A^2=e | the solution of the equation | what exactly can this equation tell us

if $A^2=E$

Lemma one

if $A^2=E$ be $A$ The characteristic value of can only be $1$ or $-1$

prove ：

set up $A$ The eigenvalue of is $\lambda$, The corresponding eigenvector is $\eta$. therefore

$$A^2\eta =AA\eta =A\lambda \eta =\lambda A\eta ={\lambda }^2\eta$$

and

$$A^2\eta =E^2\eta =\eta$$

thus

$${\lambda }^2=1$$

Lemma II

If $A^2=E$ be $\mathrm{r}(A+E)+\mathrm{r}(A-E)=n$
Some inequalities about rank

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By the lemma 12 know ,$A$ The eigenvalue of is $1$ or $-1$, And $r(A+E)+r(A-E)=n$

From the knowledge of eigenvectors, we can know , The eigenvalue $1$ The dimension of the corresponding eigenvector space is equal to $n-r(A-E)$, The eigenvalue $-1$ The dimension of the corresponding eigenvector space is equal to $n-r(A+E)$. Because the eigenvalue is only 1 and -1, therefore $A$ The number of linearly independent eigenvectors of is $n-r(A-E)+n-r(A+E)=n$ explain $A$ Diagonalize .

therefore , There are invertible matrices $P$, bring

$$P^{-1}AP=B$$

$B$ It's a diagonal matrix , Diagonal element by $1$ and $-1$ form , share $\mathrm{r}(A-E)$ individual $1$,$\mathrm{r}(A+E)$ individual $-1$.

therefore $A$ Can write $PB{P}^{-1}$ Any matrix of . among $B=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(1,1,\cdots ,1,-1,-1,\cdots ,-1)$,$\mathrm{r}(A-E)$ individual $1$,$\mathrm{r}(A+E)$ individual $-1$,$P$ Is an arbitrary invertible matrix .

in fact , set up $A=PB{P}^{-1}$, be $A^2=PB{P}^{-1}PB{P}^{-1}=PBB{P}^{-1}=PE{P}^{-1}=E$

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2022 year 6 month 22 Japan 19:09:22