Show that a given Eigenvalue belongs to Matrix without characteristic equation How do I show that the given a eigenvalue $λ=1$ is an eigenvalue to matrix
$$ T=\begin{bmatrix}-11 &9& 6\\ -8& 6& 2\\ -6& 6& 7\end{bmatrix}$$
WITHOUT using the characteristic equation... I have NO clue, please help with this :)
 A: Consider the matrix 
$$ T - \lambda = T- 1 = \begin{pmatrix} -12 & 9 & 6 \\ -8 & 5 & 2 \\ -6 & 6 & 6 \end{pmatrix} $$
Now transform it to row echelon form to get 
$$ \begin{pmatrix} -12 & 9 & 6 \\ 0 & -1 & -2 \\ 0 & \frac 32 & 3 \end{pmatrix} \leadsto \begin{pmatrix} -12 & 9 & 6 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{pmatrix}$$
As $T-1$ has rank $2$, $\dim\ker (T- 1) = 1$ and $\lambda = 1$ is an eigenvalue.
A: Write $$T-I=\pmatrix{-12&9&6\\-8 &5 & 2 \\ -6&6&6}$$
Suppose $\bf{v}=\pmatrix{a \\ b \\ c}$ satisfies $(T-I)\bf{v}=0$ (equivalently, $T\bf{v}=\bf{v}$). Then by the third component, $a=b+c$, and by the first component $12a=9b+6c$ (or $4a=3b+2c$). So $4b+4c=3b+2c$, that is $b=-2c$, and $a=-c$. So $\bf{v}=$$-c\pmatrix{1 \\ 2 \\ -1}$. Since $-8+2\cdot 5-2=0$, it follows that $\pmatrix{1 \\ 2 \\ -1}$ (or any non-zero multiple of it) is an eigenvector of $T$ with eigenvalue $1$. In particular, $1$ is an eigenvalue of $T$.
A: Or just use the definition of eigenvalue!
$\lambda= 1$ is an eigenvalue if and only if $\begin{bmatrix}-11 & 9 & 6 \\ -8 & 6 & 2 \\ -6 & 6 & 7\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$.  DO the matrix multiplication on the left and then solve the resulting equations for x, y, and z.  As long as they are not all 0, this is an eigenvalue.
