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for any integer. In the case of so . Thus the eigenvalues of are just where is an eigenvalue of .
Solved Problems for Chapter 8
If is an invertible matrix, compare the eigenvalues of and . More generally, for an
arbitrary integer, compare the eigenvalues of and .
If is an matrix and is a nonzero constant, compare the eigenvalues of and
.
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Say Then and so the eigenvalues of are just where is an eigenvalue of .
Let be invertible matrices which commute. That is, . Suppose is an eigenvector of .
Show that then must also be an eigenvector for .
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. Here it is assumed that .
Suppose is an matrix and it satisfies for some a positive integer larger than 1.
Show that if is an eigenvalue of then equals either 0 or .
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Let be the eigenvector. Then and so Hence if then and so
Show that if and , then whenever are scalars, \begin{equation*} A\left ( kX+pY\right ) =\lambda \left ( kX+pY\right ) \end{equation*}
Does this imply that is an eigenvector? Explain.
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The formula follows from properties of matrix multiplications. However, this vector might not be an eigenvector because it might equal and eigenvectors cannot equal .
Find the eigenvalues and eigenvectors of the matrix \begin{equation*} \left [ \begin{array}{rrr} 6 & 76 & 16 \\ -2 & -21 & -4 \\ 2 & 64 & 17 \end{array} \right ] \end{equation*}
One eigenvalue is
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The characteristic polynomial of this matrix is . Knowing one of the eigenvalues gives us one factor, . Use long division of polynomials to finish factoring . Now we have the following eigenvalues: , , and .
The corresponding eigenvectors are: , , and .
Find the eigenvalues and eigenvectors of the matrix \begin{equation*} \left [ \begin{array}{rrr} 3 & 5 & 2 \\ -8 & -11 & -4 \\ 10 & 11 & 3 \end{array} \right ] \end{equation*}
One eigenvalue is .
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Characteristic polynomial: .
Is it possible for a nonzero matrix to have only as an eigenvalue?
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Yes. works.
Let be the linear transformation which reflects vectors about the axis. Find a
matrix for and then find its eigenvalues and eigenvectors.
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The matrix of is .
The eigenvalues and eigenvectors are:
Let be the linear transformation which reflects all vectors in through the plane.
Find a matrix for and then obtain its eigenvalues and eigenvectors.
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The matrix of is The eigenvalues are , . Bases for the corresponding eigenspaces are:
Find the eigenvalues and eigenvectors of the matrix \begin{equation*} \left [ \begin{array}{rrr} 5 & -18 & -32 \\ 0 & 5 & 4 \\ 2 & -5 & -11 \end{array} \right ] \end{equation*}
One eigenvalue is Diagonalize if possible.
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The eigenvalues are . The eigenvectors corresponding to the eigenvalues are: Therefore this matrix is not diagonalizable.
Find the eigenvalues and eigenvectors of the matrix \begin{equation*} \left [ \begin{array}{rrr} -13 & -28 & 28 \\ 4 & 9 & -8 \\ -4 & -8 & 9 \end{array} \right ] \end{equation*}
One eigenvalue is Diagonalize if possible.
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The eigenvectors and eigenvalues are: The matrix needed to diagonalize the above matrix is and the diagonal matrix is
Find the eigenvalues and eigenvectors of the matrix \begin{equation*} \left [ \begin{array}{rrr} 89 & 38 & 268 \\ 14 & 2 & 40 \\ -30 & -12 & -90 \end{array} \right ] \end{equation*}
One eigenvalue is Diagonalize if possible.
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The eigenvectors and eigenvalues are: The matrix needed to diagonalize the above matrix is and the diagonal matrix is
Bibliography
These problems come from Chapter 7 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 359–401.
2024-09-06 23:26:00