Given a diagonalizable triangular matrix $A$, its eigenvector matrix $X$ is triangular How would I prove that given a diagonalizable triangular matrix $A$, its eigenvector matrix $X$ is triangular?
I've tried using $AX=X\Lambda$ (where $\Lambda$ is the diagonal eigenvalue matrix) and various properties of matrix multiplication, but I still have no clue where to continue from any of my starting points. Help will be much appreciated! Thanks!
EDIT: $(A-\lambda I)x=0$ has also been tried by me now, but to no avail.
 A: An intuitive understanding of this problem
There is a hacky proof below but I think this intuition has more value.
Let's take a left-triangular matrix. Fist thing to note about an $n$-dimensional left triangular matrix: the result of applying this matrix to a $n$ dimensional point $p \in \mathbb{F}^n$ is equivalent to the cumulative result of applying the upper-left submatrices to $p$. ($\mathbb{F}$ here can be complex or real).
To see this, try this $\mathbb{F}^3$ matrix:
$A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{pmatrix}$
I take a point $p \in \mathbb{F}^3$. I can directly $Ap$ to find the result of this matrix action on $p$. But note the result:
$\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} x \\ x + 2y \\ 3x + 2 + z\end{pmatrix}$.
Instead of applying the matrix in one shot, I could start with the first dimension of point $p$ and apply the upper left $1 \times 1$ matrix. This gives the result of the $1$-st dimension of the full matrix!
Now I can move to two dimensions and apply the upper left $2 \times 2$ matrix to $p_2$. The first dimension of the existing result won't change. The 2nd dimension will also be available now.
Continue doing this and note that when you have $k$ dimensions of your result, adding a dimension to the matrix and to the point, only changes the $k+1$-th dimension.
This is because of the zeros which limit the matrix's action on the "higher dimensions."
Once this point is super-clear, you can see that at each new additional dimension, we can only have an additional eigenvector. The existing eigenvectors will remain the same. Because each additional dimension $k$ gives you a new eigenvector in $k$ dimensions. So we will have eigenvectors in each dimension from $1$ all the way to $n$.
Postscript: A geoemtric understanding
A triangular matrix is actually a shear action. The 2D case is the easiest to understand:
$\begin{pmatrix} 1 & 0 \\ 2 & 1\end{pmatrix}$
You can use this tool or this tool to understand how this matrix acts.

The shearing in $n$ dimension has one invariant plane in $n-1$ dimensions i.e. the points on this plane don't rotate. They may translate but only within this plane.
You can see how each additional dimension to the input matrix is actually an additional shearing action in the new dimension. So each additional dimension results in a $1+$ dimensional invariant plane.
Inductive proof
I have an inductive proof that could work. It's not very neat and it can definitely be improved on.
$A$ is triangular, so we know that the diagonal values of $A$ are the eigenvalues of $A$.
Now, let's say that $A$ is left triangular/lower triangular. We use the characteristic equation $A - \lambda I$ now. Since we know $\lambda = a_{ii}$ for $i \in [1,n]$ ($a_{ii}$ are the diagonal elements):
$\begin{pmatrix} a_{11} & 0 & 0 & \ldots \\ a_{21} & a_{22} & 0 & \ldots
\\
\vdots \end{pmatrix} \mathbf{v} = a_{ii} \mathbf{v}$
$\begin{pmatrix} v_1 a_{11} \\ v_1a_{21} + v_2a_{22} \\ v_1a_{31} + v_2a_{32} + v_3a_{33} \\ \vdots \end{pmatrix} = \begin{pmatrix} a_{ii} v_1 \\ a_{ii}v_2 \\ a_{ii}v_3   \end{pmatrix}$
We only need this final system of equations to infer a few things:

*

*$v_1 a_{11} = a_{ii} v_1$ and $v_1 \neq 0$ only if $a_{ii} = a_{11}$. So apart from the eigenvalue of $a_{11}$ this component of the eigenvector will be zero

*A similar line of reasoning for the second row: $v_1a_{21} + v_2a_{22} = a_{ii}v_2$; say $a_{ii} \neq a_{11}$, then we already know $v_1 = 0$. But now if we want $v_2 \neq 0$, we also need $a_{ii} = a_{22}$.

*

*For any $a_{ii}$ where $i > 2$, $v_2$ = 0 (note that for $a_{11}$, $v_2 = \frac{v_1a_{21}}{a_{11} - a_{22}}$)



*extend this line of reasoning for the rest of the $v_i$ and we will find eigenvectors with

*

*$v_1 = 0$

*$v_1 = v_2 = 0$

*$v_1 = v_2 = \ldots = v_{n-1} = 0$
We can do a similar analysis for upper triangular/right triangular $A$, leading to the same conclusion.
TODO/Future work
I think we can use properties of diagonalization to show that for a triangular $A$, there is a triangular $P$, such that $P^{-1}AP = D$. Triangularity is retained through multiplication and inversion. If $A, P$ are triangular, then $P^{-1}AP$ is triangular (both multiplication and inversion retains triangularity), but we have to prove that $P^{-1}AP$ is both upper and lower triangular i.e. diagonal.
