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Linear Algebra : Eigenvalues and Eigenvectors

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Example Questions

Example Question #471 : Linear Algebra

True or false:

$\begin{bmatrix} 1\\ 2 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(2)\\ 2(1)+ 5(2) \end{bmatrix}$

$= \begin{bmatrix}4+2\\ 2+10 \end{bmatrix}$

$= \begin{bmatrix}6\\ 12 \end{bmatrix}$

Since $6 = 6 \cdot 1$ and $12 = 6 \cdot 2$ , it follows that

Av = 6v .

Therefore, such a $\lambda$ exists (and is equal to 6), and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ is indeed an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

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Example Question #21 : Eigenvalues And Eigenvectors

True or false:

$\begin{bmatrix} 1\\ 1 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(1)\\ 2(1)+ 5(1) \end{bmatrix}$

$= \begin{bmatrix}4+1\\ 2+5\end{bmatrix}$

$= \begin{bmatrix}5\\ 7\end{bmatrix}$

$5 = 5 \cdot 1$ , but $7= 7 \cdot 1$ . Therefore, there cannot exist $\lambda$ such that $A v = \lambda v$ . This means that $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ is not an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

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Example Question #21 : Eigenvalues And Eigenvectors

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

is an example of an elementary matrix. What row operation does it represent?

Possible Answers:

None of the other choices gives the correct response.

$R2 \leftrightarrow R3$

$R2+R3 \rightarrow R2$

$R2+R3 \rightarrow R3$

$R1 \leftrightarrow R4$

Correct answer:

$R2 \leftrightarrow R3$

Explanation:

An elementary matrix - a matrix which can pre-multiply a matrix of a linear system to perform a single row operation - must be obtainable by performing a single row operation on the identity matrix, which here is the four-by-four identity

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

The given matrix switches Rows 2 and 3 in the identity matrix, so the row operation that the matrix represents is the switching of Rows 2 and 3. The notation for this is $R2 \leftrightarrow R3$ .

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Example Question #481 : Linear Algebra

Suppose we have a square matrix with real-valued entries with only positive eigenvalues. Is invertible? Why or why not?

Possible Answers:

It is invertible because none of its eigenvalues is

It is not invertible because none of its eigenvalues is negative.

It is invertible because it has no negative eigenvalues.

It is not invertible because is not an eigenvalue

Correct answer:

It is invertible because none of its eigenvalues is

Explanation:

The square matrix is certainly invertible with the reason being that none of its eigenvalues is . We know that is not an eigenvalue, so the following

Av=0v

does not hold for any nonzero vector in $\mathbb{R}^n$ , i.e.

$Av\neq0$

for all nonzero $v\in \mathbb{R}^n$ . So the only vector that maps to the zero vector is the zero vector. In a square matrix, this is equivalent to the null space being the zero vector: $Null(A)=\{0\}$ . This is also equivalent to the matrix being invertible.

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Example Question #25 : Eigenvalues And Eigenvectors

A three-by-three matrix has as its three eigenvalues , , and 1. Give its characteristic equation.

Possible Answers:

$p (\lambda) = \lambda ^{3} - 6$

$p (\lambda) = \lambda ^{3} -6\lambda ^{2}+ 11\lambda -6$

$p (\lambda) = \lambda ^{3} +4\lambda ^{2} + \lambda - 6$

$p (\lambda) = \lambda ^{3} +5\lambda ^{2} - \lambda - 5$

$p (\lambda) = \lambda ^{3} -5\lambda ^{2} - \lambda + 5$

Correct answer:

$p (\lambda) = \lambda ^{3} +4\lambda ^{2} + \lambda - 6$

Explanation:

The characteristic equation of a matrix is the polynomial that has as its zeroes the eigenvalues of the matrix. Equivalently, if a matrix has a value as an eigenvalue, then its characteristic equation has as a factor the binomial expression $\lambda - a$ . Since the eigenvalues of the matrix are , , and 1, the polynomial is equal to:

$p (\lambda) =[ \lambda - (-3)][ \lambda - (-2)](\lambda - 1)$

$p (\lambda) = (\lambda+3)( \lambda +2) (\lambda - 1)$

Multiply the binomials together:

$p (\lambda) = (\lambda ^{2} + 5\lambda + 6 ) (\lambda - 1)$

$p (\lambda) = \lambda ^{3} +4\lambda ^{2} + \lambda - 6$

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Example Question #401 : Operations And Properties

A three-by-three matrix has as its three eigenvalues 2, , and -2i . Give its characteristic equation.

