# Assume that $A$ has the minimal polynomial as $(x-1)(x-2)$ and c.p. as $(x-2)^2(x-1)$

Let $$A := \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\\ \end {pmatrix}$$ Find the number of matrices similar to $$A$$ whose entries are from $$\mathbb{Z}/\mathbb{3Z}$$.

Since $$A$$ is a $$3 \times 3$$ matrix and $$A$$ has the minimal polynomial as $$(x-1)(x-2)$$ and c.p. as $$(x-2)^2(x-1)$$.Then any matrix simialr to $$A$$ will have the same minimal and characteristic polynomial.

Any matrix similar to $$A$$ will be an upper triangular matrix or a lower triangular matrix with diagonals as $$\{1,2,2\}$$ and $$\{2,2,1\}$$.

Then number of upper triangular matrix with diagonals as $$\{1,2,2\}$$ is $$3^3$$ . Similarly number of lower triangular matrix is $$3^3$$.

Then the total numbers is $$54$$. So the total number of matrices as $$2 \times 54 = 108$$.The answer is $$117$$.What am i missing?

• Why should every matrix similar to $A$ be triangular ? Commented Jan 27, 2022 at 10:11
• Not sure but every matrix similar to A must have the same minimal and characteristic polynomial polynomial and traingular matrices satisfy the condition Commented Jan 27, 2022 at 10:19
• Yes, but : 1) That does not mean that any matrix similar to $A$ must be triangular, and 2) That does not mean (a priori) that every triangular matrix with $(2,2,1)$ on the diagonal is similar to $A$... Commented Jan 27, 2022 at 10:21
• You are right that the similar matrices are in this case exactly those with the same characteristic and minimal polynomials. It's just that there are more of these than you think. For example $\begin{pmatrix} 0 & 0 & 1\\ 1& 0 &1\\ 0 & 1 & -1\end{pmatrix}$. Commented Jan 27, 2022 at 10:22
• Yes i am aware that it holds true only when $n \le 3$ Commented Jan 27, 2022 at 14:03

Let consider the map $$\begin{array}{lrcl} f : & GL_3(\mathbb{Z}/3\mathbb{Z}) & \longrightarrow & GL_3(\mathbb{Z}/3\mathbb{Z}) \\ & P & \longmapsto & PAP^{-1} \end{array}$$

We want to find the cardinality of the image of $$f$$.

• Let's begin by finding the cardinality of $$\mathcal{C} = \lbrace B \in GL_3(\mathbb{Z}/3\mathbb{Z}), BA=AB \rbrace$$.

By direct computation, one can see that a matrix $$\displaystyle{M=\begin{pmatrix} a & b & c\\ d & e & f \\ g&h&i \end{pmatrix} \in \mathcal{M}_3(\mathbb{Z}/3\mathbb{Z})}$$ commutes to $$A$$ iff $$c=f=g=h=0$$, i.e. iff $$M$$ looks like $$\displaystyle{M=\begin{pmatrix} a & b & 0\\ d & e & 0 \\ 0&0&i \end{pmatrix}}$$

For this matrix to belong to $$GL_3(\mathbb{Z}/3\mathbb{Z})$$, one needs $$i \neq 0$$ and $$ae-db \neq 0$$. So $$\mathcal{C} \simeq GL_2(\mathbb{Z}/3\mathbb{Z}) \times \mathbb{Z}/2\mathbb{Z}$$ so $$\mathrm{Card}(\mathcal{C})=48 \times 2 = 96$$.

• Now, let $$P,Q \in GL_3(\mathbb{Z}/3\mathbb{Z})$$. Then $$f(P)=f(Q)$$ iff $$PAP^{-1} = QAQ^{-1}$$, i.e. $$f(P)=f(Q) \Longleftrightarrow Q^{-1}P \in \mathcal{C}$$.

So if you fix $$P$$, then the set $$\lbrace Q \in GL_3(\mathbb{Z}/3\mathbb{Z}), f(P)=f(Q) \rbrace$$ is in bijection with $$\mathcal{C}$$, so for every $$P$$, the fiber $$f^{-1}(f(P))$$

has exactly $$96$$ elements.

This gives you directly that the cardinality of the image of $$f$$ is equal to $$\mathrm{Card}(\mathrm{Im}(f))=\frac{\mathrm{Card}(GL_3(\mathbb{Z}/3\mathbb{Z}))}{96} = \frac{11232}{96} = 117$$

so $$\boxed{\text{there are exactly }117\text{ matrices that are similar to }A}$$.

• How did you get the intuition behind the problem. This approach didn't even occur to me Commented Jan 27, 2022 at 14:05
• @ThirstForMaths Have you heard about group actions ? I did not use the formalism of group actions in my answer, but all the intuition behind it is the same as for group actions. Commented Jan 27, 2022 at 14:16