Simultaneous Diagonalization

Matrix Theory: Motivated by the fact that diagonal matrices commute and have a common eigenvector basis, we state a result on when a set of matrices can be simultaneously diagonalized. That is, a set of diagonalizable matrices has a joint eigenbasis if and only if they are a commuting set.

[MathDoctorBob]

1) If you have an invariant subspace, pick a basis for the subspace, and extend to the entire space; then the corresponding matrix is block upper triangular. I can't remember the specific video with this, but it is earlier in the playlist.

2) Every square matrix can be put into rational canonical form, which is block diagonal (blocks are companion matrices). If over C, then we can put into upper triangular form, and even better Jordan canonical form.

3) Earlier video also. (cont.)

[MathDoctorBob]

3) A straightforward induction argument shows that the char polynomial of a block-diagonal matrix is the product of the char polynomials of the diagonal blocks. That is, det(A1 * \ 0 A2) = det(A1)det(A2). You need the definition of minimal at hand to finish.

[MathDoctorBob]

Definitely. Especially where representation theory enters the picture. - Bob

[ching-tsunchou]

Do you really need to do the induction? The restriction of T_\alpha on E_i is simply \alpha_i times the identity operator, which is diagonalized in any basis. So all you need to show is that T_\beta is diagonalizable in each E_i. Did I miss something?

[josiahlocke]

I don't understand how you are using the minimal polynomial to show that T beta is diagonalizable on Ei here. How do you deduce that the MP of T beta restricted to Ei factors into distinct linears? How do you deduce that the minimal polynomial of T beta restricted to Ei divides the MP of T beta on the entire vector space?I don't see an obvious explanation in the video, could you explain?

[manughulyani]

time 8:17, theorem says factoring of minimal polynomial with different linear factors is only sufficient for diagoanlizablity (not necessary)! please verify! example identity on R^n

[maurcd]

Hey, thanks a lot for posting!

I had been looking for a complete proof of this for quite a while! Most of the proofs out there mention the induction process, but they fail to explain why, when you restrict the operators, you still get diagonalizible matrices.

[wdlang]

Simultaneous Diagonalization is important in quantum mechanics