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How can we describe the equivalence classes under the similarity relation for $2 \times 2$ matrices with respect to the field of real numbers, $\mathbb{R}$? How would the equivalence classes change if the field is $\mathbb{C}$?

I know that for ${Mat} _{1\times1}(\mathbb{R})$ each matrix has its own equivalence class, and I know that for $2\times 2$ matrices, the identity and zero matrices have their own equivalence class. But how can we describe the rest of the equivalence classes with respect to transformations and bases?

Thomas Andrews
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user67928
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2 Answers2

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I will answer part of your question.

You can determine the equivalence class of an $n\times n$ matrix by looking at its real Jordan form. In the $2\times2$ case, the real Jordan form of a matrix may take one of the following forms:

  1. $\begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix}$ for some $\lambda\ge\mu$.
  2. $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ for some $\lambda\in\mathbb{R}$.
  3. $\begin{pmatrix}a&-b\\ b&a\end{pmatrix}$ for some real number $a$ and nonzero real number $b$. Over $\mathbb{C}$, this real Jordan form is similar to the complex Jordan form $\begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}$.

In general, if $J$ is the real Jordan form of an $n\times n$ matrix $A$, the equivalence class of $A$ can be written as $\{S^{-1}AS: S\in GL_n(\mathbb{R})\}=\{S^{-1}JS: S\in GL_n(\mathbb{R})\}$. Two real matrices are similar if and only if they have the same real Jordan form. That is, they are $\mathbb{R}$-similar if and only if they are $\mathbb{C}$-similar. So, the equivalence class of a real matrix (or strictly speaking, the intersection of its equivalence class with $M_n(\mathbb{R})$) remain unchanged if you change the field from $\mathbb{R}$ to $\mathbb{C}$.

user1551
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Given a basis of $R^n$, a matrix in $M_n(R)$ describes a linear transformation with respect to that basis. When a change of basis occurs, the matrix must compensate accordingly by becoming a matrix which is similar to the old one.

So, a conjugacy class of a given matrix is just the collection of matrices which represent some $R$ linear transformation on $R^n$ in different bases.

That the zero and identity matrices are only conjugate to themselves is just a testament to the fact that the zero transformation and the identity transformation "look the same" in all bases. The same can be said of the matrices which are constant on the diagonal and zero elsewhere.

If you move from $M_n(R)$ to $M_n(C)$, then you can say the same thing about the $C$ linear transformations of $C^n$.

rschwieb
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  • In $\mathbb C$, the advantage is that you can be sure two matrices are similar by looking at their Jordan Normal Forms. Is that true over $\mathbb R$ - that is, if their (possibly complex) JNFs are the same, are they similar in $M_n(\mathbb R)$? – Thomas Andrews Mar 22 '13 at 20:57
  • When you say that the equivalence classes are collections of matrices representing linear transformations, could you be more specific? – user67928 Mar 22 '13 at 22:04
  • @user67928 I'm not really sure what to add. Given a matrix representing a linear transformation, any and all of the matrices conjugate to the matrix are matrices representing the same transformation in another base. – rschwieb Mar 25 '13 at 01:34