# Diagonalizability of a certain matrix in $\mathbb{Z}_5$

True/False: Matrix $$\ \begin{pmatrix}1&3\\3&4\end{pmatrix}$$ in $$\mathbb{Z}_5$$ is diagonalizable.

I have come to the conclusion that all the Eigenvectors for this matrix are not linearly independent, therefore it is not diagonalizable; however, I am not sure if my conclusion is correct. Would you be able to assist me in confirming my results? Thank you!

P.S. I also see that this matrix is symmetric, which should mean that it is diagonalizable, but maybe not in Zmod5?

It's not diagonalizable. Clearly both of its determinant and trace are zero. Therefore, it has only $$0$$ as its eigenvalue. If it's diagonalizable, then it would be similar to and hence must be the zero matrix itself.

Symmetric matrices can be diagonalized is a theorem that holds only over the real numbers. It's not even true over $$\mathbb C$$ (instead, conjugate symmetric matrices over $$\mathbb C$$ can be diagonalized). In fact, there are two types of theorems in linear algebra: those that are purely algebraic and hence works over almost any field, and others that use some kind of analysis such as positivity in reals. The spectral theorem belongs to the latter class.

• Thank you very much, your answer helps! Jul 25, 2022 at 15:05
• So, is the matrix nilpotent over $\mathbb{Z}_5$? Square the matrix (over $\mathbb{Z}$) and see whether you get a matrix with all its entries being divisible by $5$. Jul 25, 2022 at 15:31
• You can use Cayley-Hamilton to conclude it's nilpotent, or simply it has only zero eigenvalue. Jul 26, 2022 at 0:01

$$A=\begin{pmatrix} 1&3\\3&4\end{pmatrix}$$

Characteristics polynomial : \begin{align}\chi_A(t) &=t^2-5t-5\in \mathbb{Z}_5[t]\\&=t^2\end{align}

Hence $$0$$ is an eigenvalue of $$A$$ of multiplicity (algebraic) $$2$$.

But geometric multiplicity of $$0$$ is $$1$$ (shortage of independent eigenvectors ) i.e $$\dim E(0, A) =\dim\mathcal{N}(A) =1$$

$$\begin{pmatrix} 1&3\\3&4\end{pmatrix}\overset{R_2-3R_1}\rightarrow \begin{pmatrix} 1&3\\0&-5\end{pmatrix}= \begin{pmatrix} 1&3\\0&0\end{pmatrix}$$

• Conclusion: $A$ is not diagonalizable over $\mathbb{Z}_5$ Jul 25, 2022 at 15:55