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 [itex]A = UDU\dagger[/itex], where U is a unitary matrix and D is diagonal. Show that nay function of A defined by a power series,

[itex]f(A) = f_{0}I + f_{1}A + f_{2}A^{2} + ... + f_{n}A^{n} + ... [/itex]

can be expressed as [itex]f(A) = Uf(D)U\dagger[/itex]

The Attempt at a Solution

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

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

Thanks

 

[Fredrik]

Homework Statement

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

[itex]f(A) = f_{0}I + f_{1}A + f_{2}A^{2} + ... + f_{n}A^{n} + ... [/itex]

can be expressed as [itex]f(A) = Uf(D)U\dagger[/itex]

The Attempt at a Solution

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

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

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,

[itex]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 + ... [/itex] ?

 

[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