simple question: the eigenvectors of a matrix.

Question

\begin{align*}A=\left(\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\\end{array}\right);\end{align*}

The eigenvalues are $1$, I know one of the eigenvectors is $(1,0,0,0)$, Is that all?

The mathematica gives, why not {{1,0,0,0},{1,0,0,0},{1,0,0,0},{1,0,0,0}}?

Eigenvectors[A]

\begin{align*}\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\\end{array}\right)\end{align*}

Answer 1

I got mathematica and in the documentation of Eigenvectorit says

For an n[Cross]n matrix, Eigenvectors always returns a list of length n. The list contains each of the independent eigenvectors of the matrix, followed if necessary by an appropriate number of vectors of zeros. »

So mathematica says you have exactly one linear independent eigenvector.

Answer 2

Suppose that $Av = v$; that is, $v$ is an eigenvector for the eigenvalue $1$. Writing $v = (v_1, v_2, v_3, v_4)^T$, we find that

$$v_1 + 2v_2 + 3v_3 + 4v_4 = v_1 \implies 2v_2 + 3v_3 + 4v_4 = 0$$

$$v_2 + 2v_3 + 3v_4 = v_2 \implies 2v_3 + 3v_4 = 0$$

$$v_3 + 2v_4 = v_3 \implies 2v_4 = 0$$

Now back substituting leads to $v_2 = v_3 = v_4 = 0$, and so the eigenspace is spanned by $(1, 0, 0, 0)^T$.