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Suppose $A$ is a square diagonal matrix. Suppose $B$ is a square diagonalizable matrix (i.e. there is a basis $\beta$ for $V = \mathbb{R}^n$ with respect to which the linear transformation represented by $B$ is representable by a diagonal matrix). Can we find a basis for $V$ wrt which both $A$ and $B$ are diagonal? If so, how?

If $B$ represents some lin. transf. $T$ with resp. to the standard basis for $V$, and $B$ is diagonalizable, then there is a basis $\beta$ for $V$ with respect to which $T$ is representable by $kI$ (some diagonal matrix). If $A$ is diagonalizable to begin with (wrt standard basis), will it also be diagonal wrt $B$ in general? I want to say no. How can I prove that a basis $B'$ exists with respect to which both $A$ and $B$ are diagonal?

Nażysław Zbyłutowicz
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  • Two matrices are simultaneusly diagonalizable only if they commutes. See http://math.stackexchange.com/questions/236212/simultaneously-diagonalizable-proof – Emilio Novati Mar 31 '17 at 18:36

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I'm having trouble figuring out exactly what you're asking, but it should all be handled by the following theorem:

Take any two diagonalizable transformations $S,T:V \to V$. There exists $\mathcal B$ for which the two matrices are both diagonal (that is, $S$ and $T$ can be simultaneously diagonalized) if and only if $ST = TS$.

For instance, consider $$ A = \pmatrix{2&1\\1&2}, \qquad B = \pmatrix{2&0\\0&2} $$ they are not both diagonal, but they are both diagonalizable with $AB = BA$. If we take a base such as $\mathcal B = \{(1,1),(1,-1)\}$, we find that the matrix of these transformations relative to the basis is $$ A = \pmatrix{3&0\\0&1} \qquad B = \pmatrix{2&0\\0&2} $$ on the other hand, if we had $$ A = \pmatrix{2&1\\1&2}, \qquad B = \pmatrix{2&0\\0&5} $$ we would find that there is no such basis, since $AB \neq BA$.

Ben Grossmann
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