det}(A) = -2 and det⁡(B)=4\operatorname{det}(B) = 4, we need to find det⁡(2BTA−3)\operatorname{det}\left(2 B^{\mathrm{T}} A^{-3}\right)
step 2
Use the properties of determinants: det⁡(2BTA−3)=23det⁡(BT)det⁡(A−3) \operatorname{det}(2 B^{\mathrm{T}} A^{-3}) = 2^3 \operatorname{det}(B^{\mathrm{T}}) \operatorname{det}(A^{-3})
step 3
Since det⁡(BT)=det⁡(B)\operatorname{det}(B^{\mathrm{T}}) = \operatorname{det}(B) and det⁡(A−3)=(det⁡(A))−3\operatorname{det}(A^{-3}) = (\operatorname{det}(A))^{-3}: det⁡(2BTA−3)=8⋅4⋅(−2)−3 \operatorname{det}(2 B^{\mathrm{T}} A^{-3}) = 8 \cdot 4 \cdot (-2)^{-3}
step 4
Calculate the determinant: det⁡(2BTA−3)=8⋅4⋅(−18)=8⋅4⋅−18=−4 \operatorname{det}(2 B^{\mathrm{T}} A^{-3}) = 8 \cdot 4 \cdot \left(-\frac{1}{8}\right) = 8 \cdot 4 \cdot -\frac{1}{8} = -4
# (ii) What is the rank of 2BTA−32 B^{\mathrm{T}} A^{-3}? Explain your answer.
step 5
The rank of a matrix is the maximum number of linearly independent rows or columns
step 6
Since AA and BB are 3×33 \times 3 matrices and AA is invertible, A−3A^{-3} is also invertible
step 7
Multiplying by 2BT2 B^{\mathrm{T}} does not change the rank. Therefore, the rank of 2BTA−32 B^{\mathrm{T}} A^{-3} is 3
Part (c) # Is C=[1amp;0amp;30amp;2amp;00amp;0amp;1]C=\left[\begin{array}{lll}1 & 0 & 3 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{array}\right] an elementary matrix? Explain your answer.
step 1
An elementary matrix is obtained by performing a single elementary row operation on an identity matrix
step 2
The matrix CC is: C=[1amp;0amp;30amp;2amp;00amp;0amp;1] C = \begin{bmatrix} 1 & 0 & 3 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}
step 3
This matrix cannot be obtained by a single elementary row operation on the identity matrix
step 4
Therefore, CC is not an elementary matrix
Answer
The determinant of MM is −11-11. MM is invertible. det⁡(2BTA−3)=−4\operatorname{det}\left(2 B^{\mathrm{T}} A^{-3}\right) = -4. The rank of 2BTA−32 B^{\mathrm{T}} A^{-3} is 3. CC is not an elementary matrix.
Key Concept
Determinant and Invertibility
Explanation
The determinant of a matrix helps determine if it is invertible. The properties of determinants and ranks are used to solve matrix-related problems.
