Is an Upper Triangular Matrix with Equal Diagonal Entries Diagonalizable?

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"Let A be an upper triangular matrix with entires in a field F. Suppose that all the diagonal entries of A are equal. Show that A is diagonalizable if and only if it is diagonal."

I'm reviewing old assignments for a midterm. I remember doing this (backward direction is trivial), I can't remember how, but I remember it was easy. Any hints?
 
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If A is similar to a diagonal matrix D, what can you say about the diagonal entries of D?
 
What do you mean by similar? Do you mean they differ by a few entries? In that case the diagonal entries of D will be the same of that of A.
 
A is similar to D if there exists an invertible matrix X where X*A*X^{-1}=D.

Saying "A is similar to a diagonal matrix D" is the same thing as saying "A is diagonalizable" except I find it gives a less cumbersome way to give this diagonal matrix a name.

So what can you say about the entries of this D?
 
I don't know, but I found a way of doing it:

A is upper triangular, and the diagonal entries are all equal, therefore A has a single eingenvalue e. Suppose A is diagonalizable, then V=null(A-eI), therefore A=eI.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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