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Linear Algebra : Eigenvalues and Eigenvectors

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Example Question #41 : Eigenvalues And Eigenvectors

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

Which of the following is an eigenvalue of ?

Possible Answers:

Correct answer:

Explanation:

Find the characteristic equation of by obtaining the determinant of $\lambda I - A$ .

$\lambda I - A$

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

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

$= \begin{bmatrix} \lambda -1 &- 2 & -3 \\ -1 & \lambda +1 &- 4 \\ 2 & 4 & \lambda -1 \end{bmatrix}$

The determinant of this $3 \times 3$ matrix can be found by adding the upper-left to lower-right products:

$(\lambda - 1)(\lambda + 1) (\lambda - 1) + (-2)(-4)(2) + (-3 )(-1)(4)$

$=( \lambda^{3} - \lambda^{2} - \lambda + 1 )+ 16 +12$

$= \lambda^{3} - \lambda^{2} - \lambda + 29$

Adding the upper-right to lower-left products:

$(-3)(\lambda + 1)(2) + (-4)(4)(\lambda-1) + (\lambda - 1) (-2)(-1)$

$= (-6 \lambda -6) + (-16 \lambda+16) + (2\lambda - 2)$

$= -20 \lambda +8$

And subtracting the latter from the former:

$= \lambda^{3} - \lambda^{2} - \lambda + 29 - ( -20 \lambda +8)$

$= \lambda^{3} - \lambda^{2} +19 \lambda + 21$

The characteristic equation is

$\lambda^{3} - \lambda^{2} +19 \lambda + 21 = 0$

Factor the polynomial completely

$(\lambda + 1) (\lambda ^{2} - 2 \lambda + 21) = 0$

The solution set of this equation is comprised of the zeroes of both polynomial factors:

$\left \{ -1, 1 - i \sqrt{20}, 1 + i \sqrt{20} \right \}$

Only is a choice.

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Example Question #41 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ .

Is an eigenvalue of , and if so, what is the dimension of its eigenspace?

Possible Answers:

No.

Yes; the dimension is 2.

Yes; the dimension is 3.

Yes; the dimension is 1.

Correct answer:

No.

Explanation:

Assume that is an eigenvalue of . Then, if $b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

Ab = -2 b , or, equivalently,

(A + 2I) b =O ,

where I,O are the $3 \times 3$ identity and zero matrices, respectively.

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ , so

$A + 2 I = \begin{bmatrix} 2+2 & 3 & 6 \\ 0 & 4+2 & 4 \\ 2 & 0 & 4+2 \end{bmatrix}$

$= \begin{bmatrix} 4& 3 & 6 \\ 0 & 6 & 4 \\ 2 &6& 2\end{bmatrix}$

Changing to reduced row echelon form:

$R1 \leftrightarrow R3$

$\begin{bmatrix} 2 &6& 2 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}$

$\frac{1}{2} R1 \rightarrow R1$

$\begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}$

$-4R1+ R3 \rightarrow R3$

$\begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 0& -9 & 2 \end{bmatrix}$

$\frac{1}{6} R2 \rightarrow R2$

$\begin{bmatrix} 1 &3&1 \\ 0 & 1 & \frac{2}{3} \\ 0& -9 & 2 \end{bmatrix}$

$-3R2 + R1 \rightarrow R1$

$9 R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 8 \end{bmatrix}$

$\frac{1}{8} R3 \rightarrow R3$

$\begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 1 \end{bmatrix}$

We do not need to go further to see that this matrix will not have a row of zeroes. This means the rank of the matrix is 3, and the nullity is 0. If this happens, the tested value, in this case , is not an eigenvalue.

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Example Question #41 : Eigenvalues And Eigenvectors

A $4 \times 4$ matrix has exactly two distinct eigenvalues: 1 and . Which of the following cannot be its characteristic equation?

Possible Answers:

$\lambda ^{4} - 2 \lambda^{2} + 1 = 0$

$\lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0$

$\lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0$

Any of the four choices are possible characteristic equations.

