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?
Best Answer
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 $x_1 = 16$