\begin{bmatrix}-1&4&-5&-3&4\\-7&8&-5&7&6\\-1&3&-3&-2&3\\3&-2&1&-2&-2\\2&1&-4&-14&3\end{bmatrix}
a) Consider $M$ as a matrix with real coefficients. Compute the eigenspace of $M$ for the eigenvalue $2$. It is a subspace of $\mathbb{R}^5$?
- For this question I should use the formula lamda times identity matrix $- M$ and do the arithmetics, right? Is it required to find the null space?
b)Let $M'$ be the matrix in $\text{Mat}_{5\times 5}(F_2)$ obtained by taking the coefficients of $M$ modulo $2$. Compute the eigenspace of $M'$ for the eigenvalue $0 \in F_2$ and for the eigenvalue $1\in F_2$.
- For this question, I do not know how to obtain the $M'$ matrix of $M$ modulo $2$.
Thank you for your help.
Guide:
Yes, compute the nullspace of $2I-M$, that is solve for $v$ in $(2I-M)v=0$.
Note that all entries of $M$ are integers. If you see an odd number, replace it by $1$. If you see an even number, replace it by $0$.
For eigenvector for the eigenvalue $1 \in F_2$, we have to find the nullspace of
$$\begin{bmatrix} 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$
Let me permute some rows,
$$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix}$$
Notice that the $4$-th row is the sum of hte first two rows and we can get rid of it.
$$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix}$$
Add the third row to the first gives me
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix}$$
Adding first two rows to the fifth and then swap the $4$-th and $5$-th row gives me
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
Hence the eigenvector is $\begin{bmatrix}0 & 0 & 1 & 1 & 0 \end{bmatrix}^T.$