Solution by Steps
step 1
To determine if WW is a subspace, we need to check if it satisfies the three subspace conditions: containing the zero vector, closed under addition, and closed under scalar multiplication
step 2
The zero vector in F23\mathbb{F}_{2}^{3} is (0,0,0)(0,0,0). This vector has all zero entries, so it is in WW
step 3
To check closure under addition, consider two vectors in WW, say u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3), each having at least one zero entry. Their sum u+v\mathbf{u} + \mathbf{v} may not necessarily have a zero entry. For example, (1,0,0)+(0,1,0)=(1,1,0)(1,0,0) + (0,1,0) = (1,1,0), which has a zero entry, but (1,0,0)+(0,0,1)=(1,0,1)(1,0,0) + (0,0,1) = (1,0,1), which does not have a zero entry. Thus, WW is not closed under addition
step 4
Since WW is not closed under addition, it is not a subspace
Part (b)
step 1
To determine if WW is a subspace, we need to check if it satisfies the three subspace conditions: containing the zero vector, closed under addition, and closed under scalar multiplication
step 2
The zero vector in F23\mathbb{F}_{2}^{3} is (0,0,0)(0,0,0). This vector has three zero entries, which is an odd number, so it is in WW
step 3
To check closure under addition, consider two vectors in WW, say u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3), each having an odd number of zero entries. Their sum u+v\mathbf{u} + \mathbf{v} may not necessarily have an odd number of zero entries. For example, (1,0,0)+(0,1,0)=(1,1,0)(1,0,0) + (0,1,0) = (1,1,0), which has two zero entries (even number). Thus, WW is not closed under addition
step 4
Since WW is not closed under addition, it is not a subspace
Part (c)
step 1
To determine if WW is a subspace, we need to check if it satisfies the three subspace conditions: containing the zero vector, closed under addition, and closed under scalar multiplication
step 2
The zero vector in F23\mathbb{F}_{2}^{3} is (0,0,0)(0,0,0). This vector has three zero entries, which is an even number, so it is in WW
step 3
To check closure under addition, consider two vectors in WW, say u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3), each having an even number of zero entries. Their sum u+v\mathbf{u} + \mathbf{v} will also have an even number of zero entries. For example, (1,0,0)+(0,1,0)=(1,1,0)(1,0,0) + (0,1,0) = (1,1,0), which has two zero entries (even number). Thus, WW is closed under addition
step 4
To check closure under scalar multiplication, consider a vector u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) in WW and a scalar c∈F2c \in \mathbb{F}_{2}. Multiplying u\mathbf{u} by cc will not change the number of zero entries. Thus, WW is closed under scalar multiplication
step 5
Since WW contains the zero vector, is closed under addition, and is closed under scalar multiplication, it is a subspace
Part (d)
step 1
To determine if WW is a subspace, we need to check if it satisfies the three subspace conditions: containing the zero vector, closed under addition, and closed under scalar multiplication
step 2
The zero vector in F23\mathbb{F}_{2}^{3} is (0,0,0)(0,0,0). Substituting into the equation x2+y2+z2=0x^2 + y^2 + z^2 = 0, we get 02+02+02=00^2 + 0^2 + 0^2 = 0, which is true. Thus, the zero vector is in WW
step 3
To check closure under addition, consider two vectors in WW, say u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3). Their sum u+v\mathbf{u} + \mathbf{v} must also satisfy the equation x2+y2+z2=0x^2 + y^2 + z^2 = 0. However, in F2\mathbb{F}_{2}, the sum of squares is not necessarily zero. For example, (1,1,0)+(1,1,0)=(0,0,0)(1,1,0) + (1,1,0) = (0,0,0), which does not satisfy the equation. Thus, WW is not closed under addition
step 4
Since WW is not closed under addition, it is not a subspace
Answer
(a) Not a subspace, (b) Not a subspace, (c) Subspace, (d) Not a subspace
Key Concept
Subspace conditions in vector spaces
Explanation
To determine if a set is a subspace, it must contain the zero vector, be closed under addition, and be closed under scalar multiplication.
Solution by Steps
step 1
To determine if TT is surjective, we need to check if the rank of matrix AA is equal to the dimension of the codomain WW
step 2
The rank of AA is the number of non-zero rows in its row echelon form. From the Asksia-LL calculator result, the row-reduced form of AA is: [1amp;0amp;2amp;0amp;0amp;30amp;1amp;−10amp;4amp;0amp;−20amp;0amp;0amp;0amp;1amp;70amp;0amp;0amp;0amp;0amp;00amp;0amp;0amp;0amp;0amp;0] \left[\begin{array}{cccccc} 1 & 0 & 2 & 0 & 0 & 3 \\ 0 & 1 & -10 & 4 & 0 & -2 \\ 0 & 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] The rank of AA is 3
step 3
Since the rank of AA (3) is less than the number of rows (5), TT is not surjective
Answer
TT is not surjective.
(b) What is the nullity of TT?
step 1
The nullity of TT is the dimension of the kernel of TT. By the rank-nullity theorem, we have: nullity(T)=dim(V)−rank(T) \text{nullity}(T) = \text{dim}(V) - \text{rank}(T) where dim(V)\text{dim}(V) is the number of columns of AA
step 2
Given dim(V)=6\text{dim}(V) = 6 and rank(T)=3\text{rank}(T) = 3, we get: nullity(T)=6−3=3 \text{nullity}(T) = 6 - 3 = 3
Answer
The nullity of TT is 3.
