Find the eigenvectors and eigenvalues of A geometrically $$A=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}=\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix} \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}.$$
Scale by 2 in the $x$- direction, then scale by 2 in the $y$- direction, then projection onto the line $y = x$.
I am confused with question since it is not a textbook like question. I don't know why $A$ equals to 3 matrices. 
You could ignore the word geometrically for the sake of easiness
thanks

The geometry is what makes things easier (for me). Without the geometry, it would be a mechanical computation which I would not like doing, and might get wrong.
Note that the vector $(1,1)$ gets scaled by our two scalings to $(2,2)$, and projection on $y=x$ leaves it at $(2,2)$. So the vector $(1,1)$ is an eigenvector with eigenvalue $2$.
Now consider the vector $(-1,1)$. The two scalings send it to $(-2,2)$. Projection onto $y=x$ gives us $(0,0)$. So $(-1,1)$ is an eigenvector with eigenvalue $0$.
