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Linear Algebra : Eigenvalues and Eigenvectors

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Example Question #71 : Eigenvalues And Eigenvectors

is an involutory matrix.

True, false, or indeterminate: 0 is an eigenvalue of .

Possible Answers:

False

True

Indeterminate

Correct answer:

False

Explanation:

An eigenvalue of an involutory matrix must be either 1 or . This can be seen as follows:

Let $\lambda$ be an eigenvalue of involutory matrix . Then for some eigenvector $\textbf{v}$ ,

$A\textbf{v} = \lambda \textbf{v}$

Premultiply both sides by :

$AA\textbf{v} =A \lambda \textbf{v}$

$A^{2}\textbf{v} = \lambda A \textbf{v}$

By definition, an involutory matrix has as its square, so

$I \textbf{v} = \lambda A \textbf{v}$

$\textbf{v} = \lambda A \textbf{v}$

$\frac{1}{ \lambda }\textbf{v} =A \textbf{v}$

By transitivity,

$\lambda \textbf{v} = \frac{1}{ \lambda }\textbf{v}$

Thus, $\lambda = \frac{1}{\lambda}$ , or $\lambda^{2} = 1$

It follows that $\lambda = \pm 1$ . The statement is false.

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Example Question #451 : Operations And Properties

The trace of a singular $2 \times 2$ matrix is 12. Give its set of eigenvalues.

Possible Answers:

$\left \{ -6, 6\right \}$

$\left \{ 0, 12\right \}$

Insufficient information is given to answer the question.

$\left \{ 0, 6\right \}$

$\left \{ 12\right \}$

Correct answer:

$\left \{ 0, 12\right \}$

Explanation:

, being a singular matrix, must have 0 as an eigenvalue. Let $\lambda$ be its other eigenvalue..

The trace of a matrix is equal to the sum of its eigenvalues, so

$\lambda+ 0 = \textup{Tr}(A)= 12$ ,

and

$\lambda = 12$

The set of eigenvalues of is $\left \{ 0, 12\right \}$ .

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Example Question #451 : Operations And Properties

A $4 \times 4$ matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

True, false, or indeterminate: the matrix is singular.

Possible Answers:

True

False

Indeterminate

Correct answer:

False

Explanation:

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. Since 0 is not an element of its eigenvalue set, is nonsingular.

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Example Question #452 : Operations And Properties

A $5 \times 5$ matrix has as its set of eigenvalues $\left \{ -1000, 1000, 0, -1000i, 1000i \right \}$ .

True, false, or indeterminate: the matrix is singular.

Possible Answers:

False

Indeterminant

True

Correct answer:

True

Explanation:

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. This is seen to be the case.

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Example Question #531 : Linear Algebra

The trace of a singular $3 \times 3$ matrix is 0.

Which of the following must be true of the eigenvalues of as a result?

Possible Answers:

One eigenvalue is 0; the other two are each other's additive inverse.

One eigenvalue is 0; the other two are each other's complex conjugate.

The only eigenvalue is 0.

One eigenvalue is 0; the other two are each other's multiplicative inverse.

0 is not an eigenvalue.

Correct answer:

One eigenvalue is 0; the other two are each other's additive inverse.

Explanation:

is singular, so the matrix must have 0 as an eigenvalue.

Let $\lambda_{1}, \lambda_{2}$ be the other two eigenvalues. The sum of the eigenvalues of a matrix is equal to its trace, so

$\lambda_{1}+ \lambda_{2} + 0 = \textup{Tr}(A) = 0$

and

$\lambda_{1}+ \lambda_{2} = 0$

$\lambda_{2} = - \lambda_{1}$

It follows that one eigenvalue must be 0, and the other two must be additive inverses.

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Example Question #73 : Eigenvalues And Eigenvectors

The trace of a singular $3 \times 3$ matrix is 0; one of its eigenvalues is . What is it characteristic equation?

Possible Answers:

$\lambda ^{3}+16\lambda = 0$

$\lambda ^{3}- \lambda^{2}-16\lambda +16 = 0$

$\lambda ^{3}-16\lambda = 0$

$\lambda ^{3}- \lambda^{2}+16\lambda -16 = 0$

$\lambda ^{3}+ \lambda^{2}-16\lambda -16 = 0$

Correct answer:

$\lambda ^{3}+16\lambda = 0$

Explanation:

, being a singular matrix, must have 0 as an eigenvalue; it also has as an eigenvalue. Being $3 \times 3$ , it will have one more; call this eigenvalue $\lambda$ .

