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Let $A$ be a real symmetric matrix and $B$ a real positive-definite matrix. Is it possible to simultaneously diagonalize of $A$ and $B$? Thank you very much.

Yiorgos S. Smyrlis
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LJR
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  • Some times. For instance if they're both diagonal matrices they are simultaneously diagonalisable. – Arthur Dec 24 '14 at 09:03

2 Answers2

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If you consider $A,B$ as quadratic forms, then, of course, they are simultaneously diagonalizable. That is, there is an invertible $S$ s.t. $S^TBS=I,S^TAS=D$ where $D$ is a diagonal matrix.

Proof: diagonalize $B$ and $B^{-1/2}AB^{-1/2}$. Since $B^{-1/2}AB^{-1/2}$ is symmetric, there is an orthogonal $O$ s.t. $O^TB^{-1/2}AB^{-1/2}O=D$ where $D$ is diagonal. Finally take $S=B^{-1/2}O$.

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Yes, if and only if they have the same eigenvectors.

Equivalently, if and only if they commute.

Yiorgos S. Smyrlis
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    Not "have the same eigenvectors", but "admit a basis of common eigenvectors". Take for instance $B=I$ and $A$ not a scalar matrix. Commutation is a correct (and simpler) if and only if condition. – Marc van Leeuwen Dec 24 '14 at 10:38
  • But still, they do have the same eigenvectors! – Yiorgos S. Smyrlis Dec 25 '14 at 10:46
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    No they don't. The only reasonable meaning for having "the same eigenvectors" is having the same set of eigenvectors. One wouldn't say two polynomials $P,Q$ have the same roots unless every root of $P$ is a root of $Q$ and vice versa. A far-fetched alternative interpretation could be: having some common eigenvector (but "share an eigenvector" without plural would then be much clearer), but this does not give a necessary and sufficient condition either. Just try to define what you mean by having the same eigenvectors. – Marc van Leeuwen Dec 25 '14 at 11:30