A $\neq \pm$ I and $A^2$ = I. Prove that every vector z has a unique decomposition z = x + y where Ax = x and Ay = -y Let A be a real valued and symmetric nxn
matrix with entries such that
A $\neq \pm$ I and $A^2$ = I
(a)  Prove that there exist non-zero column vectors v and w
such that Av = v and Aw =-w
(b) Prove that every vector z has a unique decomposition z = x + y
where Ax = x and Ay = -y
My try : 
well, since $A^2$ = I we can say A can have only 1 or -1 as it's eigenvalues. Now for the 1st part (part (a)) 
there exists a column vector(the eigen vectors)  v and w corresponding to 1 and -1 respectively. 
Now, what if all the eigenvalues are 1 or all the eigenvalues are -1 ? 
I mean , how to show that A does not have all the eigenvalues equal to -1 or 1 i.e. A  at least has one eigen value of the opposite sign ?? 
Yet, I'm unable utilize the fact A $\neq \pm$ I ! 
for part (b)
i know that eigenvectors corresponding to different eigenvalues are orthogonal. and orthogonal decomposition of a vector is unique. so if there exist a z such that z=x+y and Ax = x and Ay = -y then the decomposition must be unique. but how to show that for every z it will be possible to decompose it in a way such that Ax = x and Ay = -y ?
 A: EDIT:
Since $A$ is real and symmetric, it can be diagonalized by orthogonal congruence:
$A = PDP^{-1}$
For some orthogonal matrix $P$ and diagonal matrix $D$ (that has the same eigenvalues as $A$, since they are similar).
Suppose to the contrary that all the eigenvalues are equal to 1 (or resp. -1). Then $D$ is a diagonal matrix with only one value (1 or -1), i.e $D=\pm I$.
$\pm I \neq A = P(\pm I) P^{-1} = \pm PP^{-1}=\pm I$, and that serves as the needed contradiction.
Therefore $A$ has both 1 and -1 as eigenvalues which easily proves a) like you stated.
For b), use the fact that the eigenspaces decompose $\mathbb R^n$ since they are orthogonal and their dimensions add up to $n$ - you can take an orthogonal basis for each eigenspace, and get an orthogonal subset of $\mathbb R ^ n$ (because different eigenspaces are orthogonal) which is a basis since $A$ is diagonalizable.
A: Let $X=\frac12 (A+I)$. Since $A \not = -I$ we get that $X$ is not the zero matrix so for some $z$ we get $v = Xz \not = 0$ and $Av = AXz = \frac12 A(A+I)z = \frac12 (A^2+A)z = \frac12 (I+A)z = Xz = v$. Similarly with $Y=-\frac12 (A-I)$, nonzero since $A \not = I$, for some $z$ we get $w = Yz \not = 0$ and $Aw = AYz = -w$. 
For any $z$, let $x = Xz$ and $y = Yz$, then $x+y = Xz + Yz = (X+Y)z = \frac12 (A+I-A+I)z = z$ and $Ax = AXz = Xz = x$ as before and $Ay = AYz = -Yz = -y$.
For uniqueness, if $z=x+y$ with $Ax=x$ and $Ay=-y$, then $Xz = X(x+y) = x$ since $(A+I)x = Ax+x = x+x =2x$ and thus $Xx = x$ and $(A+I)y = Ay+y = -y+y = 0$ and thus $Xy = 0$. Similarly $Yz = y$ since $Yx = 0$ and $Yy = y$. 
A: We do not need that $A$ is symmetric. We not not even ever have heard of eigenvectors. We do not even need that we work in a finte-dimensional space and that our ground field is $\Bbb R$. All we need is that $2$ is invertible in out ground field.
Given $z$, let $x=\frac 12(z+Az)$ and $y=\frac12(z-Az)$. Then clearly $z=x+y$, $Ax=\frac12(Az+A^2z)=\frac12(Az+z)=x$, $Ay=\frac12(Az-A^2z)=\frac12(Az-z)=-y$. This shows part b)
For part a), note that $x=\frac 12(z+Az)$ is automatically a vector with $x=Ax$. We just have to confirm that it does not happen that $x=0$ for all $z$. But that is just the same as saying that $A\ne -I$. Likewise, $y=\frac12(Az-z)$ will be non-zero for some $z$ as otherwise we would have $Az=z$ for all $z$, i.e., $A=I$.
