Help to determine a basis for eigenspace

Question

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?

Answer 1

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 $\begin{pmatrix}1\\0\\0\end{pmatrix}$

Answer 2

So, you should be able to see that the solution to $(A - 3I)x = 0$ is given by $$ x = \pmatrix{t\\0\\0} \quad t \in \Bbb R $$ Since this solution space is, by definition, the eigenspace associated with $3$, the set $\{(1,0,0)^T\}$ forms a basis for this eigenspace.