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
$$\left( \begin{array}{cc} 0 & 1\\ 0 & 0 \end{array} \right)$$
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)

 

[blue_raver22]

for your question. The special property of diagonal matrices is indeed an important one and can be proven through various methods. One way to prove it is by using the properties of matrix multiplication. Since diagonal matrices have all their non-diagonal elements equal to zero, any function of a diagonal matrix will also have all its non-diagonal elements equal to zero. This is because the diagonal elements are the only ones that will be involved in the multiplication process. Therefore, we can say that any function of a diagonal matrix is just a function of its diagonal elements.

Another way to prove this property is by using the definition of diagonal matrices. A diagonal matrix is defined as a matrix where the non-diagonal elements are zero, and the diagonal elements can be any real numbers. Therefore, any function of a diagonal matrix will only affect the diagonal elements, leaving the non-diagonal elements unchanged. This also proves that any function of a diagonal matrix is just a function of its diagonal elements.

In regards to your specific question about proving that $e^{iD}$ of a diagonal matrix is unitary, it can be done by using the properties of diagonal matrices and the definition of unitary matrices. A unitary matrix is a square matrix that is equal to its own conjugate transpose. So, if we can show that $e^{iD}$ satisfies this condition, then we can prove that it is unitary. Using the above proof, we know that $e^{iD}$ will only affect the diagonal elements, which means that its conjugate transpose will also only affect the diagonal elements. Since the diagonal elements of $e^{iD}$ are complex numbers, their conjugate transpose will be their complex conjugates. Therefore, we can say that $(e^{iD})^*=e^{-iD}$, and since $e^{iD}$ is equal to its conjugate transpose, it satisfies the definition of a unitary matrix.

In conclusion, the special property of diagonal matrices can be proven through various methods, including using the properties of matrix multiplication and the definition of diagonal matrices. And using these properties, we can also prove that $e^{iD}$ of a diagonal matrix is unitary. I hope this helps to clarify the proof of this important property.