Eigenvalues and Eigenvectors - Linear Algebra

Card 1 of 348

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

The eigenvalues of a three-by-three matrix are:

$\left \{ -9, 8, 10i \right \}$

Which one is the dominant eigenvalue?

Tap to reveal answer

Answer

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

.

, so is the dominant eigenvalue.

← Didn't Know|Knew It →

Question

The eigenvalues of a three-by-three matrix are:

$\left \{ -9, 8, 10i \right \}$

Which one is the dominant eigenvalue?

Tap to reveal answer

Answer

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

.

, so is the dominant eigenvalue.

← Didn't Know|Knew It →

Question

The eigenvalues of a three-by-three matrix are:

$\left \{ -9, 8, 10i \right \}$

Which one is the dominant eigenvalue?

Tap to reveal answer

Answer

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

.

, so is the dominant eigenvalue.

← Didn't Know|Knew It →

Question

The eigenvalues of a three-by-three matrix are:

$\left \{ -9, 8, 10i \right \}$

Which one is the dominant eigenvalue?

Tap to reveal answer

Answer

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

.

, so is the dominant eigenvalue.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

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.

← Didn't Know|Knew It →

Question

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

Which of the following is an eigenvalue of ?

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

, 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

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

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

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

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

← Didn't Know|Knew It →

Question

The characteristic equation of a three-by-three matrix is

$\lambda ^{3}+ 10\lambda - 57= 0$

Which of the following is its dominant eigenvalue?

Tap to reveal answer

Answer

The eigenvalues of a matrix are the zeroes of its characteristic equation.

We can try to extract one or more zeroes using the Rational Zeroes Theorem, which states that any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the leading coefficient. Since the constant is 57, and the leading coefficient of the polynomial is 1, any rational zeroes must be one of the set $\left \{ 1,3,19 , 57 \right \}$ divided by 1, with positive or native numbers taken into account. Thus, any rational zeroes must be in the set

$\left \{\pm 1, \pm 3, \pm19 ,\pm 57 \right \}$

By trial and error, it can be determined that 3 is a solution of the equation:

$\lambda ^{3}+ 10\lambda - 57= 0$

$3^{3}+ 10 (3) - 57= 0$

True.

It follows that $\lambda - 3$ is a factor. Divide $\lambda ^{3}+ 10\lambda - 57$ by $\lambda - 3$ ; the quotient can be found to be $\lambda ^{2} +3 \lambda+19$ , which is prime.

The characteristic equation is factorable as

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

We already know that $\lambda_{1}= 3$ is an eigenvalue; the other two eigenvalues are the zeroes of $\lambda ^{2} +3 \lambda+19$ , which can be found by way of the quadratic formula:

$\lambda _{2}= \frac{-3+ \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 + \sqrt{-67}}{2} =- \frac{3} {2}+ \frac{ \sqrt{67}}{2} i$

$\lambda _{3}= \frac{-3- \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 - \sqrt{-67}}{2} =- \frac{3} {2}-\frac{ \sqrt{67}}{2} i$

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. Calculate the absolute values of all three eigenvalues:

$\left |\lambda_{1} \right |= \left |3 \right | = 3$

$\left |\lambda _{2} \right| =\left | - \frac{3} {2}+ \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left ( \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

$\left |\lambda _{3} \right| =\left | - \frac{3} {2}- \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left (- \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

Therefore,

$\left |\lambda _{2} \right| = \left |\lambda _{3} \right| > \left |\lambda _{1} \right|$ .

Since no single eigenvalue has absolute value strictly greater than that of the other two, the matrix has no dominant eigenvalue.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

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.

← Didn't Know|Knew It →

Question

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

Which of the following is an eigenvalue of ?

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

, 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

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

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

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

The characteristic equation of a three-by-three matrix is

$\lambda ^{3}+ 10\lambda - 57= 0$

Which of the following is its dominant eigenvalue?

Tap to reveal answer

Answer

The eigenvalues of a matrix are the zeroes of its characteristic equation.

We can try to extract one or more zeroes using the Rational Zeroes Theorem, which states that any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the leading coefficient. Since the constant is 57, and the leading coefficient of the polynomial is 1, any rational zeroes must be one of the set $\left \{ 1,3,19 , 57 \right \}$ divided by 1, with positive or native numbers taken into account. Thus, any rational zeroes must be in the set

$\left \{\pm 1, \pm 3, \pm19 ,\pm 57 \right \}$

By trial and error, it can be determined that 3 is a solution of the equation:

$\lambda ^{3}+ 10\lambda - 57= 0$

$3^{3}+ 10 (3) - 57= 0$

True.

