Let $C$ be a $4 \times 4$ matrix with all eigenvalues $λ = 2, −1$ and eigenspaces; calculate $C^4u$

Question

Pardon for a poor title but I have this question which I am very confused:

Let $C$ be a $4 \times 4$ matrix with all eigenvalues $λ = 2, −1$ and following eigenspaces: $E_2=\begin{pmatrix} 1\\1 \\ 1\\1 \end{pmatrix}$ and $E_{-1}=\left(\begin{pmatrix} 1\\2\\ 1\\1 \end{pmatrix},\begin{pmatrix} 1\\1 \\ 1\\2 \end{pmatrix} \right)$. Calculate the $C^4u $ for $u=\left[\begin{pmatrix} 6\\8 \\ 6\\9 \end{pmatrix}\right]$ if possible, and explain if not possible.

Now my understanding was to first get the diagonal matrix of $C$ so that we can easily find the powers of C. We already have eigenvectors from the basis of eigenspaces so we can get the coordinate matrix $Q$ and we can write $D=Q^{-1}CQ$, where D is diagonal matrix consisting of eigen values.

Problems: Firstly I cant seem to prove that C is diagonalisable or not, then only we can write the above equation. For that I need to check whether the characterestic polynomial splits or not (which might as we got the eigenvalues), and then check muliplicity of the eigenvalues and verify if $\text{multiplicity}(\lambda)=\text{dim}(E_\lambda)$, and we dont have muluplicity given, so I am kinda stuck. I also need multiplicity to write the diagonal matrix.

So I need help here, as I am stuck and cant seem to go other way. Thanks in advance.

Answer 1

You cannot recover the entire matrix $C$ from what you have. You cannot hope to diagonalize $C$ either. Indeed $C$ is $4 \times 4$ and [informally put] you were given only $3$ eigenvectors, so the final eigenvalue may be anything. You can still find $C^4u$ however, iff $u$ is a linear combination of the $3$ eigenvectors you were given [as you also know the eigenvalues for each of those $3$ eigenvectors], and in fact it is..

You want to express instead $u$ as a linear combination of the $3$ eigenvectors you wrote, if it is possible, and you can check that it is indeed possible, in fact:

$$ u = \begin{pmatrix} 6 \\8 \\6 \\ 9 \end{pmatrix} = 2 \begin{pmatrix} 1 \\2 \\1 \\ 1 \end{pmatrix} + 3 \begin{pmatrix} 1 \\1 \\1 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\1 \\1 \\ 1 \end{pmatrix}.$$

Then from the above equation, as $C^4$ is a linear transformation, the following equation also holds: $$C^4 u = C^4\begin{pmatrix} 6 \\8 \\6 \\ 9 \end{pmatrix} = 2 C^4\begin{pmatrix} 1 \\2 \\1 \\ 1 \end{pmatrix} + 3 C^4\begin{pmatrix} 1 \\1 \\1 \\ 2 \end{pmatrix} + C^4\begin{pmatrix} 1 \\1 \\1 \\ 1 \end{pmatrix}.$$

However, if $v$ is an eigenvector of a matrix $M$ with eigenvalue $\lambda$, then for any positive integer $r$, $v$ is also an eigenvector of $M^r$ with eigenvalue $\lambda^r$. Thus, each of the vectors on the RHS are eigenvectors of $C$, so they are eigenvectors of $C^4$, where each of the eigenvalues is taken to the $4$-th power, so use this: $$2 C^4\begin{pmatrix} 1 \\2 \\1 \\ 1 \end{pmatrix} = 2(-1)^4\begin{pmatrix} 1 \\2 \\1 \\ 1 \end{pmatrix};$$

$$ 3 C^4\begin{pmatrix} 1 \\1 \\1 \\ 2 \end{pmatrix} = 3(-1)^4\begin{pmatrix} 1 \\1 \\1 \\ 2 \end{pmatrix};$$ $$C^4\begin{pmatrix} 1 \\1 \\1 \\ 1 \end{pmatrix} = 2^4\begin{pmatrix} 1 \\1 \\1 \\ 1 \end{pmatrix}.$$

Can you finish from here.