Similar Matrices and Jordan Form

June 15, 2019

The eigenvalues of the inverse matrix $A^{-1}$ is its inverses, because:

$\vert A\vert=\lambda_1*\lambda_2*...$

And from determinant’s property, we know the determinant of $A^{-1}$ is the inverses of the determinant of $A$

$\vert A^{-1}\vert=\frac{1}{\vert A\vert}=\frac{1}{\lambda_1*\lambda_2*...}=\frac{1}{\lambda_1}*\frac{1}{\lambda_2}*...$

What comes from this is if I know the matrix $A$ is positive definite then $A^{-1}$ will also be positive definite.

Another useful property is if $A,B$ are positive definite, then $$A+B$$ is also positive definite. We can use the property learned last note to prove this $\mathbf x^\top A\mathbf x>0$ , we have:

$\begin{align} \mathbf x^\top A\mathbf x>0\\ \mathbf x^\top B\mathbf x>0 \end{align} \tag{except x at 0}$

Then clearly

$\mathbf x^\top (A+B)\mathbf x>0$

Now redefine $A$ as arbitrary rectangular $m\times n$ matrix. Then $A^\top A$ is square and symmetric. It is a positive semidefinite matrix because:

$\mathbf x^\top A^\top A\mathbf x=(A\mathbf x)^\top(A\mathbf x)=\Vert A\mathbf x\Vert^2\geq0$

If we want it to be positive definite, to get rid of the possibility that the length of vector $A\mathbf x$ is zero, we just need to get rid of the possibility that except $\mathbf x$ is length zero ( $Ax$ has length zero only if x in null space, we exclude the 0 vector in the definition of this test, what’s left is to exclude others with free variables). So we need full column rank fo $A$ . That’s the requirement how $A^\top A$ is invertible.

Similar Matrices

Note we’re no longer expecting symmetric matrices here. But still use square matrices. Square matrices $A$ and $B$ are similar if and only if there exists an invertible matrix $M$ s.t.

$B=M^{-1}AM$

One example is

$\Lambda=S^{-1}AS$

What's special about similar matrices? They have same eigenvalues. Why?

$\begin{align} A\mathbf x&=\lambda \mathbf x\\ AMM^{-1}\mathbf x&=\lambda \mathbf x\\ M^{-1}AMM^{-1}\mathbf x&=\lambda M^{-1}\mathbf x\\ BM^{-1}\mathbf x&=\lambda M^{-1}\mathbf x\\ B\mathbf v&=\lambda\mathbf v\\ \end{align}$

This finishes the proof. And of course, the eigenvectors do not stay the same. The new eigenvectors are $M^{-1}$ times the eigenvectors of $A$ .

Bad Cases

There’re two cases if we have two eigenvalues same. One is we have a diagonal matrix, i.e., some number times the identity matrix, this matrix won’t be changed at all for $M^{-1}AM$ . So this kind of matrix has a small family. The other is just the rest.