# Find real and complex matrix

How to show that there is no real $3 \times 3$ matrix whose minimal polynomial is $x^2+1$, but there is real $2\times2$ as well as complex $3\times 3$ matrix with minimal polynomial $x^2+1$. I don't understand how to start solving this question ... Please help me

• Welcome to MathSE. Please use MathJax to format your future posts. I've helped you out with this one. – Theoretical Economist Sep 4 '18 at 17:16
• Do you know the definition of minimal polynomial? See, we need to know what you know so that we can answer your question better. – Teresa Lisbon Sep 4 '18 at 17:17
• @астонвіллаолофмэллбэрг Yes smallest degree polynomial that annihilate matrix. – user499117 Sep 4 '18 at 17:22
• A $3 \times 3$ matrix is a matrix of odd dimension. Hence, its characteristic polynomial is of odd degree, hence contains a real root, since complex roots of a real polynomial always occur in conjugate pairs, so are even in number, but here the total number of roots is odd. Now, every root of the characteristic polynomial is also a root of the minimal polynomial, so the minimal polynomial must also have a real root, which $x^2+1$ does not.(Also, you may delete the answer below) – Teresa Lisbon Sep 4 '18 at 17:28

If $A$ is a $3 \times 3$ real matrix with minimal polynomial $x^2+1$, then $$\sigma(A) = \{\text{zeroes of }x^2+1\} = \{i, -i\}$$ On the other hand, the characteristic polynomial of $A$ is a third degree polynomial with real coefficients so it has at least one real root, which would be an eigenvalue. This is a contradiction.
Notice that the real $2 \times 2$ matrix $\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$ and complex $3 \times 3$ matrix $\begin{bmatrix} i & 0 & 0 \\ 0 & i & 0\\ 0 & 0 & -i\end{bmatrix}$ both have minimal polynomials equal to $x^2+1$.