# Proof of equal eigenvalues for normal matrix iff matrix is $cI$

I can certainly show that the eigenvalues of some scalar multiplied by the identity are all equal and that said matrix is normal but how would I go about beginning a proof that a normal matrix with all equal eigenvalues implies it can only be $A = cI$ where c is some constant?

Feeling like I missed something rather obvious here...

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Know any theorems that hold Specially for normal matrices? –  Erick Wong Dec 6 '12 at 21:18

A normal matrix is diagonalizable and hence $$A = X \Lambda X^{-1}$$ Since the eigenvalues are all equal to $c$, we get that $\Lambda = c I$. Hence, $$A = X \Lambda X^{-1} = X \left(c I \right) X^{-1} = c XX^{-1} = cI$$