# 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? Jan 12, 2018 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. Jan 12, 2018 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. Jan 12, 2018 at 15:44
• I worked it out with your help and i think i understand the reasoning behind the hint. Thank you! Jan 12, 2018 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 $v^\dagger u=0$.