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
 A: 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 $0$.
A: I will try to give you as much hint as possible without spoiling the answer to the question.
Let's talk about what a matrix-vector multiplication is (or at least how it can be interpreted). A matrix $A$ can be thought of as a linear transformation of vector $u$; the linear transformation could be a rotation, a scaling, a projection, etc... or even a combination of all those. 
So if $A$ is a $2\times2$ matrix and $u$ is a vector in $\mathbb{R}^2$, then $v=Au$ is linear transformation of $u$ that I called $v$.  
For example, if I want to define a scaling operation in the $x$- direction, I could formulate it as follows: if given a vector $z=\begin{bmatrix} x \\ y \end{bmatrix}$, I would like the image of $z$ to be twice as long in the $x$- coordinate or in other words, I would like the image of $z$ (let's call it $w$) to be $w=\begin{bmatrix}2x \\ y\end{bmatrix}$. The question is, can I define a matrix $A_1$ such that $w=A_1z$? The answer is yes, that matrix will be $A_1=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$ (I will let you work out the details, but your book probably thinks about it in terms of image of unitary vectors $e_1, e_2$ etc..). 
The cool thing about matrices and linear transformation is that if I know that the matrix $A_1$ does a certain operation (scaling in the $x$- direction) and $A_2$ does a (rotation) and $A_3$ does a projection, then I can combine all these operations into one by  defining a the product $A_{\text{all}}=A_3\times A_2\times A_1$. So $A_\text{all}$ will combine the operation of $A_1$, then $A_2$ then $A_3$ (read from right to left). 
The above tells you (more or less) what could be thought of as a geometric interpretation of a matrix: it is a linear transformation.  
Now what is an eigenvector of a matrix $B$. Well by definition, if $v$ is an eigenvector of $B$ then $\lambda v = B v$. What this says is that the only geometric operation that $B$ applies to $v$ is a scaling operation. To be more technical, the eigenvectors of $B$ (if they have real components) are the axis that are left unchanged (only scaled) by whatever operation $B$ denotes. 
Now using the fact that a matrix is a linear transformation (that can be interpreted geometrically in your case) and that the eigenvectors are the vectors that are left unchanged (only scaled), can you think about the eigenvectors of your matrix
$A=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}=\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{bmatrix}\times \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\times \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$ will be? 
