How would i find the eigenvector of the following eigenvalue? Im having a really hard time with finding eigenvectors in general. Is it the same thing as computing the null space of the matrix with the eigenvalue plugged in?
Like i have this matrix:
A = {{5,0,0},{0,1,3},{0,3,1}}
$$
        \begin{matrix}
        5 & 0 & 0 \\
        0 & 1 & 3 \\
        0 & 3 & 1 \\
        \end{matrix}
$$
Ive calculated the eigenvalues and they are: 5, 4, and -2
So for finding the eigenvector of eigenvalue 5 i plug it back into the matrix and i get:
A = {{0,0,0},{0,-4,3},{0,3,-4}}
$$
        \begin{matrix}
        0 & 0 & 0 \\
        0 & -4 & 3 \\
        0 & 3 & -4 \\
        \end{matrix}
$$
Wolframalpha tells me that the eigenvector for this eigenvalue is {{1},{0},{0}}
$$
        \begin{matrix}
        1 \\
        0 \\
        0 \\
        \end{matrix}
$$
I just can't wrap my head around how one would end up with this answer.
And is there a quick and easy way to find eigenvectors without completely row reducing it?
 A: You wish to compute the eigenvalues and eigenvectors of the matrix
$$
A=
\begin{bmatrix}
5 & 0 & 0 \\
0 & 1 & 3 \\
0 & 3 & 1 
\end{bmatrix}
$$
To do so, first note that the characteristic polynomial of $A$ is
$$
\chi_A(t)
= \det(tI-A)
= \det
\begin{bmatrix}
t-5 & 0 & 0 \\
0 & t-1 & -3 \\
0 & -3 & t-1
\end{bmatrix}
= (t-5)(t-4)(t+2)
$$
This tells us that the eigenvalues of $A$ are indeed $5$, $4$, and $-2$.
Now, computing a basis for the eigenspace of $A$ corresponding to the eigenvalue $5$ amounts to computing a basis for the nullspace of 
$$
5I-A=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 4 & -3 \\
0 & -3 & 4
\end{bmatrix}
$$
To do so, note that
$$
\DeclareMathOperator{rref}{rref}\rref(5I-A)
= \rref
\begin{bmatrix}
0 & 0 & 0 \\
0 & 4 & -3 \\
0 & -3 & 4
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}
$$
This tells us that that $(4I-A)\vec x=\vec 0$ if and only if
$$
\vec x=\langle x_1,x_2,x_3\rangle =\langle x_1,0,0\rangle
=x_1\langle 1, 0, 0\rangle
$$
Hence $\{\langle 1,0,0\rangle\}$ is a basis for the nullspace of $5I-A$.
Can you repeat the process for the other two eigenvalues? In other words, can you plug the other two eigenvalues into the equation $tI-A$, compute the reduced row-echelon form of $tI-A$, and find a basis for the nullspace of $tI-A$?
A: Here I refer to the matrix after plugging in the eigenvalue of $5$.
The lower right $2\times 2$ matrix is not singular and so you can't take nonzero linear combinations of the second and third columns to get the zero vector. Since, throwing in a multiple of the first column is not going to change anything you have only to work with the first column and it turns out that if you take any multiple of the $\lbrace0,0,0\rbrace$ you will be fine. The entries of the eigenvector are saying take a multiple of the first and nothing of the second and third columns $\lbrace 1,0,0\rbrace$.
