If there is a $2\times 2$ matrix $A^2 = I$, what will the complete solution set for $A$ be?

Question

So the basic thing I started with was equating the diagonal and the other other two entries of matrix $A$ respectively which gives, $$a=\pm d, \\b=c$$ But there is this matrix $\begin{bmatrix}0 & 2\\ \frac12 & 0\end{bmatrix}$ who's square also equals $I$.

So, what error did I make? And what will the method to obtain the complete solution set be?

Please write how did you obtain the wrong conditions. Oterwise it is impossible to say what error did you make. — user, Apr 05 '23 at 13:53
Does this answer your question? Trace of a matrix $A$ with $A^2=I$ — Sourav Ghosh, Apr 05 '23 at 13:53
Did you compute $A^2$ in general and set it equal to $I$? It looks like you just guessed (incorrectly) at some solutions, without attempting to find all solutions. So your fundamental error was thinking that guessing solutions was sufficient. — Lee Mosher, Apr 05 '23 at 14:14
From How to ask a good question: "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." — jjagmath, Apr 05 '23 at 14:15
I agree with the previous comment. It's hard to pin point your the error in your calculations if you don't show your calculations. — jjagmath, Apr 05 '23 at 14:19
I've added just now. the calculation part i did. Sorry for the inconvenience. — Chatrapal Singh Rathore, Apr 05 '23 at 15:35
Be very careful. If $ax=ay$, is it valid to conclude $x=y$? — Ted Shifrin, Apr 05 '23 at 20:33
By that reasoning, since $0 \cdot 3 = 0 \cdot 4$ we conclude that $3=4$? — Lee Mosher, Apr 06 '23 at 13:54

tmwki · Answer 1 · 2023-04-06T02:09:42.847

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I don't know how you got the condition on the entries of the matrix. There are infinitely many such matrices. To see this note that $A^2=I$ implies that the minimal polynomial of the matrix is $(x-1), (x+1)$ or $(x+1)(x-1)$ since the minimal polynomial divides the annihilating polynomial $x^2-1$. This means that A is diagonalisable and hence has the form $$ A=I,\quad A=-I,\quad A=S\begin{pmatrix}1&0\\0&-1\end{pmatrix}S^{-1} $$ for any invertible matrix $S$. All these choices satisfy $A^2=I$.

edited Apr 06 '23 at 02:09

answered Apr 05 '23 at 13:39

tmwki

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$A=I$ is also possible, of course. – Levent Apr 05 '23 at 15:21
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$A=-I$ works too . The condition on the characteristic polynomial does not stay that $A$ should have exactly one eigenvalue $1$ and one eigenvalue $-1$. It can have two eigenvalues $1 $(and thus $I=A$) or two eigenvalues $-1$ (and thus $A=-I$) – Lelouch Apr 05 '23 at 15:32
I've just added the calculations i did. Sorry – Chatrapal Singh Rathore Apr 05 '23 at 15:37

If there is a $2\times 2$ matrix $A^2 = I$, what will the complete solution set for $A$ be?

1 Answers1