About matrix diagonalization in C from the characteristic polynomial.

Question

Ok the excercise is:

You have one characteristic polynomial, it's:

$\lambda^4 + \lambda^2$

Find two matrixes with this polynomial, one of them diagolalizable in C and the other one not.

so the first thing i have done is to find a matrix wich had this polynomial. I have found.

\begin{pmatrix} 0&0 &-1 &0 \\ 0&0 &0 &0 \\ 1&0 &0 &0 \\ 0&0 &0 &0 \end{pmatrix}

For ($A-\lambda I)=0$ we have:

\begin{pmatrix} -\lambda &0 &-1 &0 \\ 0&-\lambda &0 &0 \\ 1&0 &-\lambda &0 \\ 0&0 &0 &-\lambda \end{pmatrix}

wich if im not wrong gives us the $p(A)$=$\lambda^4 + \lambda^2$.

This matrix, again if im not wrong, is diagonalizable in C. Now if i need to find another one not diagonalizable i need to build another matrix from scratches or, as i suspect, there is a simple method to do it?

Answer 1

I assume that we're talking about matrices with real entries.

The easiest way to approach this problem is to build examples using either Jordan normal form or the real Jordan form. Another approach is to build block diagonal matrices where each block is a companion matrix.

That's the general method, let's look at your approach. It turns out that your example is essentially the same as $$ \pmatrix{0&-1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0} $$ which is the "real Jordan form" associated with the diagonalizable matrix here. To get this from your answer, I switch the roles of the $2$nd and $3$rd rows/columns.

One answer to the non-diagonalizable matrix with this characteristic polynomial is $$ \pmatrix{0&-1&0&0\\1&0&0&0\\0&0&0&\color{red}1\\0&0&0&0} $$ again, this is an example of a "real Jordan form". We could have also used the companion matrix $$ \pmatrix{ 0&0&0&0\\ 1&0&0&0\\ 0&1&0&-1\\ 0&0&1&0} $$

Answer 2

The first observation to make is that your polynomial factors as $X^4+X^2=X^2(X^2+1)$ and that the factors $X^2$ and $X^2+1$ are relatively prime. By the kernel decomposition theorem, for any linear operator with this characteristic polynomial, the space decomposes as a direct sum of stable subspaces for which the characteristic polynomials of the restrictions of the operators are $X^2$ respectively $X^2+1$. You can therefore assume without loss of generality that the matrix is block-diagonal with $2\times2$ diagonal blocks with those characteristic polynomials. The factor $X^2+1$ further factors over $\Bbb C$ as $(X-\mathbf i)(X+\mathbf i)$, which are distinct monic factors of degree$~1$, so that $2\times2$ will be diagonalisable over $\Bbb C$ with eigenvalues $\mathbf i,-\mathbf i$ in all cases; en example with rational entries is the companion matrix $(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix})$ of the polynomial $X^2+1$. On the other hand $X^2$ has a repeated factor, so that part may or may not be diagonalisable; it is diagonalisable only if it is zero, and non diagonalisable if it is any other nilpotent matrix, like $(\begin{smallmatrix}0&0\\1&0\end{smallmatrix})$ (which happens to be the companion matrix of the polynomial $X^2$).