The sum of the eigenvalues of a matrix is equal to its trace, so

$\lambda + 4i + 0 = \textup{Tr}(A)$

$\lambda + 4i = 0$

$\lambda = - 4i$

The set of eigenvalues is $\left \{ -4i, 0, 4i \right \}$ . The eigenvalues of a matrix are the solutions of its characteristic (polynomial) equation, which, as a consequence, is

$(\lambda + 4i )(\lambda -0 )(\lambda - 4i ) = 0$

$(\lambda )(\lambda + 4i )(\lambda - 4i ) = 0$

$(\lambda )(\lambda ^{2}+16) = 0$

$\lambda ^{3}+16\lambda = 0$

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Example Question #77 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 2 & 3 & 0 \\ 1 & -1 & 0 \\ 4 & 1 & -3\end{bmatrix}$

True or false: 0 is an eigenvalue of .

Possible Answers:

False

True

Correct answer:

False

Explanation:

A necessary and sufficient condition for a matrix to have 0 as an eigenvalue is for the matrix to have determinant 0. Find the determinant of by adding the alternating products of each entry in any row or column and the corresponding adjoint. The third column is the easiest to do this with:

$\det A= \begin{vmatrix} 2 & 3 & 0 \\ 1 & -1 & 0 \\ 4 & 1 & -3\end{vmatrix}$

$= 0 \begin{vmatrix} 1 & -1 \\ 4 & 1 \end{vmatrix}- 0 \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix}+ (-3)\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}$

$= (-3)\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}$

$= (-3) \left [ (2)(-1)- 3(1) \right ]$

= (-3)(-5)

= 15

Since $\det A \ne 0$ , 0 is not an eigenvalue of .

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Example Question #78 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 0 & 0 & 3 \\ -3 & 9 & -3 \\ 4 & -12 & -6 \end{bmatrix}$

True or false: 0 is an eigenvalue of .

Possible Answers:

False

True

Correct answer:

True

Explanation:

$\det A = \begin{vmatrix} 0 & 0 & 3 \\ -3 & 9 & -3 \\ 4 & -12 & -6 \end{vmatrix}$

$=0 \begin{vmatrix} 9 & -3 \\ -12 & -6 \end{vmatrix} +0 \begin{vmatrix} -3 & -3 \\ 4 & -6 \end{vmatrix}+ 3\begin{vmatrix} -3 & 9 \\ 4 & -12 \end{vmatrix}$

$=3\begin{vmatrix} -3 & 9 \\ 4 & -12 \end{vmatrix}$

= 3[(-3)(-12)-9(4)]

= 3(36-36)

Since $\det A = 0$ , has zero as an eigenvalue.

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Example Question #73 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 4 & 3 \\ 2 & x\end{bmatrix}$

Calculate so that has 0 as an eigenvalue.

Possible Answers:

x = -4

$x= \frac{3}{2}$

x= 0

$x= \frac{1}{4}$

$x= \frac{7}{4}$

Correct answer:

$x= \frac{3}{2}$

Explanation:

A necessary and sufficient condition for a matrix to have 0 as an eigenvalue is for the matrix to have determinant 0. Find the determinant of in terms of by taking the product of the main diagonal elements and subtracting the product of the other two:

$\det A = \begin{vmatrix} 4 & 3 \\ 2 & x\end{vmatrix} = 4x - 3(2) = 4x - 6$

Set this equal to 0 and solve for :

4x - 6= 0

4x= 6

$x= \frac{3}{2}$ .

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Example Question #73 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 4 & 3 \\ 2 & x\end{bmatrix}$

Calculate so that has 2 as an eigenvalue.

Possible Answers:

x= 6

x= 7

x= 2

x= 3

x=5

Correct answer:

x=5

Explanation:

A necessary and sufficient condition for a number $\lambda$ to be an eigenvalue of is for

$\det (A - \lambda I)= 0$

to be true. Therefore, first, find A- 2I ; this is

$A - 2I = \begin{bmatrix} 4 & 3 \\ 2 & x\end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 4 & 3 \\ 2 & x\end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2 & 3 \\ 2 & x-2 \end{bmatrix}$

Set the determinant of this matrix, which is found by taking the product of the main diagonal elements and subtracting the product of the other two, equal to 0:

$\begin{vmatrix} 2 & 3 \\ 2 & x-2 \end{vmatrix} = 0$

2(x-2)-3(2)=0

2x-4-6=0

2x-10=0

2x=10

x=5

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Linear Algebra : Eigenvalues and Eigenvectors

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #71 : Eigenvalues And Eigenvectors

Example Question #451 : Operations And Properties

Example Question #451 : Operations And Properties

Example Question #452 : Operations And Properties

Example Question #531 : Linear Algebra

Example Question #73 : Eigenvalues And Eigenvectors

Example Question #77 : Eigenvalues And Eigenvectors

Example Question #78 : Eigenvalues And Eigenvectors

Example Question #73 : Eigenvalues And Eigenvectors

Example Question #73 : Eigenvalues And Eigenvectors

All Linear Algebra Resources

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