Help to determine a basis for eigenspace Please find a basis for the eigenspace corresponding to eigenvalue=3 for the following matrix:
$$
\pmatrix{3&1&0\\0&3&1\\0&0&3}
$$
[3  1  0]
[0  3  1]
[0  0  3]
I have already calculated  [A-(lambda)I] and the result is the following augmented matrix:
 x1  x2  x3
[0   1    0    0]
[0   0    1    0]
[0   0    0    0]

From here I can see that:
* x1 is a free variable
* x2 y x3 are lead variables.
¿How is the basis determined?

Apparently $\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\in\ker (A-3I)$ iff $x_2=x_3=0$. So the eigenspace is spanned by $\begin{pmatrix}1\\0\\0\end{pmatrix}$
