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Linear Algebra : Eigenvalues and Eigenvectors

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Example Questions

Example Question #1 : Eigenvalues And Eigenvectors

Find the Eigen Values for Matrix .

$A=\begin{bmatrix} 5 & -3 \\ 1& -2 \end{bmatrix}$

Possible Answers:

$\lambda=\frac{1- \sqrt{69}}{2}$

$\lambda=\frac{1+ \sqrt{69}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

There are no Eigen Values

$\lambda=\frac{1+ \sqrt{68}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

$\lambda=\frac{1+ \sqrt{69}}{2}$

Correct answer:

$\lambda=\frac{1+ \sqrt{69}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

Explanation:

The first step into solving for eigenvalues, is adding in a $-\lambda$ along the main diagonal.

$A=\begin{bmatrix} 5-\lambda & -3 \\ 1& -4-\lambda \end{bmatrix}$

Now the next step to take the determinant.

$\det(A)=(5-\lambda)(-4-\lambda)-(-3)(1)$

$\det(A)=(5-\lambda)(-4-\lambda)+3$

Now lets FOIL, and solve for $\lambda$ .

$(5-\lambda)(-4-\lambda)+3=-20-5\lambda+4\lambda+\lambda^2+3$

$=\lambda^2-\lambda-17$

Now lets use the quadratic equation to solve for $\lambda$ .

$\lambda=\frac{1\pm\sqrt{(1)^2-4(1)(-17)}}{2(1)}$

$\lambda=\frac{1\pm\sqrt{1+68}}{2}$

$\lambda=\frac{1\pm\sqrt{69}}{2}$

So our eigen values are

$\lambda=\frac{1+ \sqrt{69}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

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Example Question #1 : Eigenvalues And Eigenvectors

Find the eigenvalues for the matrix $\begin{bmatrix} 2 & 1\\ 3& 4\end{bmatrix}$

Possible Answers:

1, 5

2, 4

2, 5

1, 3

Correct answer:

1, 5

Explanation:

The eigenvalues, $\lambda$ , for the matrix are values for which the determinant of $\begin{bmatrix} 2- \lambda & 1\\ 3 & 4 - \lambda \end{bmatrix}$ is equal to zero. First, find the determinant:

$(2-\lambda)(4-\lambda) - (3)(1) = 8 - 4 \lambda - 2 \lambda + \lambda ^2 - 3 = \lambda ^2 - 6 \lambda + 5$

Now set the determinant equal to zero and solve this quadratic:

$\lambda^2 - 6 \lambda + 5 = 0$ this can be factored:

$(\lambda - 5 ) (\lambda - 1 ) = 0$

The eigenvalues are 5 and 1.

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Example Question #3 : Eigenvalues And Eigenvectors

Which is an eigenvector for $A = \begin{bmatrix} 2 &1 \\3 &4 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 1\\3 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} -3\\1 \end{bmatrix}$

Possible Answers:

both $\mathbf{u}$ and $\mathbf{v}$

$\mathbf{u}$

$\mathbf{v}$

neither one is an Eigenvector

Correct answer:

$\mathbf{u}$

Explanation:

To determine if something is an eignevector, multiply times A:

$\begin{bmatrix} 2 &1 \\3 & 4\end{bmatrix} \begin{bmatrix} 1\\ 3\end{bmatrix} = \begin{bmatrix} 15\\ 5\end{bmatrix}$

Since this is equivalent to $5\begin{bmatrix} 1\\ 3\end{bmatrix}$ , $\begin{bmatrix} 1\\ 3\end{bmatrix}$ is an eigenvector (and 5 is an eigenvalue).

$\begin{bmatrix} 2 & 1\\ 3& 4\end{bmatrix} \begin{bmatrix} -3\\1 \end{bmatrix} = \begin{bmatrix} -5\\ -5\end{bmatrix}$

This cannot be re-written as $\mathbf{v}$ times a scalar, so this is not an eigenvector.

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Example Question #4 : Eigenvalues And Eigenvectors

Find the eigenvalues for the matrix $\begin{bmatrix} -3 &6 \\ 2& 1 \end{bmatrix}$

Possible Answers:

5, -1

5, 3

3, -5

3, -1

-3, 2

Correct answer:

3, -5

Explanation:

The eigenvalues are scalar quantities, $\lambda$ , where the determinant of $\begin{bmatrix} -3-\lambda & 6 \\ 2& 1-\lambda \end{bmatrix}$ is equal to zero.

First, find an expression for the determinant:

$(-3-\lambda)(1-\lambda) - 12$

$=-3 - \lambda + 3 \lambda + \lambda ^2 - 12 = \lambda ^2 + 2 \lambda -15$

Now set this equal to zero, and solve:

$\lambda^2 + 2 \lambda -15 = 0$ this can be factored (or solved in another way)

$(\lambda+ 5) (\lambda-3) = 0$

The eigenvalues are -5 and 3.

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Example Question #3 : Eigenvalues And Eigenvectors

Which is an eigenvector for $A = \begin{bmatrix} -3 & 6\\ 2& 1\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -3\\1 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 1\\ 1\end{bmatrix}$ ?

