Prove $(\lambda I-A)^{-1}=\sum_{i} \Big(\frac{v_iu_i}{\lambda-\lambda_i}\Big)$. $u_i, v_i$ are left and right eigenvectors for eigenvalue $\lambda_i$ If a real matrix $A$ has distinct eigenvalues $\lambda_i$, and $u_i$ and $v_i$ are left and right eigenvectors corresponding to eigenvalue $\lambda_i$, then prove
$(\lambda I-A)^{-1}=\sum_{i} \Big(\frac{v_iu_i}{\lambda-\lambda_i}\Big)$.
$v_i$ is right eigenvector $\Rightarrow$ $Av_i=\lambda_iv_i$, whereas $u_i$ is left eigenvector $\Rightarrow$ $u_iA=\lambda_iu_i$. 
Taking transpose $(u_iA)^T=A^Tu_i^T=\lambda_iu_i^T$. Therfore, $u_i^T$ is the right eigenvector of $A^T$ with same eigenvalue $\lambda_i$.
Also, $|\lambda I -A|=|(\lambda I -A )^T|=|\lambda I -A^T|$.
I don't know how to proceed further. Any help would be appreciated.  
 A: There's likely a prettier way to do this but what follows should, at least, be straightforward. 
First of all, what do we know about $(\lambda I - A)^{-1}$? Really, just the equation you've supposed and that 
\begin{equation}
(\lambda I - A) (\lambda I - A)^{-1} = I
\end{equation}
So let's use these. 
\begin{align}
(\lambda I - A) (\lambda I - A)^{-1} &= I \ \Rightarrow \\
(\lambda I - A) \sum_{i} \frac{v_i u_i}{\lambda - \lambda_i} &= I \ \Rightarrow \\
\sum_{i} \frac{(\lambda v_i - A v_i) u_i}{\lambda - \lambda_i} &= I \ \Rightarrow \\
\sum_{i} \frac{(\lambda - \lambda_i ) v_i u_i}{\lambda - \lambda_i} &= I \ \Rightarrow \\
\sum_{i} v_i u_i &= I 
\end{align}
Now, all we have to do is prove that $\sum_{i} v_i u_i = I$. To do this, we note that since the $\lambda_i$ are distinct, $u_i , v_i$ span $\mathbb{R}^n$. So we can write any vector in $\mathbb{R}^n$ as a linear combination of $\left\{v_i\right\}$. Hence, it suffices to show that 
\begin{equation}
\left( \sum_{i} v_i u_i \right) v_j = v_j 
\end{equation}
and 
\begin{equation}
\left( \sum_{i} v_i u_i \right) v_j =  \sum_{i} v_i \left(u_i \cdot v_j \right) 
\end{equation}
now, to examine $\left(u_i \cdot v_j \right)$, we look at 
\begin{align}
A v_j &=  \lambda_j v_j  \ \Rightarrow \\
u_i \left( A v_j \right) &=  \lambda_j \left( u_i \cdot v_j \right)  \ \Rightarrow \\
\left(u_i  A\right) v_j  &=  \lambda_j \left( u_i \cdot v_j \right)  \ \Rightarrow \\
\left(\lambda_i u_i \right) v_j  &=  \lambda_j \left( u_i \cdot v_j \right)  \ \Rightarrow \\
0  &=  \left[ \lambda_j - \lambda_i \right]\left( u_i \cdot v_j \right)  
\end{align}
 and since $\lambda_j \neq \lambda_i$ for $j \neq i$, we have that $u_i \cdot v_j = 0$ for $i \neq j$, and we can pick, say, $v_j$ so that $u_i \cdot v_i = 1$. Hence, we have that 
\begin{equation}
\left( \sum_{i} v_i u_i \right) v_j = v_j 
\end{equation}
and we are done!
A: Since $A$ has distinct eigenvalues, it is diagonalizable, i.e., $A = Q \Lambda Q^{-1}$. Thus, we can write
$$
(\lambda I - A)^{-1} = (\lambda I - Q \Lambda Q^{-1})^{-1} = Q(\lambda I - \Lambda )^{-1}Q^{-1}. 
$$
One can also show that the $i$'th column of $Q$ and the $i$'th row of $Q^{-1}$ correspond to the right eigenvector and left eigenvector of the $i$'th eigenvalue of $A$, respectively. Therefore, we get
$$
\left[
\begin{array}{cc}
v_1 & v_2 & \cdots & v_n \end{array}
\right] 
\text{diag} \Big(\frac{1}{\lambda - \lambda_1}, \ \cdots \ ,\frac{1}{\lambda - \lambda_n} \Big)
\left[
\begin{array}{cc}
u_1 \\
\vdots \\
u_n
\end{array}
\right] =
\sum_i \Big( \frac{1}{\lambda - \lambda_i} \Big) v_iu_i.
$$
The last equality is sometimes called a dyadic expansion of the matrix product, and can be easily proven (see here).
