# 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.

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 $$(\lambda I - A) (\lambda I - A)^{-1} = I$$

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

$$\left( \sum_{i} v_i u_i \right) v_j = v_j$$

and

$$\left( \sum_{i} v_i u_i \right) v_j = \sum_{i} v_i \left(u_i \cdot v_j \right)$$

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

$$\left( \sum_{i} v_i u_i \right) v_j = v_j$$

and we are done!

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).