Show A = UDU(dagger) can be written as f(A) = Uf(D)U(dagger)

[Smazmbazm]

Homework Statement

A diagonalisable square matrix A can be written $A = UDU\dagger$, where U is a unitary matrix and D is diagonal. Show that nay function of A defined by a power series,

$f(A) = f_{0}I + f_{1}A + f_{2}A^{2} + ... + f_{n}A^{n} + ...$

can be expressed as $f(A) = Uf(D)U\dagger$

The Attempt at a Solution

Not sure where to start. I know one can repeat a linear operator to get results such as $A^{0} = 1, A^{1} = A, A^{2} = AA$ but how do you show that for D? Are we simply doing the same? $D^{0} = 1, D^{1} = D, D^{2} = DD$

So $f(A) = U * f_{0}I + f_{1}D + f_{2}D^{2} + ... + f_{n}D^{n} + ... * U\dagger$ ?

Thanks

 

[Fredrik]

Homework Statement

A diagonalisable square matrix A can be written $A = UDU\dagger$, where U is a unitary matrix and D is diagonal. Show that nay function of A defined by a power series,

$f(A) = f_{0}I + f_{1}A + f_{2}A^{2} + ... + f_{n}A^{n} + ...$

can be expressed as $f(A) = Uf(D)U\dagger$

The Attempt at a Solution

Not sure where to start. I know one can repeat a linear operator to get results such as $A^{0} = 1, A^{1} = A, A^{2} = AA$ but how do you show that for D? Are we simply doing the same? $D^{0} = 1, D^{1} = D, D^{2} = DD$

So $f(A) = U * f_{0}I + f_{1}D + f_{2}D^{2} + ... + f_{n}D^{n} + ... * U\dagger$ ?

Thanks

That last line is messed up in some way. If you meant to start the right-hand side with $U(\dots$ and end it with $\dots)U^\dagger$, then you have just written down the equality you're supposed to prove. But yes, that thing in the middle is $f(D)$. So it can't be a bad idea to try to use that (the definition of $f(D)$) to rewrite $Uf(D)U^\dagger$ in some way.

 

[Smazmbazm]

Should it look like this,

$f(A) = U*f_{0}I*U\dagger + U*f_{1}D*U\dagger + U*f_{2}D^{2}*U\dagger + ... + U*f_{n}D^{n}*U\dagger + ...$ ?

 

[Fredrik]

I assumed that you you were trying to write down the equality $f(A)=Uf(D)U^\dagger$, with $f(D)=1+f_0 D+\cdots+f_n D^n+\cdots$. You obviously can't assume that the former equality holds, since that's what you want to prove, but you can start
$$Uf(D)U^\dagger=U\left(1+f_0 D+\cdots+f_n D^n+\cdots\right)U^\dagger,$$ to see if you can end up with $f(A)$. The right-hand side above can be written the way you just wrote it, but I wouldn't use $*$ for multiplication. It's standard to not use any symbol at all, and if you want to use one, it should be $\circ$, since multiplication of linear operators is just composition of functions.

 

[Smazmbazm]

Ok got it. Thanks for your help, Fredrik