I have a matrix
$$A=\begin{bmatrix}2&1&3\\0&1&-2\\0&0&3\end{bmatrix}.$$
And obtained the eigenvalues $\{1, 2, 3\}$.
For $\lambda = 2$, I get this matrix after row reduction: $$\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\0 & 0 & 0\\\end{bmatrix}$$
Here is what I tried:
$$\left\{\begin{array}{c}y = 0\\
z = 0\end{array}\right.$$
but what now? Aren't I supposed to express it with a free variable, $x$ ? But how can I if its all $0$ ?
I successfully obtained the vector for eigenvalue $\lambda= 3$.
Best Answer
For triangular matrices as $ A=\begin{bmatrix} 2 & 1 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 3\end{bmatrix} $
the equation $Av=\lambda v$ can be presented as a set of equations
$\left \{ \begin{array}{c}2x+y+3z = \lambda x \\ y-2z=\lambda y \\ 3z=\lambda z \end{array} \right.$
which is relatively easy to solve if we start from $z$ from the third equation , next $y$ from the second one and $x$ from the first one (always consider possible zero and non-zero value - first calculated non-zero value can be taken arbitrarily as for example equal to $1$).