(c) Find a non-zero linear combination of the basis vectors in the kernel of TT.
step 1
To find a non-zero linear combination in the kernel of TT, we need to find a non-trivial solution to Ax=0A\mathbf{x} = \mathbf{0}. From the Asksia-LL calculator result, the null space of AA is: {(−2x−3z,10x−4y+2z,x,y,−7z,z):x,y,z∈R} \{(-2x - 3z, 10x - 4y + 2z, x, y, -7z, z) : x, y, z \in \mathbb{R}\}
step 2
Choosing x=1x = 1, y=0y = 0, and z=0z = 0, we get: (−2(1)−3(0),10(1)−4(0)+2(0),1,0,−7(0),0)=(−2,10,1,0,0,0) (-2(1) - 3(0), 10(1) - 4(0) + 2(0), 1, 0, -7(0), 0) = (-2, 10, 1, 0, 0, 0)
Answer
A non-zero linear combination in the kernel of TT is −2b1+10b2+b3-2\mathbf{b}_1 + 10\mathbf{b}_2 + \mathbf{b}_3.
(d) Find a basis for the image of TT.
step 1
The basis for the image of TT is given by the pivot columns of AA in its row echelon form. From the row-reduced form of AA, the pivot columns are the first, second, and fifth columns
step 2
Therefore, the basis for the image of TT is: {[10000],[01000],[00100]} \left\{\left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right], \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right], \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right]\right\}
Answer
The basis for the image of TT is {c1,c2,c3}\left\{\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3\right\}.
(e) What is the rank of BB?
step 1
The rank of BB is the same as the rank of AA because BB is another representation of the same linear transformation TT with respect to different bases
step 2
From the previous steps, we know the rank of AA is 3. Therefore, the rank of BB is also 3
Answer
The rank of BB is 3.
Key Concept
Linear Transformations and Matrix Rank
Explanation
The rank of a matrix representing a linear transformation determines the dimension of the image of the transformation, while the nullity determines the dimension of the kernel. The rank-nullity theorem relates these concepts to the dimension of the domain.
Generated Graph
Solution by Steps
step 1
To determine if TT is invertible, we need to check if the determinant of the matrix [T][T] is non-zero
step 2
The matrix [T][T] is given by: [T]=(1amp;2−2amp;−4) [T] = \begin{pmatrix} 1 & 2 \\ -2 & -4 \end{pmatrix} We calculate the determinant using the formula for a 2x2 matrix: det(A)=ad−bc \text{det}(A) = ad - bc where A=(aamp;bcamp;d) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
step 3
Substituting the values from [T][T]: det([T])=(1)(−4)−(2)(−2)=−4+4=0 \text{det}([T]) = (1)(-4) - (2)(-2) = -4 + 4 = 0 Since the determinant is zero, the matrix [T][T] is not invertible
(c) Determine if TT is invertible
step 1
The characteristic polynomial of the matrix representation [T]C,B[T]_{\mathcal{C}, \mathcal{B}} is given as −λ5+6λ-\lambda^5 + 6\lambda
step 2
To determine if TT is invertible, we need to check if λ=0\lambda = 0 is a root of the characteristic polynomial
step 3
Solving the characteristic polynomial: −λ5+6λ=0 -\lambda^5 + 6\lambda = 0 Factoring out λ\lambda: λ(−λ4+6)=0 \lambda(-\lambda^4 + 6) = 0 This gives us the roots: λ=0orλ4=6 \lambda = 0 \quad \text{or} \quad \lambda^4 = 6 Since λ=0\lambda = 0 is a root, the matrix is not invertible
(d) Determine if TT is invertible
step 1
Given that Nullity⁡(T)≥rank⁡(T)\operatorname{Nullity}(T) \geq \operatorname{rank}(T) and V≠{0}V \neq \{\mathbf{0}\}, we need to determine if TT is invertible
step 2
For TT to be invertible, the nullity of TT must be zero (i.e., Null⁡(T)={0}\operatorname{Null}(T) = \{\mathbf{0}\})
step 3
Since Nullity⁡(T)≥rank⁡(T)\operatorname{Nullity}(T) \geq \operatorname{rank}(T), it implies that there are non-zero vectors in the null space of TT. Therefore, TT is not invertible
Answer
(a) TT is not invertible because the determinant of [T][T] is zero.
