Matrices with all zero eigenvalues

[Leo321]

If I have a matrix for which all eigenvalues are zero, what can be said about its properties?
If I multiply two such matrices, will the product also have all zero eigenvalues?

Thanks

 

[HallsofIvy]

All eigenvalue of an n by n matrix, A, are 0 if and only if $A^n v= 0$ for all vectors, v.

If $$A= \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$ and $$B= \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$$, then A and B both have all eigenvalues 0 but AB does not.

 

[Leo321]

Thanks!
How do you show that Av=0 for all vectors v?
I am not sure I understand the meaning of a matrix with all-zero eigenvalues. Obviously you can't decompose it to a diagonal representation.

 

[HallsofIvy]

Thanks!
How do you show that Av=0 for all vectors v?

You don't- it isn't necessarily true. For example, take
$$A= \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$ and
$$v= \begin{bmatrix}0 \\ 1\end{bmatrix}$$. The only matrix A, such that "Av= 0 for all vectors v" is, of course, the 0 matrix.

What is true, as I said before, is that $A^nv= 0$ for all vectors v, where A is an n by n matrix.

I am not sure I understand the meaning of a matrix with all-zero eigenvalues. Obviously you can't decompose it to a diagonal representation.

Not quite obvious!
$$\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$
is such a "diagonal representation".

An n by n matrix can be "diagonalized" if and only if there exist n independent eigenvectors. If that is not true, then the matrix can be put into "Jordan Normal Form" which has its eigenvalues along the main diagonal and possibly "1"s on the diagonal above the main diagonal- with 0 elsewhere.
If A is 3 by 3 then it can be reduced to one of these three forms:
$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
or
$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$
depending upon whether the "eigenspace" of its eigenvectors has dimension 3, 2, or 1, respectively.

 

[Leo321]

Thanks

 

[trambolin]

I think you are struggling with the question rather than its answer. The question says whenever I have a matrix (say n-by-n), I compute the eigenvalues with the characteristic equation and I obtain $\lambda^n=0$. Then I have one more matrix as such. And when I multiply these two matrices and compute the eigenvalues, can I get the same zero eigenvalues? Actually HallsofIvy provided you such matrices for a counterexample. my suggestion is that you work out the eigenvalues of that example for both A,B and also AB.

For your original question, let me poison you with some more terminology : http://en.wikipedia.org/wiki/Nilpotent_matrix" . You can read the properties from the link.