# Finding the second Eigenvector of a repeated Eigenvalue

The matrix $A= \left(\begin{matrix} 2 & -3 & 6\\0 & 5 & -6\\0 & 1 & 0\end{matrix}\right)$ for which I am trying to find the Eigenvalues and Eigenvectors. I have repeated Eigenvalues of $\lambda_1 = \lambda_2 = 2$ and $\lambda_3 = 3$.

After finding the matrix substituting for $\lambda_1$ and $\lambda_2$, I get the matrix $\left(\begin{matrix} 0 & 1 & -2\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$ as the row reduced echelon form.

So far, I have managed to obtain one of the eigenvectors, which is $\left(\begin{matrix} 0\\2\\1\end{matrix}\right)$. This was done by setting the third column of the RREF as the free parameter.

The solutions show that there is a second eigenvector for this eigenvalue, which is $\left(\begin{matrix} 1\\0\\0\end{matrix}\right)$.

How can I obtain this second eigenvector?

• Your pivot is in the second column, so the dependent variable is $x_2$ and $x_1$ and $x_3$ are free. The second eigenvector comes from choosing $x_1=1,x_3=0$. – Ian Oct 17 '15 at 3:22

Let's solve $(A - 2I) v = 0$ for vectors $v$.

The equation resolves to:

$$\begin{pmatrix} 0 & -3 & 6 \\ 0 & 3 & -6 \\ 0 & 1 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0$$

which imposes the condition $v_2 = 2 v_3$, with no condition on $v_1$. The general solution is, for scalars $a$ and $b$:

$$\begin{pmatrix} a \\ 2b \\ b \end{pmatrix} = a \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+ b \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$$

Thus, $(1,0,0)^t$ and $(0,2,1)^t$ are the required eigenkets.

Look at your matrix. It gives you the lone equation $y - 2z = 0$ So any scalar multiple of $[0\quad 2\quad 1]^t$ is an eigenvector for $\lambda = 2$. But here, $x$ is allowed to do whatever it wants, so any multiple of $[1\quad 0 \quad 0]^t$ will be an eigenvector for $\lambda = 2$. These two vectors form a basis for this eigenspace.