(c) TT is not invertible because λ=0\lambda = 0 is a root of the characteristic polynomial.
(d) TT is not invertible because Nullity⁡(T)≥rank⁡(T)\operatorname{Nullity}(T) \geq \operatorname{rank}(T).
Key Concept
Invertibility of a Linear Transformation
Explanation
A linear transformation TT is invertible if and only if its matrix representation has a non-zero determinant, its characteristic polynomial does not have zero as a root, and its nullity is zero.
Solution by Steps
step 1
The set SS contains 4 elements: {◯,◯,□,□}\{\bigcirc, \bigcirc, \square, \square\}
step 2
The power set P(S)\mathcal{P}(S) contains all subsets of SS
step 3
The number of subsets of a set with nn elements is 2n2^n
step 4
Since SS has 4 elements, P(S)\mathcal{P}(S) has 24=162^4 = 16 elements
step 5
The dimension of P(S)\mathcal{P}(S) as a vector space over F2\mathbb{F}_2 is 16
Answer
The dimension of P(S)\mathcal{P}(S) is 16.
(b) Show that TT is surjective.
step 1
To show that TT is surjective, we need to show that for every element in F22\mathbb{F}_2^2, there is a subset X∈P(S)X \in \mathcal{P}(S) such that T(X)T(X) maps to that element
step 2
Consider the elements of F22\mathbb{F}_2^2: {[0,0],[0,1],[1,0],[1,1]}\{[0,0], [0,1], [1,0], [1,1]\}
step 3
For [0,0][0,0]: Choose X={}X = \{\} (empty set). It has an even number of black elements and an even number of circles
step 4
For [0,1][0,1]: Choose X={◯}X = \{\bigcirc\}. It has an even number of black elements and an odd number of circles
step 5
For [1,0][1,0]: Choose X={□}X = \{\square\}. It has an odd number of black elements and an even number of circles
step 6
For [1,1][1,1]: Choose X={◯,□}X = \{\bigcirc, \square\}. It has an odd number of black elements and an odd number of circles
step 7
Since we can find a subset XX for each element in F22\mathbb{F}_2^2, TT is surjective
Answer
TT is surjective.
(c) List all the elements in the null-space of TT.
step 1
The null-space of TT consists of all subsets X∈P(S)X \in \mathcal{P}(S) such that T(X)=[0,0]T(X) = [0,0]
step 2
This means XX must have an even number of black elements and an even number of circles
step 3
The subsets of SS that satisfy this condition are: {}\{\}, {◯,◯}\{\bigcirc, \bigcirc\}, {□,□}\{\square, \square\}, {◯,◯,□,□}\{\bigcirc, \bigcirc, \square, \square\}
Answer
The elements in the null-space of TT are {}\{\}, {◯,◯}\{\bigcirc, \bigcirc\}, {□,□}\{\square, \square\}, {◯,◯,□,□}\{\bigcirc, \bigcirc, \square, \square\}.
(d) Find the matrix representation of TT with respect to the ordered basis of singleton sets B={{◯},{◯},{□},{□}}\mathcal{B}=\{\{\bigcirc\},\{\bigcirc\},\{\square\},\{\square\}\} and the standard ordered basis S\mathcal{S} of F22\mathbb{F}_{2}^{2}.
step 1
The ordered basis B\mathcal{B} is {{◯},{◯},{□},{□}}\{\{\bigcirc\}, \{\bigcirc\}, \{\square\}, \{\square\}\}
step 2
Apply TT to each basis element:
step 3
T({◯})=[0,1]T(\{\bigcirc\}) = [0,1]
step 4
T({◯})=[0,1]T(\{\bigcirc\}) = [0,1]
step 5
T({□})=[1,0]T(\{\square\}) = [1,0]
step 6
T({□})=[1,0]T(\{\square\}) = [1,0]
step 7
The matrix representation of TT with respect to B\mathcal{B} and S\mathcal{S} is:
step 8
[0amp;0amp;1amp;11amp;1amp;0amp;0]\begin{bmatrix} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \end{bmatrix}
Answer
The matrix representation of TT is [0amp;0amp;1amp;11amp;1amp;0amp;0]\begin{bmatrix} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \end{bmatrix}.
