Given All Eigenvalues and Eigenspaces, Compute a Matrix Product

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)

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
$C^4=\begin{bmatrix} 21 \\ 23 \\ 21 \\ 24 \end{bmatrix}.$

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