Proving the special property of diagonal matrix?

[Seydlitz]

Is it possible to prove the fact that any function of diagonal matrix is just a function of its element?

I don't know how I could express the proof. I can prove that a multiplication of diagonal matrix will just be the multiplication of its element using summation notation, or diagonal matrix is automatically symmetrical matrix, but for the more general case I'm at lost. The book that I'm reading also assume directly that it's the case. Indeed it's the case, but is it just a definition that is unprovable?

I'm asking this because I want to prove that $e^{iD}$ of diagonal matrix is unitary using other method than series expansion and the fact that diagonal matrix is just very important for eigenvalues. If I use the fact that the exponent of diagonal matrix is just the exponent of its element, then proof is straightforward. But then you might think this is difficult to justify without proof:

$$(e^{iD})_{ij}=e^{iD_{ij}}$$

Thanks

 

[maajdl]

One approach is to use a Taylor expansion.
Another is to assume that -in the end- everything end up in expressions involving the four elementary operations.
Taking it as a definition in not that bad, as this is very close to the two other approaches I mentioned.

 

[Seydlitz]

That actually makes sense to me, thanks! :)

 

[Office_Shredder]

It's not true that any function of a diagonal matrix will give a diagonal matrix; for example the function which sends every matrix to
\left(<br /> \begin{array}{cc}<br /> 0 &amp; 1\\<br /> 0 &amp; 0<br /> \end{array}<br /> \right)
clearly outputs non-diagonal matrices.

It's a special property that if you define your function with a Taylor series, and the constant term is diagonal, that every diagonal matrix will get mapped to a diagonal matrix (whose entries are the appropriate Taylor series)