Originally Posted by

**linearalgebradunce** do you mean the polynomial I found, - lambda cubed + 5lambda squared - 6lambda. (sorry i couldnt find the symbols)

That's it. Factoring the polynomial, we get

$\displaystyle

-\lambda^3+5\lambda^2-6\lambda=-\lambda(\lambda^2-5\lambda+6)=-\lambda(\lambda-2)(\lambda-3)$

so the eigenvalues are 0, 2 and 3.

however i have also read that for a triangular square matrix the eigenvalues are the diagonal values?

I didn't know that but that's right.

i then cannot find the eigenvectors?! ive been attempting this question for a number of days now, please help.

Let's find the eigenvector $\displaystyle v=\begin{bmatrix}x\\y\\z\end{bmatrix}$ whose corresponding eigenvalue is 2.

We know that $\displaystyle Bv=2v

\Longleftrightarrow \begin{bmatrix} 0& 1& 1\\

0& 2& 2\\

0& 0& 3

\end{bmatrix}

\cdot

\begin{bmatrix}

x\\

y\\

z

\end{bmatrix}

=2\begin{bmatrix}x \\ y \\ z \end{bmatrix}

\Longleftrightarrow \begin{bmatrix} y +z\\ 2y+2z\\ 3z \end{bmatrix}

=\begin{bmatrix}2x \\ 2y \\ 2z \end{bmatrix} $ thus we have to solve $\displaystyle \begin{cases}

y+z=2x&(1)\\

2y+2z=2y&(2)\\

3z=2z&(3)\\

\end{cases}

$

From (3) we get that $\displaystyle z=0$, (2) is useless (0=0) and (1) tells us that $\displaystyle y=2x$ so $\displaystyle v=\begin{bmatrix}x\\2x\\0 \end{bmatrix}$ for $\displaystyle x\neq 0$. We can, for example, choose $\displaystyle x=1$ to get $\displaystyle v=\begin{bmatrix}1\\2\\0 \end{bmatrix}$.