Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
Problem 485
Let
\[A=\begin{bmatrix}
1 & -14 & 4 \\
-1 &6 &-2 \\
-2 & 24 & -7
\end{bmatrix} \quad \text{ and }\quad \mathbf{v}=\begin{bmatrix}
4 \\
-1 \\
-7
\end{bmatrix}.\]
Find $A^{10}\mathbf{v}$.
You may use the following information without proving it.
The eigenvalues of $A$ are $-1, 0, 1$. The eigenspaces are given by
\[E_{-1}=\Span\left\{\, \begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix} \,\right\}, \quad E_{0}=\Span\left\{\, \begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} \,\right\}, \quad E_{1}=\Span\left\{\, \begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix} \,\right\}.\]
(The Ohio State University, Linear Algebra Final Exam Problem)
Sponsored Links
Contents
Solution.
Since $A$ has three distinct eigenvalues, the eigenvectors
\[\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}, \begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} , \begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix} \]
form a basis of $\R^3$.
Thus, we can express the vector $\mathbf{v}$ as a linear combination
\[\mathbf{v}=x\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}+y\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix}+z\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix}\]
for some scalars $x,y,z$.
We determine the values of $x, y, z$ by Gauss-Jordan elimination.
The augmented matrix of the system is
\begin{align*}
\left[\begin{array}{rrr|r}
3 & -2 & -4 & 4 \\
-1 &1 & 2 & -1 \\
-5 & 4 & 7 & -7
\end{array} \right].
\end{align*}
Applying the elementary row operations, we obtain
\begin{align*}
\left[\begin{array}{rrr|r}
3 & -2 & -4 & 4 \\
-1 &1 & 2 & -1 \\
-5 & 4 & 7 & -7
\end{array} \right]
\xrightarrow[\text{Then } -R_1]{R_1\leftrightarrow R_2}
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
3 & -2 & -4 & 4 \\
-5 & 4 & 7 & -7
\end{array} \right]\\[6pt]
\xrightarrow[R_3+5R_1]{R_2-3R_1}
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 1 & 2 & 1 \\
0 & -1 & -3 & -2
\end{array} \right]
\xrightarrow[\text{Then } -R_3]{\substack{R_1+R_2\\R_3+R_2}}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & 1 & 1
\end{array} \right]\\[6pt]
\xrightarrow{R_2-2R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{array} \right].
\end{align*}
So we have
\[x=2, y=-1, z=1,\]
and the linear combination is
\[\mathbf{v}=2\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}-\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix}+\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix}.\]
Recall that if $\lambda$ is an eigenvalue and $\mathbf{u}$ is a corresponding eigenvector of $A$, then we have
\[A^{10}\mathbf{u}=\lambda^{10}\mathbf{u}.\]
Using this property, we compute
\begin{align*}
A^{10}\mathbf{v}&=A^{10}\left(\,2\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}-\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix}+\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix} \,\right)\\[6pt]
&=2A^{10}\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}-A^{10}\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix}+A^{10}\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix}\\[6pt]
&=2(-1)^{10}\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}-0^{10}\begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix}+(1)^{10}\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix}\\[6pt]
&=2\begin{bmatrix}
3 \\
-1 \\
-5
\end{bmatrix}+\begin{bmatrix}
-4 \\
2 \\
7
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
2 \\
0 \\
-3
\end{bmatrix}.
\end{align*}
In summary, we obtain
\[A^{10}\mathbf{v}=\begin{bmatrix}
2 \\
0 \\
-3
\end{bmatrix}.\]
Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)
This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).
The other problems can be found from the links below.
- Find All the Eigenvalues of 4 by 4 Matrix
- Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
- Diagonalize a 2 by 2 Matrix if Diagonalizable
- Find an Orthonormal Basis of the Range of a Linear Transformation
- The Product of Two Nonsingular Matrices is Nonsingular
- Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
- Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
- Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
- Idempotent Matrix and its Eigenvalues
- Diagonalize the 3 by 3 Matrix Whose Entries are All One
- Given the Characteristic Polynomial, Find the Rank of the Matrix
- Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ (This page)
- Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$
Add to solve later
Sponsored Links
4 Responses
[…] Compute $A^{10}mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ […]
[…] Compute $A^{10}mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ […]
[…] Compute $A^{10}mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ […]
[…] Compute $A^{10}mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$ […]