Why does commuting matrices have same eigenvectors?

[netheril96]

I googled for a proof,but didn't find one.
Could anyone give me a link to a proof?

 

[Mark44]

Is this a homework or test problem? If so, you're not likely to get much help without showing what you have tried.

 

[netheril96]

Is this a homework or test problem? If so, you're not likely to get much help without showing what you have tried.

Neither
I just read a book on quantum physics which mentioned this theorem.I want a rigorous proof.

 

[quasar987]

First off, this is QM so let's specialize to hermitian matrices and recall that those always have a complete set of eigenvectors. Suppose A and B are two nxn commuting hermitian matrices over C. And suppose v is an eigenvector of A with eigenvalue a: Av=av. Then, A(Bv)=(AB)v=(BA)v=B(Av)=B(av)=a(Bv). That is to say, if v is an eigenvector of A with eigenvalue, then Bv is also an eigenvector of A with eigenvalue a. Another way to say this is that the eigenspace V_a of A corresponding to the eigenvalue a is stable under B, meaning $B(V_a)\subset V_a$.

What this means is that if you write the matrix B with respect to a basis of C^n of the form (b_1,...,b_n) where each b_i is an eigenvector of A and where adjacent vectors correspond to the same eigenvalue, then B will be block diagonal. And each block is itself a hermitian matrix, so for instance if the first block is kxk, then it has k linearly independant eigenvectors. And if (w1,...,wk) is any such eigenvector, then (w1,...,wk,0,...0) is an eigenvector of B. And since it lives in the eigenspace of the first eigenvalue of A, then it is also an eigenvector of A. And so like that, for each mxm block of B we find m linearly independant eigenvectors of B which are also eigenvectors of A.

 

[blue_raver22]

The reason why commuting matrices have the same eigenvectors is due to a fundamental property of linear transformations. When two matrices commute, it means that they can be multiplied in any order without changing the result. This also means that they have the same eigenvectors, as eigenvectors are transformed by a matrix in the same way regardless of the order of multiplication.

To prove this, we can start by considering the definition of eigenvectors and eigenvalues. An eigenvector of a matrix A is a non-zero vector x such that Ax = λx, where λ is a scalar known as the eigenvalue.

Now, let's consider two commuting matrices A and B. This means that AB = BA. Let x be an eigenvector of A with eigenvalue λ. This means that Ax = λx. We can then multiply both sides by B, giving us B(Ax) = B(λx). Using the commutativity of A and B, we can rearrange the equation to get (BA)x = (λB)x. But since AB = BA, this can also be written as A(Bx) = (λB)x. This means that Bx is also an eigenvector of A with eigenvalue λ.

Similarly, we can show that Bx is an eigenvector of B with the same eigenvalue λ. This is because Bx = B(Ax) = (BA)x = (AB)x = A(Bx).

Therefore, we have shown that if x is an eigenvector of A, then it is also an eigenvector of B, and vice versa. This is why commuting matrices have the same eigenvectors.

As for a link to a proof, here is one that outlines a similar approach to the one described above: https://math.stackexchange.com/questions/249156/why-do-commuting-matrices-have-the-same-eigenvectors. I hope this helps!