Videos for Section 4.7


Suppose that $A$ and $B$ are diagonalizable operators on a vector space $V$. We can find a basis of eigenvectors of $A$, making calculations involving $A$ really simple, but then $B$ is likely to be ugly. Or we can diagonalize $B$, in which case $A$ will probably be ugly.

When can we have our cake and eat it too? We'd like to find a basis $\mathcal{B}=\{{\bf b}_1, \ldots, {\bf b}_n\}$, such that each vector ${\bf b}_j$ is an eigenvector of $A$ (say, with eigenvalue $\lambda_j$) and is also an eigenvector of $B$ (say, with eigenvalue $\mu_j$). Finding such a basis is called simultaneously diagonalizing $A$ and $B$.


Theorem: Two diagonalizable operators $A$ and $B$ can be simultaneously diagonalized if, and only if, they commute. (That is, if and only if $AB=BA$.)


In the first video we go over what this means, work out an example, and prove the theorem in the special case that all of the eigenvalues of $A$ have multiplicity 1.

In the second video, we prove the theorem in general. The proof is constructive, and shows you how to actually find the eigenvectors, and I work out an example (which is also found in the book) in detail.