linear algebra - Find eigenvalues and eigenvectors of this matrix

Problem: Let $$\begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*}$$ Compute all the eigenvalues and eigenvectors of $A$.

Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me $$\begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*}$$ after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $\lambda_2 = 4$ with multiplicity $1$.

Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: $$\begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*}$$ I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?

Answer

You are right, and to show how this applies more generally, look at the eigenspace corresponding to the other eigenvalue, $\lambda = 0$. This has multiplicity 3, so we expect exactly 3 linearly independent vectors in this space. The main equation looks like $A \vec{x} = 0 \vec{x} = \vec{0}$, in other words,
$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
It's not hard to see that we have 4 identical equations, so the only constraint is $x + y + z + w = 0$, so we define $w = -x-y-z$ and any vector in the eigenspace now looks like

$$\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ -x-y-z \end{pmatrix} = x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$$
and the three vectors which form the basis of the eigenspace are now specified.