# eigenvectors of a triangular matrix

I'm aware that the diagonal of a triangular matrix is its set of eigenvalues, but I'm not sure how best to compute the eigenvectors.

For example, if we consider $A=\begin{pmatrix} 1&2&3\\0&4&5\\0&0&6\\ \end{pmatrix}$ then clearly the eigenvalues are $1,4,6$. Using wolfram I see that ''the'' eigenvectors are $(16,25,10), (2,3,0),(1,0,0)$.
Here is my attempt, I guess I would plug in $6$ for the equation $(A-\lambda I)x=0$ then just hack away:

$$(1-5)x_1 + 2x_2 +3x_3 = 0$$

But I guess in doing I would set $x_1=1$ and then I would solve for $x_2$ in terms of $x_3$ from the first equation: $$-5+2x_2+3x_3 = 0$$ So that $x_2=\frac{1}{2}(-3x_2 + 5)$ so that from the second equation we $$-2(\frac{1}{2}(-3x_2 + 5))=0$$ so that $x_3=\frac{-5}{4}$ and hence $x_2=-\frac{5}{2}$.... we scale it all up by a factor of $4$ now.
Therefore, my first eigenpair is $(6,\begin{pmatrix} 4\\-5\\-10 \end{pmatrix})$

For $\lambda = 4$ we start with the first equation and get that $$-3x_1+2x_2+3x_3$$ To make this eigenvector linearly independent we let $x_3=0$ right off the start and see that $x_1 = \frac{2}{3}x_2$ so that if we let $x_2=3$ then $x_3=2$. And for the final eigenvector, we want it linearly independent from the others so we can easily see that $(1,0,0)$ satisfies all three equations and is linearly independent. I'm not sure this is correct and if it is, that it is the most efficient way (by hand). Any suggestions?

• I realize I just made some mistakes at the top that were probably leading me to being confused. Commented Oct 16, 2016 at 20:56

Okay let's get the eigenvector $x$ corresponding to $\lambda$

$$(A - \lambda I)x = 0$$ Denoting $x_i$ the $i^{th}$ entry of $x$, we get from the first row: $$x_{1} = \frac{1}{5}(2x_2 + 3 x_3)$$ and the second row: $$2x_2 = 5x_3$$ Note that this is 3 unknowns in two equations, so you have one free parameter, say $x_3$ (without any loss of generality). Set $x_3 = 10$, you get $x_2 = 25$ and $x_1 = 16$

Dont bother with $x_1,x_2,...$. Just calculate with matrices if they are small and simple (like if they have integer entries).

$A=\begin{pmatrix} 1&2&3\\0&4&5\\0&0&6\\ \end{pmatrix}$

$A-6I=\begin{pmatrix} -5&2&3\\0&-2&5\\0&0&0\\ \end{pmatrix}$

Now use the last two columns to get the 2nd row to zero, so set $v=\begin{pmatrix} x\\5\\2 \end{pmatrix}$ (keep in mind to always use integers to make your life easy). Now you want the first row to be zero so you calculate $2*5+3*2=16$ so you have $x=\frac{16}{5}$ so $v=\begin{pmatrix} \frac{16}{5}\\5\\2 \end{pmatrix}$ and rescaling gives you $w=\begin{pmatrix} 16\\25\\10 \end{pmatrix}$ (again rescale to keep your calculations simple, nobody likes to calculate with alot of fractions).

Keep in mind if you have $A=(v_1\ v_2\ v_3)$ where $v_i$ are the columns of the matrix you have $Ay=y_1v_1+y_2v_2+y_3v_3$ where $y=\begin{pmatrix} y_1\\y_2\\y_3 \end{pmatrix}$ to make your calculations simple.

Note that, for any triangular matrix, a vector with all elements zero except the first will be an eigenvector. There will be a second eigenvector with all elements zero except the first two, etc.