# Thread: Similarity of Matrices

1. ## Similarity of Matrices

Is there any way to tell if a matrix $\displaystyle A$ is similar to a matrix $\displaystyle B$ other than finding an explicit $\displaystyle S$ such that $\displaystyle B=S^{-1}AS$?

The reason is that I am trying to find conjugacy classes of $\displaystyle SL_2(\mathbb{Z}/3\mathbb{Z})$ in the easiest possible way.

2. Similar matrix - Wikipedia, the free encyclopedia

Jordan normal form - Wikipedia, the free encyclopedia

Matrices A and B are similar iff their Jordan normal forms are equal (this is because every matrix in an algebraically closed field has a unique Jordan form).

If you know the minimal polynomial of a matrix, then it's easier to calculate its jordan form:

Let $\displaystyle F$ be an algebraically-closed field, $\displaystyle V$ a vector space over $\displaystyle F$,
$\displaystyle A \in F^{n \times n}$ represent a linear operator $\displaystyle T:V \rightarrow V,$
$\displaystyle m_A = (x-c_1)^{r_1} \cdot (x-c_2)^{r_2} \cdot \ldots \cdot (x-c_k)^{r_k}$ be the minimal polynomial of A, where each $\displaystyle r_i$ is the geometrical multiplicity of $\displaystyle c_i$, and let $\displaystyle w_i = Ker(T-c_i I)^{r_i}.$

Then by the primary decomposition theorem, $\displaystyle V = w_1 \oplus w_2 \oplus \ldots \oplus w_k,$ and after a few more steps, we get that each $\displaystyle T_i = T|_{w_i}$ is represented by a Jordan block matrix:

$\displaystyle J_i = \begin{pmatrix} c_i & 0 & 0 & 0 & \ldots & 0 \\ 1 & c_i & 0 & 0 & \ldots & 0 \\ 0 & 1 & c_i & 0 & \ldots & 0 \\ \vdots & & \ddots & \ddots \\ \vdots & & & \ddots & \ddots \\ 0 & 0 & 0 & \ldots & 1 & c_i \end{pmatrix}, J_i \in F^{r_i \times r_i}$

And the Jordan canonical form of A is the direct sum of its Jordan blocks.

3. Originally Posted by redsoxfan325
Is there any way to tell if a matrix $\displaystyle A$ is similar to a matrix $\displaystyle B$ other than finding an explicit $\displaystyle S$ such that $\displaystyle B=S^{-1}AS$?

The reason is that I am trying to find conjugacy classes of $\displaystyle SL_2(\mathbb{Z}/3\mathbb{Z})$ in the easiest possible way.

In general, other poster already told you: two square matrices over some field are simmilar iff their Jordan forms (over some extension field of the original one that contains all the eigenvalues of either of the matrices) are equal.

In you case, though, it's way easier since you're dealing with 2 x 2 matrices, and then two matrices here are simmilar iff they have the same characteristic and minimal polynomials, which is far from being true in other cases (well, also with 3x3 matrices it is true, but with 4 x 4 there already counterexamples)

Tonio

4. Originally Posted by tonio
In general, other poster already told you: two square matrices over some field are simmilar iff their Jordan forms (over some extension field of the original one that contains all the eigenvalues of either of the matrices) are equal.

In you case, though, it's way easier since you're dealing with 2 x 2 matrices, and then two matrices here are simmilar iff they have the same characteristic and minimal polynomials, which is far from being true in other cases (well, also with 3x3 matrices it is true, but with 4 x 4 there already counterexamples)

Tonio
Hmm, are you sure that there are counter examples with 4x4 matrices? I'm pretty sure I remember seeing somewhere that the smallest order matrix you could use to contradict this would be a 7x7.

5. Originally Posted by Defunkt
Similar matrix - Wikipedia, the free encyclopedia

Jordan normal form - Wikipedia, the free encyclopedia

Matrices A and B are similar iff their Jordan normal forms are equal (this is because every matrix in an algebraically closed field has a unique Jordan form).

