Given All Eigenvalues and Eigenspaces, Compute a Matrix Product

Ohio State University exam problems and solutions in mathematics

Problem 189

Let $C$ be a $4 \times 4$ matrix with all eigenvalues $\lambda=2, -1$ and eigensapces
\[E_2=\Span\left \{\quad \begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix} \quad\right \} \text{ and } E_{-1}=\Span\left \{ \quad\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix},\quad \begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix} \quad\right\}.\]

Calculate $C^4 \mathbf{u}$ for $\mathbf{u}=\begin{bmatrix}
6 \\
8 \\
6 \\
9
\end{bmatrix}$ if possible. Explain why if it is not possible!

(The Ohio State University Linear Algebra Exam Problem)
 
LoadingAdd to solve later

Solution.

Let
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix} \mathbf{v}_3=\begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix}.\] The vector $\mathbf{v}_1$ is an eigenvector corresponding to eigenvalue $\lambda=2$ and the vectors $\mathbf{v}_2, \mathbf{v}_2$ are eigenvectors corresponding to $\lambda=-1$.


If the vector $\mathbf{u}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, then we can compute $C\mathbf{u}$ and hence $C^4\mathbf{u}$.
So let us first determine whether
\[x_1 \mathbf{v}_1+x_2\mathbf{v}_2+x_3 \mathbf{v}_3=\mathbf{u}\] has a solution or not.

We use Gauss-Jordan elimination method to find the solution.
The augmented matrix for this system is
\[\left[\begin{array}{rrr|rrr}
1 & 1 & 1 & 6 \\
1 &2 & 1 & 8 \\
1 & 1 & 1 & 6 \\
1 & 1 & 2 & 9
\end{array}\right].\] By elementary row operations, this matrix reduces to
\[\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 \\
0 &1 & 0 & 2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 0
\end{array}\right].\] Thus the solution is
\[x_1=1, x_2=2, x_3=3\] and we have the linear combination
\[ \mathbf{v}_1+2\mathbf{v}_2+3 \mathbf{v}_3=\mathbf{u}.\]


Now we have
\begin{align*}
C^4\mathbf{u}&= C^4\mathbf{v}_1+2C^4\mathbf{v}_2+3C^4 \mathbf{v}_3\\
&=2^4\mathbf{v}_1+2(-1)^4\mathbf{v}_2+3(-1)^4 \mathbf{v}_3\\
&=16\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}+2\begin{bmatrix}
1 \\
2 \\
1 \\
1
\end{bmatrix}+3\begin{bmatrix}
1 \\
1 \\
1 \\
2
\end{bmatrix}\\
&=\begin{bmatrix}
21 \\
23 \\
21 \\
24
\end{bmatrix}.
\end{align*}
Therefore we obtain
\[C^4=\begin{bmatrix}
21 \\
23 \\
21 \\
24
\end{bmatrix}.\]
LoadingAdd to solve later

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Linear algebra problems and solutions
Two Eigenvectors Corresponding to Distinct Eigenvalues are Linearly Independent

Let $A$ be an $n\times n$ matrix. Suppose that $\lambda_1, \lambda_2$ are distinct eigenvalues of the matrix $A$ and let...

Close