It follows that $\lambda - 3$ is a factor. Divide $\lambda ^{3}+ 10\lambda - 57$ by $\lambda - 3$ ; the quotient can be found to be $\lambda ^{2} +3 \lambda+19$ , which is prime.

The characteristic equation is factorable as

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

We already know that $\lambda_{1}= 3$ is an eigenvalue; the other two eigenvalues are the zeroes of $\lambda ^{2} +3 \lambda+19$ , which can be found by way of the quadratic formula:

$\lambda _{2}= \frac{-3+ \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 + \sqrt{-67}}{2} =- \frac{3} {2}+ \frac{ \sqrt{67}}{2} i$

$\lambda _{3}= \frac{-3- \sqrt{3^{2}-4(1)(19)}}{2} = \frac{-3 - \sqrt{-67}}{2} =- \frac{3} {2}-\frac{ \sqrt{67}}{2} i$

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. Calculate the absolute values of all three eigenvalues:

$\left |\lambda_{1} \right |= \left |3 \right | = 3$

$\left |\lambda _{2} \right| =\left | - \frac{3} {2}+ \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left ( \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

$\left |\lambda _{3} \right| =\left | - \frac{3} {2}- \frac{ \sqrt{67}}{2} i \right |$

$= \sqrt{\left ( - \frac{3} {2} \right )^{2}+\left (- \frac{ \sqrt{67}}{2} \right ) ^{2}}$

$= \sqrt{\frac{9}{4}+\frac{67}{4}}$

$= \sqrt{\frac{76}{4}}$

$= \sqrt{19}$

$\approx 4.4$

Therefore,

$\left |\lambda _{2} \right| = \left |\lambda _{3} \right| > \left |\lambda _{1} \right|$ .

Since no single eigenvalue has absolute value strictly greater than that of the other two, the matrix has no dominant eigenvalue.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

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.

← Didn't Know|Knew It →

Question

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

Which of the following is an eigenvalue of ?

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

, 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

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

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

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

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.

← Didn't Know|Knew It →

Question

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

Which of the following is an eigenvalue of ?

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

, 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

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

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

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

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

← Didn't Know|Knew It →

Question

$A = \begin{bmatrix} 0 & 0 & i \ 0 &1 & 0 \- i & 0 & 0 \end{bmatrix}$

Give the set of eigenvalues of ; if an eigenvalue has multiplicity greater than 1, repeat the eigenvalue that many times.

Tap to reveal answer

Answer

$\lambda$ is an eigenvalue of if it is a solution of the characteristic polynomial equation

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

Set this equation:

$\lambda I - A = \lambda \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 0 & i \ 0 &1 & 0 \- i & 0 & 0 \end{bmatrix}$

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

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

$\det (\lambda I - A )= \begin{vmatrix} \lambda & 0 & -i \ 0 & \lambda -1 & 0 \ i & 0 &\lambda \end{vmatrix}$

The determinant can be found most easily by adding the products of each entry in one of the rows or columns to its corresponding cofactor. Since the second row has only one nonzero entry, we use this one:

$\det (\lambda I - A )= 0 C_{21}+ (\lambda- 1)C_{22} + 0 C_{23}$

$\det (\lambda I - A )= (\lambda- 1)C_{22}$

$C_{22}=-1^{2+2} M_{22} = M_{22}$ ,

the minor formed by striking out row 2 and column 2:

$\begin{vmatrix} \lambda & \times & -i \ \times & \times& \times \ i & \times &\lambda \end{vmatrix}$

$M_{22} = \begin{vmatrix} \lambda & -i \ i &\lambda \end{vmatrix}$

Take the upper-left to lower-right product, and subtract the upper-right to lower-left product:

$M_{22} = \lambda( \lambda ) - (-i ) (i )= \lambda ^{2}- 1 = (\lambda + 1 ) (\lambda - 1 )$

Thus,

$C _{22} = (\lambda + 1 ) (\lambda - 1 )$ ,

and

$\det (\lambda I - A )= (\lambda- 1) (\lambda + 1 ) (\lambda - 1 )$

This makes and 1 the solutions of the characteristic equation, the latter with multiplicity 2. It follows that has as its eigenvalues the set $\left \{ -1, 1, 1\right \}$ .

← Didn't Know|Knew It →

0

Knew It