Operations and Properties - Linear Algebra

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Question

is an involutory matrix.

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

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Answer

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 :

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

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

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

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Question

Find the Eigen Values for Matrix .

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

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Answer

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 $\lambda$ .

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

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

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

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Answer

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:

this can be factored:

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

The eigenvalues are 5 and 1.

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Question

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

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Answer

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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Question

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

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Answer

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:

this can be factored (or solved in another way)

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

The eigenvalues are -5 and 3.

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Question

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

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Answer

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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Question

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

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Answer

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$

this can be solved by factoring:

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

The solutions are -2 and -7

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Question

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 $\mathbf{v} = \begin{bmatrix} 2\ 3\end{bmatrix}$

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Answer

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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

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

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Answer

$\begin{align}&$\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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Question

$\begin{align}&$\text{Find any and all eigenvalues for the matrix }$\&A=\begin{bmatrix}18&-6\-12&-10\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$A=\begin{bmatrix}18&-6\-12&-10\end{bmatrix}\&$\text{We can represent that equation as follows }$\begin{bmatrix}18 - \lambda&-6\-12&- \lambda - 10\end{bmatrix}\&$\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:}$\&\lambda ^{2} - 8\lambda - $252\&\lambda_{1}$$=20.37;\lambda_{2}$=-12.37\end{align}$

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Question

$\begin{align}&$\text{Calculate the eigenvalues for the matrix }$\&\begin{bmatrix}11&10\-5&3\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$\begin{bmatrix}11&10\-5&3\end{bmatrix}\&$\text{We can write a matrix of the form }$\begin{bmatrix}11 - \lambda&10\-5&3 - \lambda\end{bmatrix}\&$\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} - 14\lambda + $83\&\lambda_{1}$$=7.00+5.83i;\lambda_{2}$=7.00-5.83i\end{align}$

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Question

$\begin{align}&$\text{Find any and all eigenvalues for the matrix }$\&A=\begin{bmatrix}-20&-7\-14&12\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$A=\begin{bmatrix}-20&-7\-14&12\end{bmatrix}\&$\text{We can represent that equation as follows }$\begin{bmatrix}- \lambda - 20&-7\-14&12 - \lambda\end{bmatrix}\&$\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:}$\&8\lambda + \lambda ^{2} - $338\&\lambda_{1}$$=-22.81;\lambda_{2}$=14.81\end{align}$

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Question

$\begin{align}&$\text{Find any and all eigenvalues for the matrix }$\&A=\begin{bmatrix}-2&-17&-11\17&-14&13\2&20&-17\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$A=\begin{bmatrix}-2&-17&-11\17&-14&13\2&20&-17\end{bmatrix}\&$\text{We can represent that equation as follows }$\begin{bmatrix}- \lambda - 2&-17&-11\17&- \lambda - 14&13\2&20&- \lambda - 17\end{bmatrix}\&$\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 (351\lambda + 33\lambda ^{2} + \lambda ^{3} + $9359)\&\lambda_{1}$$=-0.83+17.26i;\lambda_{2}$$=-0.83-17.26i;\lambda_{3}$=-31.33\end{align}$

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Question

$\begin{align}&$\text{Calculate the eigenvalues for the matrix }$\&\begin{bmatrix}-4&-10\12&-3\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$\begin{bmatrix}-4&-10\12&-3\end{bmatrix}\&$\text{We can write a matrix of the form }$\begin{bmatrix}- \lambda - 4&-10\12&- \lambda - 3\end{bmatrix}\&$\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:}$\&7\lambda + \lambda ^{2} + $132\&\lambda_{1}$$=-3.50+10.94i;\lambda_{2}$=-3.50-10.94i\end{align}$

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Question

$\begin{align}&$\text{Calculate the eigenvalues for the matrix }$\&\begin{bmatrix}-15&14&5\-6&1&-4\-17&-11&-15\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$\begin{bmatrix}-15&14&5\-6&1&-4\-17&-11&-15\end{bmatrix}\&$\text{We can write a matrix of the form }$\begin{bmatrix}- \lambda - 15&14&5\-6&1 - \lambda&-4\-17&-11&- \lambda - 15\end{bmatrix}\&$\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{Knowing this, we can expand and solve:}$\&(-1)\cdot (320\lambda + 29\lambda ^{2} + \lambda ^{3} - $992)\&\lambda_{1}$$=-15.74+12.27i;\lambda_{2}$$=-15.74-12.27i;\lambda_{3}$=2.49\end{align}$

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Question

$\begin{align}&$\text{Find any and all eigenvalues for the matrix }$\&A=\begin{bmatrix}-7&16\-5&-16\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$A=\begin{bmatrix}-7&16\-5&-16\end{bmatrix}\&$\text{We can represent that equation as follows }$\begin{bmatrix}- \lambda - 7&16\-5&- \lambda - 16\end{bmatrix}\&$\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:}$\&23\lambda + \lambda ^{2} + $192\&\lambda_{1}$$=-11.50+7.73i;\lambda_{2}$=-11.50-7.73i\end{align}$

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Question

$\begin{align}&$\text{Find the eigenvalues for the matrix }$\&A=\begin{bmatrix}3&-18&-11\-6&13&-20\-19&-14&6\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\text{Eigenvalues of a matrix A, usually denoted as }$\lambda$\text{, are values which satisfy}$\&det(A-\lambda I)$\text{, where I is the identity matrix. For our matrix }$A=\begin{bmatrix}3&-18&-11\-6&13&-20\-19&-14&6\end{bmatrix}\&$\text{We can define a matrix of the form }$\begin{bmatrix}3 - \lambda&-18&-11\-6&13 - \lambda&-20\-19&-14&6 - \lambda\end{bmatrix}\&$\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{Using this, we can solve for our eigenvalues:}$\&(-1)\cdot (\lambda ^{3} - 22\lambda ^{2} - 462\lambda + $11735)\&\lambda_{1}$$=-22.30;\lambda_{2}$$=22.15+5.95i;\lambda_{3}$=22.15-5.95i\end{align}$

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Question

$\begin{align}&$\text{Calculate the eigenvalues for the matrix }$\&\begin{bmatrix}-13&-5&5\11&-17&18\11&-1&-3\end{bmatrix}\end{align}$

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Answer

$\begin{align}&$\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 }$\begin{bmatrix}-13&-5&5\11&-17&18\11&-1&-3\end{bmatrix}\&$\text{We can write a matrix of the form }$\begin{bmatrix}- \lambda - 13&-5&5\11&- \lambda - 17&18\11&-1&- \lambda - 3\end{bmatrix}\&$\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{Knowing this, we can expand and solve:}$\&(-1)\cdot (329\lambda + 33\lambda ^{2} + \lambda ^{3} + $1172)\&\lambda_{1}$$=-7.08+3.47i;\lambda_{2}$$=-7.08-3.47i;\lambda_{3}$=-18.84\end{align}$

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