Normal matrix that isn't diagonalizable; counterexample?

[algebrat]

I've been reading that the diagonalizable matrices are normal, that is, they commute with their adjoint: $M^*M=MM^*$, where $M^*$ is the conjugate transpose of $M$.

So a matrix is diagonalizable if and only if it is normal, see: http://en.wikipedia.org/wiki/Normal_matrix

But from Boyce's section on repeated eigenvalues for systems of first-order differential equations, we have the example $$M=\begin{pmatrix}1&-1\\1&3\end{pmatrix}.$$ Which if we check $MM^*$, we get $$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}
=\begin{pmatrix}2&-2\\-2&8\end{pmatrix}.$$ The last is symmetric, so clearly $MM^*=M^*M$.

However, Boyce does the problem, finds repeated eigenvalue of $r=2$, only one eigenvector, for $MP=PJ$, with $J$ not diagonal, that is, $$MP=
\begin{pmatrix}1&-1\\1&3\end{pmatrix}
\begin{pmatrix}1&0\\-1&-1\end{pmatrix}=
\begin{pmatrix}2&1\\-2&-3\end{pmatrix}=
\begin{pmatrix}1&0\\-1&-1\end{pmatrix}
\begin{pmatrix}2&1\\0&2\end{pmatrix}
=PJ$$

So we've found that $M$ is normal, but it is not diagonalizable. What is going on?

 

[algebrat]

Oh, $P$ is not unitary, as the normal matrix theorem describes. I'll have to think about that. It seems like if normal matrix theorem gets you a better $J$ matrix with fewer allowable $P$ matrices, then what Boyce did should have gotten [STRIKE]an orthonormal P matrix with[/STRIKE] $J$ diagonal.

 

[micromass]

Your M isn't normal.

You concluded that M is normal because $MM^*$ is symmetric. But $MM^*$ is always symmetric/hermitan:

$$(MM^*)^*=M^{**}M^*=MM^*$$

Don't make the common mistake that

$$(MM^*)^*=M^*M$$

this is false.

Also, note that a matrix being normal means that it is diagonalizable by orthogonal/unitary operators. A matrix can be diagonalizable, without being normal.

 

[algebrat]

Aha, thank you for catching both of my mistakes micromass. So the matrix wasn't normal, which would have been diagonalizable, and diagonalizable does not imply normal.

 

[skiller]

I'm no expert on matrices, but this line looks wrong to me:

$$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}
=\begin{pmatrix}2&-2\\-2&8\end{pmatrix}$$

Shouldn't it be:
$$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}3&1\\-1&1\end{pmatrix}
=\begin{pmatrix}4&0\\0&4\end{pmatrix}$$
?

As I said, I'm not too clued up on adjuncts and such like, so I'm asking more for clarification of my own thinking rather than pointing out any error. Please ignore if I'm talking rubbish.

 

[micromass]

No, that part was fine.

The transpose (which is the adjoint for real matrices) switches the rows and columns.

For example, in M, the first row is (1 -1). So that should be the first column of M*. The second row is (1 3) and that should be the second column of M*

So M* is indeed

$$\left(\begin{array}{cc}1 & 1\\ -1 & 3\end{array}\right)$$

In particular, the diagonal entries always remain fixed.

For complex matrices, the adjoint is a bit more complicated. There, it involves not only transposing (=switching rows and columns) but you also need to take the complex conjugate of the entries.

 

[skiller]

No, that part was fine.

Ahh... I understand now, thanks. I think I was confusing "adjugate" with "adjoint".

I must brush up on my matrices!

 

[micromass]

Ahh... I understand now, thanks. I think I was confusing "adjugate" with "adjoint".

I must brush up on my matrices!

Aaah, that makes sense. There are indeed two definitions of adjoint which are not equivalent (hence the reason why we call one adjugate instead of adjoint). I should have known you meant the adjugate when I saw that your multiplication resulted in a diagonal matrix.
But yeah, be careful with the term adjoint!

 

[skiller]

But yeah, be careful with the term adjoint!

Yeah, I've since discovered the ambiguity!

However, this still isn't correct:

$$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}
=\begin{pmatrix}2&-2\\-2&8\end{pmatrix}$$

It should be:
$$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}
=\begin{pmatrix}2&-2\\-2&10\end{pmatrix}$$

Although I don't suppose this affects the original question; just a minor quibble.

 

[micromass]

Ah yes, it should indeed be 10!

 

[algebrat]

Ah yes, it should indeed be 10!

So you all claim that $1\oplus1+3\oplus3=10$ and not $8$, hmmm? (JK! But that was my propogating mistake originally.)

 

[algebrat]

Aaah, that makes sense. There are indeed two definitions of adjoint which are not equivalent (hence the reason why we call one adjugate instead of adjoint). I should have known you meant the adjugate when I saw that your multiplication resulted in a diagonal matrix.
But yeah, be careful with the term adjoint!

Right, like micromass said, adjoint matrix usually refers to conjugate transpose, and this generalizes to... I think it's called adjoint operator. The basic idea is that the adjoint is what appears in the dual space, that is, for an inner product, $x\cdot(Ay)=(A^*x)\cdot y.$

Then, on a higher plane, they have something called adjoint functor, which I am pretty sure is similar, but never on the same level. What I believe they have in common is that the adjoint is sort of a mirror image in a dual world. Then I think there's a different adjoint (maybe 2) in Lie algebras, having to do with conjugacy. It used to drive me nuts trying to figure out if any of these things are the same, but I think it was just an overuse of the word adjoint, which means something like the object closest too perhaps.

So, I guess they decided to rename your adjoint matrix (in sense of finding inverse), so in some older books it is called adjoint, in others it is called... You all named adjunct and adjugate as candidates.