Possible Answers:

Neither is an eigenvector

$\mathbf{u}$

Both $\mathbf{u}$ and $\mathbf{v}$

$\mathbf{v}$

Correct answer:

Both $\mathbf{u}$ and $\mathbf{v}$

Explanation:

To determine if something is an eigenvector, multiply by the matrix A:

$\begin{bmatrix} -3 & 6\\ 2& 1\end{bmatrix} \begin{bmatrix} -3\\ 1\end{bmatrix} = \begin{bmatrix} 15\\ -5 \end{bmatrix}$

This is equivalent to $-5\begin{bmatrix} -3\\ 1\end{bmatrix}$ so this is an eigenvector.

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

This is equivalent to $3\begin{bmatrix} 1\\ 1\end{bmatrix}$ so this is also an eigenvector.

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Example Question #2 : Eigenvalues And Eigenvectors

Determine the eigenvalues for the matrix $\begin{bmatrix} -5 &2 \\ 3 &-4 \end{bmatrix}$

Possible Answers:

7, -2

-7, -2

2,7

-7,2

Correct answer:

-7, -2

Explanation:

The eigenvalues are scalar quantities $\lambda$ where the determinant of $\begin{bmatrix} -5-\lambda & 2\\ 3& -4-\lambda\end{bmatrix}$ is equal to zero. First, write an expression for the determinant:

$(-5-\lambda) (-4-\lambda) -6 = 0$

$20 + 4\lambda+5 \lambda + \lambda^2 - 6 = 0$

$\lambda^2 + 9 \lambda +14 = 0$ this can be solved by factoring:

$(\lambda+2) (\lambda+7) = 0$

The solutions are -2 and -7

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Example Question #1 : Eigenvalues And Eigenvectors

Which is an eigenvector for the matrix $A = \begin{bmatrix} -5 &2 \\ 3& -4\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -2\\ 3\end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 2\\ 3\end{bmatrix}$

Possible Answers:

Both $\mathbf{u}$ and $\mathbf{v}$

$\mathbf{u}$

Neither one is an eigenvector

$\mathbf{v}$

Correct answer:

$\mathbf{v}$

Explanation:

To determine if a vector is an eigenvector, multiply with A:

$\begin{bmatrix} -5 & 2\\ 3& -4 \end{bmatrix} \begin{bmatrix} -2\\ 3\end{bmatrix} = \begin{bmatrix} 16\\-18 \end{bmatrix}$ . This cannot be expressed as an integer times $\mathbf{u}$ , so $\mathbf{u}$ is not an eigenvector

$\begin{bmatrix} -5 & 2\\ 3& -4 \end{bmatrix} \begin{bmatrix} 2\\ 3\end{bmatrix} = \begin{bmatrix} -4\\-6 \end{bmatrix}$ This can be expressed as $-2\begin{bmatrix} 2\\ 3\end{bmatrix}$ , so $\mathbf{v}$ is an eigenvector.

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Example Question #2 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}6&-13\\8&-19\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-3.27;\lambda_{2}=-17.73$

$\lambda_{1}=1.73;\lambda_{2}=-12.73$

$\lambda_{1}=9.73;\lambda_{2}=-4.73$

$\lambda_{1}=0.73;\lambda_{2}=-13.73$

Correct answer:

$\lambda_{1}=0.73;\lambda_{2}=-13.73$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}6&-13\\8&-19\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}6 - \lambda &-13\\8&- \lambda - 19\end{bmatrix} \\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&13\lambda + \lambda ^{2} - 10\\&\lambda_{1}=0.73;\lambda_{2}=-13.73\end{align*}$

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Example Question #9 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}0&-2\\6&9\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-4.37;\lambda_{2}=1.37$

$\lambda_{1}=8.63;\lambda_{2}=14.37$

$\lambda_{1}=4.63;\lambda_{2}=10.37$

$\lambda_{1}=1.63;\lambda_{2}=7.37$

Correct answer:

$\lambda_{1}=1.63;\lambda_{2}=7.37$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}0&-2\\6&9\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}-\lambda&-2\\6&9 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&\lambda ^{2} - 9\lambda + 12\\&\lambda_{1}=1.63;\lambda_{2}=7.37\end{align*}$

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Example Question #3 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-7&3&-11\\10&-10&0\\8&16&19\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=18.29+5.19i;\lambda_{2}=18.29-5.19i;\lambda_{3}=-7.59$

$\lambda_{1}=9.29+5.19i;\lambda_{2}=9.29-5.19i;\lambda_{3}=-16.59$

$\lambda_{1}=11.29+5.19i;\lambda_{2}=11.29-5.19i;\lambda_{3}=-14.59$

$\lambda_{1}=13.29+5.19i;\lambda_{2}=13.29-5.19i;\lambda_{3}=-12.59$

Correct answer:

$\lambda_{1}=9.29+5.19i;\lambda_{2}=9.29-5.19i;\lambda_{3}=-16.59$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&3&-11\\10&-10&0\\8&16&19\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&3&-11\\10&- \lambda - 10&0\\8&16&19 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 195\lambda + 1880)\\&\lambda_{1}=9.29+5.19i;\lambda_{2}=9.29-5.19i;\lambda_{3}=-16.59\end{align*}$

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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 #1 : Eigenvalues And Eigenvectors

Example Question #1 : Eigenvalues And Eigenvectors

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