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Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a linear operator whose corresponding matrix $A$ (with respect the canonic bases in both the domain and the codomain) is symmetric. I have to show that $f$ is the orthogonal projection on $Im(f)$ if and only if $A^2=A$. Can you help me?

TheWanderer
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Hint: If $A^2=A$ then $A(A-1) \in I(A)$ so the matrix is diagonalizable with eigenvalues $0,1$. Let's now represent this diagonalized matrix:

$$J=\begin{pmatrix} 1 \\ &\ddots \\ &&1 \\ &&&0 \\ &&&&\ddots \\&&&&&0\end{pmatrix}$$

What does this matrix to your vector when you do $J(v), v\in V$

Alberto Andrenucci
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    What do you mean by I(A)? I solved the exercise in this way: if $f$ is the orthogonal projection on $Im(f)$, then $A^2=A$ (since it is a projection). On the other hand, I only must prove that for any $v \in V$, $v-Av \in Ker(f)$. In this way, using the orthogonality of $Ker(f)$ and $Im(f)$ and the fact that $Av \in Im(f)$ for any $v \in V$, I conclude that $<Av,w-Aw>=0$. Is it correct? – TheWanderer Jun 21 '17 at 13:56
  • $I(A)$ is the ideal generated by your matrix. In this matrix you can clearly see that the first $k$ columns, with the number $1$, represent the image.. and the other ones the kernel. What can you conclude? – Alberto Andrenucci Jun 21 '17 at 13:58