If you know the minimal polynomial of a matrix, then it's easier to calculate its jordan form:

Let $\displaystyle F$ be an algebraically-closed field, $\displaystyle V$ a vector space over $\displaystyle F$,
$\displaystyle A \in F^{n \times n}$ represent a linear operator $\displaystyle T:V \rightarrow V,$
$\displaystyle m_A = (x-c_1)^{r_1} \cdot (x-c_2)^{r_2} \cdot \ldots \cdot (x-c_k)^{r_k}$ be the minimal polynomial of A, where each $\displaystyle r_i$ is the geometrical multiplicity of $\displaystyle c_i$, and let $\displaystyle w_i = Ker(T-c_i I)^{r_i}.$

Then by the primary decomposition theorem, $\displaystyle V = w_1 \oplus w_2 \oplus \ldots \oplus w_k,$ and after a few more steps, we get that each $\displaystyle T_i = T|_{w_i}$ is represented by a Jordan block matrix:

$\displaystyle J_i = \begin{pmatrix} c_i & 0 & 0 & 0 & \ldots & 0 \\ 1 & c_i & 0 & 0 & \ldots & 0 \\ 0 & 1 & c_i & 0 & \ldots & 0 \\ \vdots & & \ddots & \ddots \\ \vdots & & & \ddots & \ddots \\ 0 & 0 & 0 & \ldots & 1 & c_i \end{pmatrix}, J_i \in F^{r_i \times r_i}$

And the Jordan canonical form of A is the direct sum of its Jordan blocks.
That's interesting! The definition of "Jordan normal form" that I have always used has the diagonal of "1"s above the main diagonal!

Of course, the give the same concepts.

6. Two matrices are "similar" if they represent the same linear transformation, written in different bases. That is the same as saying that they have the same eigenvalues and the eigenvectors corresponding to those eigenvalues are the same.

7. Originally Posted by Defunkt
Hmm, are you sure that there are counter examples with 4x4 matrices? I'm pretty sure I remember seeing somewhere that the smallest order matrix you could use to contradict this would be a 7x7.

$\displaystyle \left(\begin{array}{cccc}2&1&0&0\\0&2&0&0\\0&0&2&0 \\0&0&0&2\end{array}\right)\ncong \left(\begin{array}{cccc}2&1&0&0\\0&2&0&0\\0&0&2&1 \\0&0&0&2\end{array}\right)$

And both matrices above have the same char. pol. $\displaystyle (x-2)^4$ and the same min. pol. $\displaystyle (x-2)^2$

What's important to remember here? The power at which every single irreducible factor of the min. pol. is raised only tells us what's the maximum size of a Jordan block corresponding to that eigenvalue and that there's at least one Jordan block wrt this eigenvalue of this size, but it does NOT necessarily tells us how many blocks of this size are there (for that we already need the dimension of the corresponding eigenspace)...unless the matrix is of order at most 3 x 3.

Tonio

8. Originally Posted by tonio
$\displaystyle \left(\begin{array}{cccc}2&1&0&0\\0&2&0&0\\0&0&2&0 \\0&0&0&2\end{array}\right)\ncong \left(\begin{array}{cccc}2&1&0&0\\0&2&0&0\\0&0&2&1 \\0&0&0&2\end{array}\right)$

And both matrices above have the same char. pol. $\displaystyle (x-2)^4$ and the same min. pol. $\displaystyle (x-2)^2$

What's important to remember here? The power at which every single irreducible factor of the min. pol. is raised only tells us what's the maximum size of a Jordan block corresponding to that eigenvalue and that there's at least one Jordan block wrt this eigenvalue of this size, but it does NOT necessarily tells us how many blocks of this size are there (for that we already need the dimension of the corresponding eigenspace)...unless the matrix is of order at most 3 x 3.

Tonio
Ah, yes, of course, that makes sense. The example I was referring to was when each of the matrices' eigenvalues also have the same algebraic and geometrical multiplicities.