CALCU-下午 5:10 6月3日周一50 \%MAST10007Semester 2,2023L

Solution by Steps

step 1

To determine if

S

is linearly independent or dependent, we need to examine the row-reduced form of matrix

A

step 2

The row-reduced form of

A

is given as

B = \begin{pmatrix} 1 &amp; 0 &amp; -1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 2 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 2 &amp; -2 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix}

step 3

Since

B

has a row of zeros, the rank of

A

is less than the number of columns

step 4

Therefore, the set

S

is linearly dependent

Answer

The set

S

is linearly dependent.

(b) Does

S

span

\mathcal{P}_{3}

?

step 1

To determine if

S

spans

\mathcal{P}_{3}

, we need to check if the rank of

A

is equal to the dimension of

\mathcal{P}_{3}

step 2

The dimension of

\mathcal{P}_{3}

is 4

step 3

The rank of

A

(or

B

) is 3, as there are 3 non-zero rows in

B

step 4

Since the rank is less than 4,

S

does not span

\mathcal{P}_{3}

Answer

The set

S

does not span

\mathcal{P}_{3}

.

(c) Find a basis

\mathcal{B}

for

\operatorname{span}(S)

consisting of polynomials in

\mathcal{S}

.

step 1

To find a basis for

\operatorname{span}(S)

, we need to select the linearly independent vectors from

S

step 2

From the row-reduced form

B

, the pivot columns correspond to the linearly independent vectors

step 3

The pivot columns in

B

are the 1st, 2nd, and 4th columns

step 4

Therefore, the basis

\mathcal{B}

is

\{ \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4 \}

Answer

The basis

\mathcal{B}

for

\operatorname{span}(S)

is

\{ \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4 \}

.

(d) What is the dimension of

\operatorname{span}(S)

?

step 1

The dimension of

\operatorname{span}(S)

is equal to the number of vectors in the basis

\mathcal{B}

step 2

From part (c), the basis

\mathcal{B}

has 3 vectors

Answer

The dimension of

\operatorname{span}(S)

is 3.

(e) For each polynomial in

\mathcal{S}

that is not in

\mathcal{B}

, find the coordinate vector with respect to the basis

\mathcal{B}

.

step 1

The polynomials not in

\mathcal{B}

are

\mathbf{p}_3

and

\mathbf{p}_5

step 2

To find the coordinate vector of

\mathbf{p}_3

, express it as a linear combination of

\mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_4

step 3

Solve the system:

c_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

\mathbf{p}_5

Answer

The coordinate vectors are found by solving the linear combinations.

(f) Is the set

\{ \mathbf{p}_1, \mathbf{p}_3, \mathbf{p}_5 \}

linearly independent or linearly dependent?

step 1

To determine if

\{ \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

\{ \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

A

is Hermitian if

A = A^\dagger

, where

A^\dagger

is the conjugate transpose of

A

step 2

Given

A = \begin{pmatrix} 2 &amp; -i \\ i &amp; 2 \end{pmatrix}

, we find

A^\dagger

by taking the transpose and then the complex conjugate:

step 3

A^\dagger = \begin{pmatrix} 2 &amp; i \\ -i &amp; 2 \end{pmatrix}

step 4

Since

A = A^\dagger

,

A

is a Hermitian matrix

(b) Find the eigenvalues of

A

.

step 1

To find the eigenvalues, solve the characteristic equation

\det(A - \lambda I) = 0

step 2

A - \lambda I = \begin{pmatrix} 2 - \lambda &amp; -i \\ i &amp; 2 - \lambda \end{pmatrix}

step 3

\det(A - \lambda I) = (2 - \lambda)(2 - \lambda) - (-i)(i) = \lambda^2 - 4\lambda + 3 = 0

step 4

Solving

\lambda^2 - 4\lambda + 3 = 0

gives

\lambda_1 = 3

and

\lambda_2 = 1

(c) Find the eigenspace corresponding to each eigenvalue of

A

.

step 1

For

\lambda_1 = 3

, solve

(A - 3I)\mathbf{v} = 0

:

step 2

(A - 3I) = \begin{pmatrix} -1 &amp; -i \\ i &amp; -1 \end{pmatrix}

step 3

Solving

\begin{pmatrix} -1 &amp; -i \\ i &amp; -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

gives

v_1 = \begin{pmatrix} -i \\ 1 \end{pmatrix}

step 4

For

\lambda_2 = 1

, solve

(A - I)\mathbf{v} = 0

:

step 5

(A - I) = \begin{pmatrix} 1 &amp; -i \\ i &amp; 1 \end{pmatrix}

step 6

Solving

\begin{pmatrix} 1 &amp; -i \\ i &amp; 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

gives

v_2 = \begin{pmatrix} i \\ 1 \end{pmatrix}

(d) Find a unitary matrix

U

and a diagonal matrix

D

such that

A = U D U^{-1}

.

step 1

The eigenvectors form the columns of

U

:

U = \begin{pmatrix} -i &amp; i \\ 1 &amp; 1 \end{pmatrix}

step 2

Normalize the columns of

U

to make it unitary:

step 3

U = \frac{1}{\sqrt{2}} \begin{pmatrix} -i &amp; i \\ 1 &amp; 1 \end{pmatrix}

step 4

The diagonal matrix

D

contains the eigenvalues:

D = \begin{pmatrix} 3 &amp; 0 \\ 0 &amp; 1 \end{pmatrix}

step 5

Verify

A = U D U^{-1}

:

step 6

U^{-1} = U^\dagger = \frac{1}{\sqrt{2}} \begin{pmatrix} i &amp; 1 \\ -i &amp; 1 \end{pmatrix}

step 7

A = \frac{1}{\sqrt{2}} \begin{pmatrix} -i &amp; i \\ 1 &amp; 1 \end{pmatrix} \begin{pmatrix} 3 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} i &amp; 1 \\ -i &amp; 1 \end{pmatrix} = \begin{pmatrix} 2 &amp; -i \\ i &amp; 2 \end{pmatrix}

Answer

(a)

A

is a Hermitian matrix.

(b) The eigenvalues are

\lambda_1 = 3

and

\lambda_2 = 1

.

(c) The eigenspaces are spanned by

v_1 = \begin{pmatrix} -i \\ 1 \end{pmatrix}

for

\lambda_1 = 3

and

v_2 = \begin{pmatrix} i \\ 1 \end{pmatrix}

for

\lambda_2 = 1

.

(d)

U = \frac{1}{\sqrt{2}} \begin{pmatrix} -i &amp; i \\ 1 &amp; 1 \end{pmatrix}

and

D = \begin{pmatrix} 3 &amp; 0 \\ 0 &amp; 1 \end{pmatrix}

such that

A = 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.

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