# Similarity of $3\times 3$ matrices

I am considering $$3\times 3$$ matrices, one of them is diagonalizable i.e

$$A=\left[\begin{array}{ccc} 3&0&4\\ 0&-1&0\\ -2&0&-3 \end{array}\right]= \left[\begin{array}{ccc} -1&0&-2\\ 0&1&0\\ 1&0&1 \end{array}\right] \left[\begin{array}{ccc} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{array}\right] \left[\begin{array}{ccc} -1&0&-2\\ 0&1&0\\ 1&0&1 \end{array}\right]^{-1}=P_1DP_1^{-1}.$$ The second one is not diagonalizable, and we have $$B=\left[\begin{array}{ccc} 1&0&0\\ 0&-1&1\\ 0&0&-1 \end{array}\right]= \left[\begin{array}{ccc} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}\right] \left[\begin{array}{ccc} -1&1&0\\ 0&-1&0\\ 0&0&1 \end{array}\right] \left[\begin{array}{ccc} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}\right]^{-1}=P_2JP_2^{-1}.$$ Matrices $$A$$ and $$B$$ are called similar if there exists an invertible matrix P such that $$B=P^{-1}AP$$.

In my example $$A$$ and $$B$$ are not similar, because one of them is diagonalizable and the second is not diagonalizable. I am not sure, that my explaination is proper. I would be grateful for your advices, how to explain that this matrices are not similar.

• You say that $B$ is not diagonalizable, but then write it as $B=PJP^{-1}$ and then $B\sim J$? Jan 5, 2023 at 12:56
• $J$ is Jordan matrix Jan 5, 2023 at 13:03
• But the original matrix $B$ is already a Jordan matrix. What else do you want? Jan 5, 2023 at 13:04
• oh.... I did not noticed it... (too much work.) But i am not sure, that my explaination about similarity is proper. Jan 5, 2023 at 13:10
• Well, $J$ is Jordan matrix, I didn't read correctly the matrix. Now, in order you see if $A$ and $B$ are similar you can see if we can find a matrix $P$ such that $AP=PB$. You can use brute force defining $P=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$. Jan 5, 2023 at 13:20

Since $$A$$ is diagonalizable matrix and $$B$$ is a Jordan matrix which is not a diagonal matrix, $$A$$ and $$B$$ are not similar.

Or you can say that$$\dim\{v\in\Bbb R^3\mid A.v=-v\}=2,$$whereas$$\dim\{v\in\Bbb R^3\mid B.v=-v\}=1,$$which also proves that $$A$$ and $$B$$ are not similar.

• thank you very much for explaination Jan 5, 2023 at 13:40

By contradiction, suppose that $$A\sim B$$, then there exists a matrix $$P$$ regular such that $$AP=PB$$. Let $$P=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$$ be a matrix such that $$\boxed{\det(P)\not=0}$$.

A bit of algebra shows that

• $$AP=\begin{pmatrix}3&0&4\\0&-1&0\\-2&0&-3\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}3a+4g&3b+4h&3c+4i\\-d&-e&-f\\-2a-3g&-2b-3h&-2c-3i \end{pmatrix}$$.
• $$PB=\begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1 \end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}a&b&c\\g-d&h-e&i-f\\-g&-h&-i \end{pmatrix}$$

But since we supposed that $$AP=PB$$ that is true if and only if $$a=0,\quad b=0,\quad c=0,\quad g=0,\quad h=0,\quad i=0$$ for $$d,e,f\in {\bf R}$$, but thus $$\det(P)=\det\begin{pmatrix}0&0&0\\d&e&f\\0&0&0\end{pmatrix} \implies \boxed{\det(P)=0}$$ and it is a contradiction.

Therefore, $$A$$ is not similar to the matrix $$B$$.

• We consider $AP=PB$, i think. Jan 6, 2023 at 13:21
• Of course, I fixed this just right now. Jan 6, 2023 at 14:03