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I have a small confusion. Suppose there are two $n \times n$ matrices $A$ and $B$ such that $A$ does not has $n$ independent eigenvectors. The $A$ is not diagonalizable. But $A$ and $B$ commute and I can find a matrix that diagonalizes $B$. Doesn't this violate that $A$ is diagonalizable because the same matrix also diagonalizes $A$ which diagonalizes $B$.

Two such matrices are:

$A$ $$ \begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{matrix} $$

$B$ $$ \begin{matrix} 2 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -1 & 2 \\ \end{matrix} $$

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  • $\begingroup$ Sorry, but I don't see what the problem is here. $\endgroup$ – Lord Shark the Unknown Oct 2 '18 at 17:17
  • $\begingroup$ The problem is that A is not diagonalizable but the fact that A commutes with B allows me to diagonalize A using the matrix which diagonalizes B. $\endgroup$ – Ankur Singh Oct 2 '18 at 17:23
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    $\begingroup$ But your $A$ has three independent eigenvectors. Why did you say it didn't? $\endgroup$ – Lord Shark the Unknown Oct 2 '18 at 17:25
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    $\begingroup$ If the rank of an $n\times n$ matrix is $n-k$, that just gives us $k$ independent eigenvectors to the eigenvalue $0$: the basis of the nullspace. $\endgroup$ – Misha Lavrov Oct 2 '18 at 17:34
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    $\begingroup$ real symmetric matrices are always diagonalizable. They are also congruence diagonalizable, in that there is a real matrix $P$ with $\det P = 1$ and $P^T A P = D$ is diagonal. The diagonal entries of $D$ will not usually be eigenvalues of $A,$ but do obey Sylvester Inertia. $\endgroup$ – Will Jagy Oct 2 '18 at 17:55
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In your example, matrices $A$ and $B$ are both diagonalizable (and both have $n$ independent eigenvectors), so it's not an instance of the thing you're describing:

  • $A$ has eigenvector $(1,0,1)$ to the eigenvalue $2$, and eigenvectors $(0,1,0)$ and $(1,0,-1)$ to the eigenvalue $0$.
  • $B$ has eigenvector $(1,-2,-1)$ to the eigenvalue $-1$, eigenvector $(1,1,-1)$ to the eigenvalue $2$, and eigenvector $(1,0,1)$ to the eigenvalue $3$.

(Also, since $A$ and $B$ are both symmetric in this example, we know in advance that they should be diagonalizable.)

But in general, no: just because $A$ commutes with $B$ and $B$ is diagonalizable, doesn't mean that $A$ is diagonalizable (in the same basis that diagonalizes $B$, or otherwise). For instance, any matrix (diagonalizable or otherwise) commutes with the zero matrix and the identity matrix.

Also, the Jordan form of a matrix lets us write it as $D + N$ in some basis, where $D$ is diagonal, $N$ is nilpotent (and therefore not diagonalizable in general) and $D$ commutes with $N$, giving us a whole slew of counterexamples.

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  • $\begingroup$ The eigenvalues are 2 and 0 with multiplicity 1 and 2 respectively. $\endgroup$ – Ankur Singh Oct 2 '18 at 17:40
  • $\begingroup$ Thanks for the correction. $\endgroup$ – Misha Lavrov Oct 2 '18 at 17:40

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