Can we represent matrix $\ A_{2x2}$ like this $\ A=\lambda_1R_1+\lambda_2R_2$ Let $\ A\in{M_{2x2}}$ with $\ \lambda_1,\lambda_2,\ \lambda_1 \neq \lambda_2$ as eigenvalues
Prove that matrix$\ A$ can be represented like: $\ A_{2x2}$ like this $\ A=\lambda_1R_1+\lambda_2R_2$ 
We know that $\ R_1^2=R_1,\ R_2^2=R_2,\ R_1+R_2=I,\ R_1R_2=0$
I have tried finding some relation between matrices $\ R_1,R_2$ by taking two matrices $\ R_1=\pmatrix{a&b\\c&d}$ and $\ R_2=\pmatrix{x&y\\z&w}$ and solving system of equations that we get from their relations that are given to us in the problem statement but that wasn't very useful to me.
 A: Edit: the previous answer contained a mistake: the assumption that the eigenvectors are orthogonal, which they don't have to be for this to work. Thanks to Joppy for pointing it out!
OK, so the correct things to say are the following: let $v_1,v_2$ be the eigenvectors corresponding to $\lambda_1,\lambda_2$ respectively. Then they are linearly independent, since everything in the span of $v_1$ has eigenvalue $\lambda_1\neq \lambda_2$ and vice versa. Thus every vector admits a unique decomposition $v=\alpha_1 v_1 + \alpha_2v_2$.
Now consider the operators $R_i$ which project such a vector $v$ to $\alpha_iv_i$. Then clearly $R_i^2=R_i$ since once you're in the span of $v_i$, $R_i$ does nothing. Moreover, $R_1+R_2=I$ by construction, and finally, composing $R_1R_2$ is the zero operator, since once you project on the span of $v_2$, your $v_1$-component is $0$. 
Finally, $A=\lambda_1R_1 + \lambda_2R_2$. This uses the principle that 'a matrix is what a matrix does' (which is pretty useful in linear algebra in general). Namely, to show that $A$ equals this other matrix, it suffices to show that this other matrix acts like $A$ on the two independent eigenvectors, which is straightforward.
A: You’ve described a pair of projections. The idea behind this decomposition is that since $A$ is diagonalizable, you can decompose $\mathbb R^2$ into the direct sum of the eigenspaces $E_1$ and $E_2$ of $A$ that correspond to the eigenvalues $\lambda_1$ and $\lambda_2$. This lets you decompose $A$ into a linear combination of projections onto those subspaces.  
We know that we can write any vector $\mathbf v\in\mathbb R^2$ as $\mathbf v_1+\mathbf v_2$, where $\mathbf v_1\in E_1$ and $\mathbf v_2\in E_2$. By the definition of eigenvector, $A\mathbf v=\lambda_1\mathbf v_1+\lambda_2\mathbf v_2$. Let $R_1$ and $R_2$ be linear maps (as yet unknown) such that $R_1\mathbf v=\mathbf v_1$ and $R_2\mathbf v=\mathbf v_2$. It should be clear that $R_1+R_2=I$. Then $A\mathbf v=\lambda_1R_1\mathbf v+\lambda_2R_2\mathbf v$, i.e., $A=\lambda_1R_1+\lambda_2R_2$. This gives us the decomposition of $A$ that we wanted.  
Suppose that $\mathbf v\in E_2$. Then its component $\mathbf v_1$ in the above direct-sum decomposition is zero, which means that $R_1\mathbf v=0$. But for any vector $\mathbf v$, $R_2\mathbf v\in E_2$, therefore $R_1R_2=0$. By symmetry, we also have $R_2R_1=0$.  
Now examine $A^2\mathbf v$. We have $$\begin{align}A^2\mathbf v&=(\lambda_1R_1+\lambda_2R_2)^2\mathbf v\\&=\lambda_1^2R_1^2\mathbf v+\lambda_1\lambda_2(R_1R_2+R_2R_1)\mathbf v+\lambda_2^2R_2^2\mathbf v\\&=\lambda_1^2R_1^2\mathbf v+\lambda_2^2R_2^2\mathbf v\end{align}$$ but we also know that $$A^2\mathbf v=A^2(\mathbf v_1+\mathbf v_2)=\lambda_1^2\mathbf v_1+\lambda_2^2\mathbf v_2=\lambda_1^2R_1\mathbf v+\lambda_2^2R_2\mathbf v.$$ Comparing these two expansions, we can see that one way to make them equal is to require that $R_1^2=R_1$, $R_2^2=R_2$. These conditions say that these maps are projections and by the way we’ve defined them, they’re projections onto the eigenspaces $E_1$ and $E_2$, respectively.  
We now have enough to construct $R_1$ and $R_2$ explicitly. Recall that by definition $E_1=\ker(A-\lambda_1I)$ and $E_2=\ker(A-\lambda_2I)$ so these seem like possible starting points for our two projections. We have $$\begin{align}(A-\lambda_1I)\mathbf v&=A\mathbf v-\lambda_1\mathbf v\\&=(\lambda_1\mathbf v_1+\lambda_2\mathbf v_2)-(\lambda_1\mathbf v_1+\lambda_1\mathbf v_2)\\&=(\lambda_2-\lambda_1)\mathbf v_2\in E_2\end{align}$$ so the kernel and image of this map match what we want for $R_2$. To end up with $\mathbf v_2$, divide by $\lambda_2-\lambda_1$, i.e., $$R_2={A-\lambda_1I\over\lambda_2-\lambda_1}$$ and again by symmetry, $$R_1={A-\lambda_2I\over\lambda_1-\lambda_2}.$$ To check that these are in fact projections, we compute $$\begin{align}(A-\lambda_1)^2\mathbf v&=(A-2\lambda_1A+\lambda_1^2I)\mathbf v\\&=(\lambda_1^2\mathbf v_1+\lambda_2^2\mathbf v_2)-2(\lambda_1^2\mathbf v_1+\lambda_1\lambda_2\mathbf v_2)+(\lambda_1^2\mathbf v_1+\lambda_1^2\mathbf v_2)\\&=(\lambda_2^2-2\lambda_1\lambda_2+\lambda_2^2)\mathbf v_2\\&=(\lambda_2-\lambda_1)^2\mathbf v_2\\&=(\lambda_2-\lambda_1)^2R_2\mathbf v\end{align}$$ therefore $R_2^2=R_2$. I’ll leave verifying the rest to you.  
Going back to the expansion of $A^2$ shows why this decomposition is useful. It’s easy to see from it that $A^n=\lambda_1^nR_1+\lambda_2^nR_2$. (You can also show this directly by applying the Binomal theorem and noting that all mixed terms vanish.) Thus, we have a simple way to compute powers of $A$ without ever having to find an eigenvector or perform any matrix multiplications, which we’d need to do if computing powers by diagonalizing $A$ instead. It turns out that, at least for $2\times2$ matrices, there are similar short cuts for computing powers (and exponentials) of these matrices for the cases of repeated eigenvalues and complex eigenvalues as well.