Possible Answers:

$p (\lambda) = \lambda ^{3}- 2 \lambda ^{2}-4 \lambda +8$

$p (\lambda) = \lambda ^{3}- 2 \lambda ^{2}+ 4 \lambda -8$

$p (\lambda) = \lambda ^{3}- 8$

$p (\lambda) = \lambda ^{3} +8$

$p (\lambda) = \lambda ^{3}+ 2 \lambda ^{2}-4 \lambda -8$

Correct answer:

$p (\lambda) = \lambda ^{3}- 2 \lambda ^{2}+ 4 \lambda -8$

Explanation:

$p (\lambda) =(\lambda - 2)(\lambda - 2i)[\lambda - (-2i)]$

$p (\lambda) =(\lambda - 2)(\lambda - 2i)(\lambda +2i)$

Multiply the binomials together:

$p (\lambda) =(\lambda - 2) [(\lambda ^{2}-( 2i) ^{2}]$

$p (\lambda) =(\lambda - 2) ( \lambda ^{2}- 2^{2}i ^{2} )$

$p (\lambda) =(\lambda - 2) ( \lambda ^{2}+4 )$

$p (\lambda) = \lambda ^{3}- 2 \lambda ^{2}+ 4 \lambda -8$

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Example Question #27 : Eigenvalues And Eigenvectors

The characteristic equation of a three-by-three matrix is $\lambda ^{3} - 216$ .

All of the following are eigenvalues of the matrix except for:

Possible Answers:

$-3 -3i \sqrt{3}$

$-3 + 3i \sqrt{3}$

Correct answer:

Explanation:

The characteristic equation of a matrix is the polynomial that has as its zeroes the eigenvalues of the matrix. Therefore, to find the eigenvalues, set the characteristic equation equal to zero and solve for $\lambda$ :

$\lambda ^{3} - 216 = 0$

Factor $\lambda ^{3} - 216$ as the difference of two cubes, as follows:

$\lambda ^{3} - 6^{3} = 0$

$(\lambda - 6) (\lambda ^{2} + \lambda \cdot 6 + 6 ^{2}) = 0$

$(\lambda - 6) (\lambda ^{2} + 6 \lambda + 36) = 0$

Find the zeroes of both factors.

$\lambda - 6 = 0$

$\lambda = 6$

$\lambda ^{2} + 6 \lambda + 36 = 0$

Apply the quadratic formula:

$\lambda = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ , where a = 1, b = 6, c = 36

$\lambda = \frac{-6 \pm \sqrt{6^{2}-4 (1)(36)}}{2 (1)}$

$\lambda = \frac{-6 \pm \sqrt{36-144}}{2 }$

$\lambda = \frac{-6 \pm \sqrt{-108}}{2 }$

$\lambda = \frac{-6 \pm 6i \sqrt{3}}{2 }$

$\lambda = -3 \pm 3i \sqrt{3}$

The three eigenvalues of the matrix are $-3 + 3i \sqrt{3}$ , $-3 -3i \sqrt{3}$ , and 6. Therefore, the one choice that does not given an eigenvalue is .

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Example Question #28 : Eigenvalues And Eigenvectors

A four-by-four matrix has as its set of eigenvalues $\left \{ -6, 6, -4i, 4i \right \}$ . Give its characteristic equation.

Possible Answers:

$p(\lambda) = \lambda^{4} -52\lambda^{2} +576$

$p(\lambda) = \lambda^{4} -20 \lambda^{2} -576$

$p(\lambda) = \lambda^{4}+52\lambda^{2} +576$

$p(\lambda) = \lambda^{4} +20 \lambda^{2} -576$

Correct answer:

$p(\lambda) = \lambda^{4} -20 \lambda^{2} -576$

Explanation:

The characteristic equation of a matrix is the polynomial that has as its zeroes the eigenvalues of the matrix. Equivalently, if a matrix has a value as an eigenvalue, then its characteristic equation has as a factor the binomial expression $\lambda - a$ . Since the set of eigenvalues of the matrix is $\left \{ -6, 6, -4i, 4i \right \}$ , the polynomial is equal to:

$p(\lambda) = [\lambda - (-6)] (\lambda-6) [\lambda-(-4i)] (\lambda-4i)$

$p(\lambda) =(\lambda+6) (\lambda-6)(\lambda+4i) (\lambda-4i)$

Multiply these binomials out:

$p(\lambda) =(\lambda^{2} -6^{2}) [\lambda^{2} - (4i)^{2}]$

$p(\lambda) =(\lambda^{2} -6^{2}) [ \lambda^{2} - 4^{2} i^{2}]$

$p(\lambda) =(\lambda^{2} -36 ) [\lambda^{2} - 16(-1)]$

$p(\lambda) =(\lambda^{2} -36 ) ( \lambda^{2} +16 )$

$p(\lambda) = \lambda^{4} -20 \lambda^{2} -576$

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Example Question #21 : Eigenvalues And Eigenvectors

True or false:

$\begin{bmatrix} - \sqrt{2}\\ 0\\ \sqrt{2} \end{bmatrix}$ is an eigenvector of $\begin{bmatrix} 1& -2&3 \\ 0& 1& 0\\ 4& 2& 2 \end{bmatrix}$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 1& -2&3 \\ 0& 1& 0\\ 4& 2& 2 \end{bmatrix}$ and $v= \begin{bmatrix} - \sqrt{2}\\ 0\\ \sqrt{2} \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

A v

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

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

$= \begin{bmatrix} - \sqrt{2}+0 + 3 \sqrt{2}\\ 0+0+0\\ -4 \sqrt{2}+ 0+ 2\sqrt{2}\end{bmatrix}$

$=\begin{bmatrix} 2\sqrt{2}\\ 0\\ -2 \sqrt{2} \end{bmatrix}$

$2 \sqrt{2} = -2 (-\sqrt{2})$

$0 = -2 \cdot 0$

$-2 \sqrt{2} = -2 ( \sqrt{2})$

Therefore, there exists a number $\lambda$ (equal to ) such that $A v = \lambda v$ . This makes

$v= \begin{bmatrix} - \sqrt{2}\\ 0\\ \sqrt{2} \end{bmatrix}$

an eigenvector of

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

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Example Question #22 : Eigenvalues And Eigenvectors

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

Which of the following is a consequence of the Cayley-Hamilton Theorem?

(Note: and refer to the two-by-two identity and zero matrices, respectively.)

Possible Answers:

Any eigenvector must be a linear combination of the vectors $\begin{bmatrix} 4\\ 1 \end{bmatrix}$ and $\begin{bmatrix} 3\\ -1 \end{bmatrix}$ .

$A^{2} -4 A+7I =O$

Any eigenvalue must fall between and 4 inclusive.

$A^{2} -4 A-I =O$

Any eigenvalue must be in the set $\left \{ -1, 1, 3, 4 \right \}$ .

Correct answer:

$A^{2} -4 A+7I =O$

Explanation:

The Cayley-Hamilton Theorem states that a square matrix is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of $\lambda I-A$ . This matrix is

$\lambda I-A$

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

Multiply $\lambda$ by each element in the identity:

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

Subtract elementwise:

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

Take the determinant by taking the product of the upper-left and lower-right entries, and subtracting the product of the other two:

$\begin{vmatrix} \lambda -1 & -4\\ 1& \lambda -3 \end{vmatrix}$

$= \left ( \lambda -1\right )\left ( \lambda -3\right ) - (-4)(1)$

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

$= \lambda ^{2} -4 \lambda+7$

The characteristic equation is

$\lambda ^{2} -4 \lambda+7 = 0$

By the Cayley-Hamilton Theorem, the matrix is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

$A^{2} -4 A+7I =O$

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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 #471 : Linear Algebra

Example Question #21 : Eigenvalues And Eigenvectors

Example Question #21 : Eigenvalues And Eigenvectors

Example Question #481 : Linear Algebra

Example Question #25 : Eigenvalues And Eigenvectors

Example Question #401 : Operations And Properties

Example Question #27 : Eigenvalues And Eigenvectors

Example Question #28 : Eigenvalues And Eigenvectors

Example Question #21 : Eigenvalues And Eigenvectors

Example Question #22 : Eigenvalues And Eigenvectors

All Linear Algebra Resources

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