# Find matrix from minimal polynomial?

Find a $3 \times 3$ matrix $A$ whose minimal polynomial is $m_A(x) = x^2$.

From the Cayley-Hamilton theorem, $m_A(x) = 0 \implies A^2 = 0$. So, A is nilponent so, its characteristic polynomial is $P_A(x)=x^3$,since $m_A(x)=x^2$. That means that $A \neq0$. I don't know what to do from here. Any tips?

I'd try a matrix with zeroes everywhere except in one entry (this one, not in the main diagonal).

• So, it's just a matter of testing? I'm not missing something? – Alex Matt Jan 12 '18 at 15:41
• Well, you need a matrix $A$ with a triple zero eigenvalue, such that $A\neq 0$ but $A^2=0$. This example looks fine. There could be more examples, though. – ajotatxe Jan 12 '18 at 15:44
• @AlexMatt There is some knowledge behind this hint (specifically about triangular matrices with zero diagonal) but that's best learned, in my opinion, by exploring it on your own. – Arthur Jan 12 '18 at 15:44
• I worked it out with your help and i think i understand the reasoning behind the hint. Thank you! – Alex Matt Jan 12 '18 at 15:50

$$A=\begin{matrix} [0& 0&1] \\ [0& 0&0] \\ [0&0&0] \end{matrix}$$ is one of the desired matrix.

There is a matrix $A$ that is in Jordan normal form and satisfies $A^2=0$:

$$A=\left[\begin{array}{cc|c}0&1&0\\0&0&0\\ \hline 0&0&0\end{array}\right]$$

Here's a way to determine all such matrices systematically.

First, the minimal polynomial is invariant under similarity transformations. Every matrix is similar to a matrix Jordan form, therefore it suffices to look at matrices in Jordan form. Note that we can also sort the Jordan blocks from largest to smallest through similarity transformations. For $3\times 3$ matrices, it is quite easy to list all possible Jordan forms: $$J_1 = \pmatrix{\lambda_1 & 0 & 0\\0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3}, \quad J_2 = \pmatrix{\lambda_1 & 1 & 0\\0 & \lambda_1 & 0\\ 0 & 0 & \lambda_2}, \quad J_3 = \pmatrix{\lambda_1 & 1 & 0\\0 & \lambda_1 & 1\\ 0 & 0 & \lambda_1}$$ Now as you already deduced, the characteristic polynomial is $x^3$, therefore all eigenvalues must be zero, and the matrices reduce to $$J_1 = \pmatrix{0 & 0 & 0\\0 & 0 & 0\\ 0 & 0 & 0}, \quad J_2 = \pmatrix{0 & 1 & 0\\0 & 0 & 0\\ 0 & 0 & 0}, \quad J_3 = \pmatrix{0 & 1 & 0\\0 & 0 & 1\\ 0 & 0 & 0}$$ Now as you also correctly concluded, the matrix cannot be the zero matrix; this excludes $J_1$. Also, it is easily checked that of the remaining matrices only $J_2$ has square $0$.

Therefore a matrix has the minimal polynomial $x^2$ if and only if it is similar to $J_2$.

The question remains how a matrix that is similar to $J_2$ looks like. What we immediately see is that it is of rank $1$ and has trace $0$. Since both rank and trace are invariant under similarity transformations, we can conclude that any matrix that is similar to $J_2$ must be of the form (I use $^\dagger$ to denote the conjugate transpose): $$A = uv^\dagger \text{ with } \operatorname{tr} A = v^\dagger u = 0, u\ne 0, v\ne 0$$ On the other hand, assume that $A$ has this form. Then since the two vectors are orthogonal one can an unitary transformation that transforms $u$ to $\pmatrix{a\\0\\0}$ and $v$ to $\pmatrix{0\\b\\0}$. Thus $A$ is transformed to $\pmatrix{0 & ab^* & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$, which can then be further transformed to $J_2$ by the similarity transformation $TAT^{-1}$ with $T=\operatorname{diag}(1/\sqrt{ab^*},\sqrt{ab^*},1)$.

Therefore the minimal polynomial of $A$ is $x^2$ if and only if $A=uv^\dagger$ for some non-zero vectors $u,v$ with $v^\dagger u=0$.