1

How do we prove the above proposition?
$(AA)(ij)=1$ for $i=j$ , $0$ for $i\ne j $.
So, $[AA]_{ij} = \sum_k [A]_{ik} [A]_{kj}$.
How can I proceed from this point?

Martin Sleziak
  • 56,060
bigboss
  • 21
  • 5
    What about $$ \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix}$$? – Watson Mar 08 '16 at 21:45
  • 1
    Over an arbitrary field $\mathbb{K}$ of characteristic not equal to $2$, there are exactly $n+1$ matrices, up to conjugacy, of dimension $n\times n$ which satisfy the condition. If $\text{char},\mathbb{K}=2$, then there are exactly $\left\lfloor \frac{n}{2}\right\rfloor+1$ matrices, up to conjugacy, of dimension $n\times n$ satisfying the condition. – Batominovski Mar 08 '16 at 22:08
  • Probably a duplicate of http://math.stackexchange.com/questions/106070/if-a2-i-identity-matrix-then-a-pm-i and http://math.stackexchange.com/questions/44341/finding-number-of-matrices-whose-square-is-the-identity-matrix – Martin Sleziak Mar 16 '16 at 05:56

3 Answers3

2

Suppose $A$ is diagonal matrix then $A^2$ is diagonal with diagonal elements squares of diagonal elements of $A$...

Consider

$A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$

$A=\begin{bmatrix} 1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$

and many more..

Michael Hardy
  • 1
1

This is not generally true. It only implies that $A = A^{-1}$, i.e., that $A$ is its own inverse.

Michael Hardy
  • 1
nippon
  • 445
1

This is not necessarily true. You should attempt to find a counter example in 2x2 matrices.

OttR - A. Yu
  • 43