$\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$

Correct answer:

$\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$

Explanation:

A $4 \times 4$ matrix has four not necessarily distinct eigenvalues, $\lambda_{1} , \lambda_{2} , \lambda_{3} , \lambda_{4}$ , which are the solutions of its characteristic polynomial equation

$(\lambda - \lambda_{1}) (\lambda - \lambda_{2}) (\lambda - \lambda_{3} ) (\lambda - \lambda_{4})=0$

If there are only two distinct eigenvalues, 1 and , then one of the following three situations happens:

1 has multiplicity 3 and has multiplicity 1, in which case $\lambda_{1} = \lambda_{2} = \lambda_{3}=1 , \lambda_{4} = -1$ ,

and the characteristic equation is

$(\lambda - 1) (\lambda - 1) (\lambda - 1 ) (\lambda +1)=0$

$\lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0$

1 has multiplicity 2 and has multiplicity 2, in which case $\lambda_{1} = \lambda_{2} = 1 , \lambda_{3}= \lambda_{4} = -1$ ,

and the characteristic equation is

$(\lambda - 1) (\lambda - 1) (\lambda + 1 ) (\lambda +1)=0$

$\lambda ^{4} - 2 \lambda^{2} + 1 = 0$

1 has multiplicity 1 and has multiplicity 3, in which case $\lambda_{1} = 1 , \lambda_{2} = \lambda_{3}= \lambda_{4} = -1$

and the characteristic equation is

$(\lambda - 1) (\lambda + 1) (\lambda + 1 ) (\lambda +1)=0$

$\lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0$

Of the four equations given as choices, only $\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$ cannot be a characteristic equation of the matrix.

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Example Question #41 : Eigenvalues And Eigenvectors

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

Which of the following statements follows from the Cayley-Hamilton Theorem?

Possible Answers:

has two distinct real eigenvalues.

and $A^{2}$ have the same eigenvalues.

Every vector in $\mathbb{R}^{2}$ is an eigenvector of .

None of the other choices gives a correct response.

has one eigenvalue of multiplicity 2.

Correct answer:

None of the other choices gives a correct response.

Explanation:

By the Cayley-Hamilton Theorem, a matrix is a solution of its own characteristic polynomial equation. None of the choices addresses the characteristic equation of .

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Example Question #41 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ .

Is 2 an eigenvalue of , and if so, what is the dimension of its eigenspace?

Possible Answers:

Yes; the dimension is 3.

Yes; the dimension is 1.

No.

Yes; the dimension is 2.

Correct answer:

Yes; the dimension is 1.

Explanation:

Assume that 2 is an eigenvalue of . Then, if $b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

Ab = 2 b , or, equivalently,

(A - 2I) b =O ,

where I,O are the $3 \times 3$ identity and zero matrices, respectively.

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ , so

$A - 2 I = \begin{bmatrix} 2-2 & 3 & 6 \\ 0 & 4-2 & 4 \\ 2 & 0 & 4-2 \end{bmatrix}$

$= \begin{bmatrix} 0& 3 & 6 \\ 0 & 2 & 4 \\ 2 & 0 & 2\end{bmatrix}$

Changing to reduced row echelon form:

$R1 \leftrightarrow R3$

$\begin{bmatrix}2 & 0 & 2 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\frac{1}{2}R1 \rightarrow R1$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\frac{1}{2}R2 \rightarrow R2$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 3 & 6 \end{bmatrix}$

$-3R2+ R3 \rightarrow R3$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 0 & 0 \end{bmatrix}$

The matrix is in reduced row echelon form. There is one column which does not contain leading 1s, so 2 is indeed an eigenvalue. The eigenspace is the set of all vectors (x, y, z) such that x= -z , y = -2z - that is, the set of vectors t(-1,- 2, 1) . The eigenspace has dimension 1.

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Example Question #41 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} a & 4 & -2 \\ 1 & 3 & 0 \\ - 6 & 4 & a \end{bmatrix}$

Evaluate so that the sum of the eigenvalues of is 10.

Possible Answers:

a = -6.5

a = 7

a = 3.5

The sum of the eigenvalues of is 10 regardless of the value of .

a = -13

Correct answer:

a = 3.5

Explanation:

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

$A = \begin{bmatrix} {\color{Red}\textbf{a} } & 4 & -2 \\ 1 & {\color{Red}\textbf{3}} & 0 \\ - 6 & 4 & {\color{Red}\textbf{a}} \end{bmatrix}$

Set the trace equal to 10 and solve for :

$\textup{Tr} (A ) = 10$

a+ a + 3 = 10

2 a+ 3 = 10

2a = 7

a = 3.5

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Example Question #41 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 0.1 & 0.4 & 0. 7 \\ 0. 5 & 0. 2 & 0.3 \\ 0.4 &0.4 & 0 \end{bmatrix}$

True or false: 1 is an eigenvalue of .

