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I just gave a proof for this question. Here's my follow up question: Let $A \in \ \mathbb{M}_n(\mathbb{F})$ where F is a field and there exists $n\in N$ where $A^n$= I. In the case where n=1,2, $A^1$=I and $A^2$=I.

Here's my question: In general, if A is it's own inverse, then does it necessarily follow A=I? In other words, is I the only matrix which is it's own inverse? My gut reaction is to say no, but it would probably be fairly tedious to construct a matrix multiplication formula which produces the subset $S\subset \mathbb{M}_n(\mathbb{R})$ where S = {A | AA =I }. Is there such a subset in general? We know the set's nonempty since $I\in S$. Are there any others?

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    $\begingroup$ Is $1$ the only number which is its own inverse? $\endgroup$ – Jason DeVito Mar 10 '15 at 22:51
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    $\begingroup$ I'm not sure I understand your first paragraph. What's wrong with $\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}$? $\endgroup$ – Git Gud Mar 10 '15 at 22:51
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    $\begingroup$ @JasonDeVito MULTIPLICATIVE inverse in the field of real numbers,yes. But these are matrices and don't follow the same algebraic structures in general. $\endgroup$ – Mathemagician1234 Mar 10 '15 at 22:53
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    $\begingroup$ @Mathemagician1234: There's another element in the field of real numbers which is its own inverse. Using that, you can easily build many examples of matrices which are their own inverses. $\endgroup$ – Jason DeVito Mar 10 '15 at 22:54
  • $\begingroup$ @Mathemagician1234 If you have a field, you have a multiplication, a $1$ and a $-1$. $\endgroup$ – Git Gud Mar 10 '15 at 22:55
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You are looking for involutory matrices. To answer the question: no, there are other matrices that are their own inverses.

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$$\begin{pmatrix}\cos\theta & \sin\theta \\ \sin\theta &-\cos\theta\end{pmatrix}^{-1}=\begin{pmatrix}\cos\theta & \sin\theta \\ \sin\theta &-\cos\theta\end{pmatrix}$$ for any $\theta\in\mathbb{R}$.

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    $\begingroup$ any reflection matrix $\pmatrix{\cos t & \sin t\\\sin t & -\cos t}$ would do. $\endgroup$ – abel Mar 10 '15 at 22:57
  • $\begingroup$ you are welcome. $\endgroup$ – abel Mar 10 '15 at 23:11
  • $\begingroup$ facepalm Right in front of my face and didn't see it. $\endgroup$ – Mathemagician1234 Oct 12 '16 at 5:08
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If $A^2=I$, then $(A-I)(A+I)=0$. Suppose that $J$ is the Jordan Canonical Form of $A$, then since $A=SJS^{-1}$, we have $$ \begin{align} 0 &=S^{-1}(A^2-I)S\\ &=S^{-1}AAS-S^{-1}IS\\ &=S^{-1}ASS^{-1}AS-I\\ &=J^2-I \end{align} $$ Thus, $J$ must be a diagonal matrix all of whose diagonal elements are $+1$ and $-1$.

Therefore, $A^2=I$ if and only if $A$ is diagonalizable and has eigenvalues $+1$ and $-1$.

Easy examples are diagonal matrices whose diagonal elements are $+1$ and $-1$, but any similar matrices will also work.

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Well, you can actually create examples of such matrices very easily. All you need is a linear transformation which is it's own inverse. Just choose a basis and swap some extries (make sure to do disjoint swaps), for example, say $T : \mathbb{R}^2 \to \mathbb{R}^2$, such that $T(e_1) = e_2, T(e_2) = e_1$, the corresponding matrix will be a $2 \times 2$ matrix with $1$s in bottom left and top right entries and zeroes elsewhere.

A harder question would be does there exist a basis with respect to which this linear transformation is a diagonal matrix all of whose diagonal elements are $+1$ and $−1$. The answer is yes, since then such a involutory matrix has eigenvalues $+1$ and $-1$.

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