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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.
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