Possible Answers:

True

False

Correct answer:

True

Explanation:

Examine the columns of :

$A= \begin{bmatrix} {\color{Red} 0.1} & {\color{DarkGreen} 0.4} & {\color{Blue} 0. 7} \\ {\color{Red} 0. 5} & {\color{DarkGreen} 0. 2} &{\color{Blue} 0.3} \\ {\color{Red} 0.4 }&{\color{DarkGreen} 0.4} & {\color{Blue} 0} \end{bmatrix}$

The entries of each column add up to 1:

Column 1: 0.1 + 0.5 + 0.4 = 1

Column 2: 0.4+ 0.2 + 0.4 = 1

Column 3: 0.7+0.3 + 0 = 1

It follows that could be a stochastic matrix for a state system; one of the properties of such a matrix is that one of its eigenvalues must be 1.

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Example Question #41 : Eigenvalues And Eigenvectors

A $4 \times 4$ real matrix has as two of its eigenvalues and 2-3i . Give its characteristic equation.

Possible Answers:

$\lambda ^{4}+\lambda ^{2}-36 \lambda +52 = 0$

Insufficient information is provided to answer the question.

$\lambda ^{4}+\lambda ^{2}+36 \lambda +52 = 0$

$\lambda ^{4}+4 \lambda ^{3}+9 \lambda ^{2}-16 \lambda -52 = 0$

$\lambda ^{4}-4 \lambda^{3}+9 \lambda^{2}+16 \lambda-52 = 0$

Correct answer:

Insufficient information is provided to answer the question.

Explanation:

A $4 \times 4$ matrix will have four eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since 2-3i is a zero, so it 2+3i .

We now know three of the zeroes, and since only one eigenvalue is unknown, it must be a real value. However, there is no restriction on this zero. Therefore, we have no way of determining the fourth zero - and, consequently, no way of figuring out the characteristic equation with any certainty.

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Example Question #42 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 7 & a \\ -3 & b \end{bmatrix}$ for some $a,b \in \mathbb{C}$ .

has as its eigenvalues $4+ \sqrt{3}$ and $4 - \sqrt{3}$ . Which of the following is equal to ?

Possible Answers:

a= 4

a=0

a= 2

a= -4

a = -2

Correct answer:

a= 2

Explanation:

The sum of the eigenvalues of a matrix is equal to the trace of a matrix - the sum of its diagonal elements; the product of a matrix is equal to its determinant.

Therefore, to find , it is necessary to first find . The trace of is equal to diagonal sum b+7 , so set that equal to the sum of the given eigenvalues and solve:

$b+7 = (4+ \sqrt{3}) + (4- \sqrt{3})$

b+7 = 8

b =1

The matrix is therefore

$A = \begin{bmatrix} 7 & a \\ -3 & 1 \end{bmatrix}$

can be found by setting the determinant - the upper-left to lower-right product minus the upper-right to lower-left product - equal to the product of the eigenvalues:

$7(1)-a(-3) = (4+ \sqrt{3}) (4- \sqrt{3})$

3a+ 7 = 13

3a = 6

a= 2 ,

the correct choice.

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Example Question #43 : Eigenvalues And Eigenvectors

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

Which of the following holds for the eigenvectors of ?

Possible Answers:

All eigenvectors of are scalar multiples of $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

Every vector in $\mathbb{R}^{3}$ is an eigenvector of .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ .

Correct answer:

All eigenvectors of are scalar multiples of $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

Explanation:

is a lower triangular matrix - all elements above its main diagonal are equal to 0. Such a matrix has as its eigenvalues its diagonal elements. All three diagonal elements in are 1, so 1 is the only eigenvalue of .

To find the eigenvector(s) of corresponding to $\lambda = 1$ , first, find the matrix

$A - \lambda I = A - I$ .

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

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

Perform the Gauss-Jordan elimination process to get this matrix in reduced row-echelon form:

$R1 \leftrightarrow R2$

$R2 \leftrightarrow R3$

$\begin{bmatrix} 2 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2} R1 \rightarrow R1$

$\begin{bmatrix} 1 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$-3R1 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 \\ 0& 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2} R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 \\ 0&1 & 0\\ 0 & 0 & 0 \end{bmatrix}$

This matrix is interpreted as

x = 0 , y = 0 , arbitrary.

The eigenvectors must all take the form

$\begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}$

for some scalar - equivalently, all eigenvectors of are scalar multiples of

$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

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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 #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #42 : Eigenvalues And Eigenvectors

Example Question #43 : Eigenvalues And Eigenvectors

All Linear Algebra Resources

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