# Finding the Jordan Form of matrices with mostly 0

I have been studying linear algebra and came across this question which I'm not quite sure how to solve. The problem asks whether the matrices are similar. I wanted to find the Jordan form of 3 matrices and compare to see if they are similar: $$A=\begin{bmatrix} 0 & 1 & 1 & 0\cr 0 & 0 & 0& 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , } B=\begin{bmatrix} 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , } C=\begin{bmatrix} 0 & 0 & 1 & 1\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\end{bmatrix}$$

I know that you have to find the eigenvalues from the characteristic polynomial but I'm not sure what the next step is from there. Also, since the matrices mostly contain $$0$$s is there a shortcut?

Any help would be greatly appreciated!

• These matrices are upper triangular, so you already know what the characteristic polynomial will be Commented Dec 15, 2020 at 22:11
• $A$ is not similar to $B$ or $C$ since $A^2 \neq 0$, but $B^2,C^2$ are zero. $B,C$ are similar because $(B - \lambda I)^k,(C - \lambda I)^k$ have the same rank for all eigenvalues $\lambda$ and all positive integers $k$ (notably zero for $k \geq 2$). Commented Dec 15, 2020 at 22:14

The magic words are "sparse matrices" (matrices with "most" entries equal to zero, with some specific measure for "most), but here the really important fact is that they are all strictly upper triangular, not that they are sparse.

1. All three matrices are upper triangular, so the eigenvalues are just the constants in the diagonal. In this case, each of them has a single eigenvalue $$\lambda=0$$, with algebraic multiplicity $$4$$.

2. The dimension of the eigenspace corresponding to $$\lambda$$ is the nullity. From the Rank-Nullity Theorem/Dimension Theorem, it is easier to get the ranks. It is easy to see that in all three cases the rank is $$2$$, so the nullity is $$2$$. That means that the eigenspace has dimension $$2$$.

3. The dimension of the eigenspace is equal to the number of blocks in the Jordan canonical form corresponding to $$\lambda=0$$. Thus, you have only two possible Jordan Canonical forms for these matrices: either two $$2\times 2$$ blocks, or one $$3\times 3$$ block and one $$1\times 1$$ block.

4. You have two $$2\times 2$$ blocks if and only if the square of the matrix is $$0$$. By thinking about what happens to the standard basis upon two applications of the given matrix, it is easy to verify that $$A^2\neq 0$$ (see what happens to $$\mathbf{e}_4$$), and that $$B^2=C^2=0$$.

The number of Jordan blocks of size $$n$$ for the eigenvalue $$\lambda$$ is $$2\dim\ker(A-\lambda I)^n-\dim\ker(A-\lambda I)^{n+1}-\dim\ker(A-\lambda I)^{n-1}$$

For the matrix $$A$$, we have $$\dim\ker A=2$$, $$\dim\ker A^2=3$$ and $$\dim\ker A^3=4$$. This determines completely its Jordan form, which is $$J(0,1)\oplus J(0,3)$$.

For the matrix $$B$$, we have $$\dim\ker B=2$$ and $$\dim \ker B^2=4$$, therefore its Jordan form is $$J(0,2)\oplus J(0,2)$$.

For the matrix $$C$$, we have $$\dim \ker C=2$$, $$\dim \ker C^2=3$$ and $$\dim\ker C^3=4$$, therefore it is similar to $$A$$.