If $A=A^2$, and if $v$ is a nonzero vector in $Col (A)$, then $v$ is an eigenvector of $A$ with eigenvalue $1$. Let $A ∈ M_n (\mathbb{R})$ and suppose $A = A^2$.
Prove: if $v$ is a nonzero vector in $\operatorname{Col} (A)$, then $v$ is an eigenvector of $A$ with eigenvalue $1$.
My attempt:
Let $v \in \operatorname{Col}(A)$, and let $A=[a_1 ~ a_2 ~...~a_n]$, then there exists some $c_1,...,c_n \in \mathbb{R}$ such that $v=c_1a_1+...+c_na_n$, then:
$$Av=A(c_1a_1+...+c_na_n)=A(A[c_1,...,c_n])=(A \cdot A)[c_1,...,c_n]=A^2[c_1,...,c_n]=A[c_1,...c_n]=v$$
So $Av=v$, which implies $\lambda=1$.
Is that correct?
Thanks!
 A: Your proof is quite correct but you could shorten it by calling the unknown vector as $w$. As $v$ is in the column space, there exists a vector $w$ such that $Aw = v$. As you did, just apply $A$ on both sides to obtain $v = Av$ which shows your claim.
Note that every eigenvector always lies in the column space so you have proved a double implication in this case.
You can also prove a weaker existence statement in this setup where $A≠0$ has eigenvectors of eigenvalue $1$.
First, we have $A = A^2$ which rearranges to give $A(A - I) = 0$ where $I$ is the identity matrix and $0$ denotes the zero matrix of appropriate size.
Taking determinants on both sides and using the multiplicative property, we get $\det(A)\det(A-I) = 0$.
If $\det(A-I)$ is non-zero, then it has an inverse. We multiply by the inverse on both sides $$A(A-I)(A-I)^{-1} = 0$$ and this gives us that $A$ is the zero matrix. Bummer.
Thus, $\det(A-I)$ must be zero. This implies that $A-I$ does not have full rank (not injective) and there must exist an $n$-dimensional vector $v$ such that it maps to zero (in other words, lies in $\ker A$). We now have $(A-I)v = 0$ and thus $$Av = v.$$ Note that, a vector lies in $Col(A)$ if its in the image of $A$. As $v$ is clearly of the form of a vector that comes out of action under $A$, it lies in $Col(A)$, thus completing our proof.
As an extra, you can show that if $A$ is not the identity matrix, then $\det(A)$ must be zero. This will show that $A$ is a projection onto a smaller subspace.
A: Hint: We are given that $A(Au) = A^{2}u = Au$ for all $u$ in $\mathbf{R}^{n}$.

 If $v = Au$ is non-zero for some $u$, then....