Key Concept
Linear Transformation and Vector Space
Explanation
The problem involves understanding the dimension of a power set, proving surjectivity of a linear transformation, identifying elements in the null space, and finding the matrix representation of the transformation.
Solution by Steps
step 1
Given the matrix A=[2amp;1−1amp;11amp;2]A = \left[\begin{array}{cc}2 & 1 \\ -1 & 1 \\ 1 & 2\end{array}\right], we need to find the orthogonal projection matrix PP onto the column space of AA
step 2
The orthogonal projection matrix PP is given by P=A(ATA)−1ATP = A(A^T A)^{-1} A^T
step 3
First, compute ATAA^T A: AT=[2amp;−1amp;11amp;1amp;2] A^T = \left[\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 2\end{array}\right] ATA=[2amp;1−1amp;11amp;2]T[2amp;1−1amp;11amp;2]=[6amp;33amp;6] A^T A = \left[\begin{array}{cc}2 & 1 \\ -1 & 1 \\ 1 & 2\end{array}\right]^T \left[\begin{array}{cc}2 & 1 \\ -1 & 1 \\ 1 & 2\end{array}\right] = \left[\begin{array}{cc}6 & 3 \\ 3 & 6\end{array}\right]
step 4
Next, compute (ATA)−1(A^T A)^{-1}: (ATA)−1=[6amp;33amp;6]−1=127[6amp;−3−3amp;6]=[29amp;−19−19amp;29] (A^T A)^{-1} = \left[\begin{array}{cc}6 & 3 \\ 3 & 6\end{array}\right]^{-1} = \frac{1}{27} \left[\begin{array}{cc}6 & -3 \\ -3 & 6\end{array}\right] = \left[\begin{array}{cc}\frac{2}{9} & -\frac{1}{9} \\ -\frac{1}{9} & \frac{2}{9}\end{array}\right]
step 5
Now, compute P=A(ATA)−1ATP = A(A^T A)^{-1} A^T: P=[2amp;1−1amp;11amp;2][29amp;−19−19amp;29][2amp;−1amp;11amp;1amp;2] P = \left[\begin{array}{cc}2 & 1 \\ -1 & 1 \\ 1 & 2\end{array}\right] \left[\begin{array}{cc}\frac{2}{9} & -\frac{1}{9} \\ -\frac{1}{9} & \frac{2}{9}\end{array}\right] \left[\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 2\end{array}\right] P=[59amp;19amp;4919amp;59amp;−4949amp;−49amp;59] P = \left[\begin{array}{ccc}\frac{5}{9} & \frac{1}{9} & \frac{4}{9} \\ \frac{1}{9} & \frac{5}{9} & -\frac{4}{9} \\ \frac{4}{9} & -\frac{4}{9} & \frac{5}{9}\end{array}\right]
step 6
(a) The null-space of PP is the set of all vectors that are mapped to the zero vector by PP. Since PP is a projection matrix, its null-space is the orthogonal complement of the column space of AA. The image of PP is the column space of AA
step 7
(b) The rank of PP is the dimension of the column space of AA, which is 2. The nullity of PP is the dimension of the null-space of PP, which is 1
step 8
(c) To find the singular value decomposition (SVD) of AA, we need to find matrices UU, Σ\Sigma, and VV such that A=UΣVTA = U \Sigma V^T. This involves finding the eigenvalues and eigenvectors of ATAA^T A and AATA A^T
Answer
(a) The null-space of PP is the orthogonal complement of the column space of AA. The image of PP is the column space of AA.
(b) The rank of PP is 2, and the nullity of PP is 1.
(c) The singular value decomposition of AA involves finding matrices UU, Σ\Sigma, and VV such that A=UΣVTA = U \Sigma V^T.
Key Concept
Orthogonal Projection Matrix
Explanation
The orthogonal projection matrix PP projects vectors onto the column space of AA. The rank of PP is the dimension of the column space, and the nullity is the dimension of the orthogonal complement. The SVD of AA provides a factorization into orthogonal matrices and a diagonal matrix of singular values.
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