Finding standard matrix and proving questions Let $A$ be an $n\times n$ matrix. For each $w∈\Bbb R$, we define a linear transformation $T_w:\Bbb R^n\to\Bbb R^n$ such that $T_w(u) = Au - wu$ for $u∈\Bbb R^n$.
a) Write down the standard matrix for $T_w$.
b) For any $w$, $l∈\Bbb R$, show that
$(A-wI)(A-lI) = (A-lI)(A-wI)$.
c) Suppose $A$ is diagonalizable and the eigenvalues of $A$ are $w_1$, $w_2$, ... , $w_k$. If $v$ is an eigenvector of A, say $Av=w_iv$ for some $i$, show that $(A-w_1I)(A-w_2I)...(A-w_kI)v=0$.
What I have done so far:
By letting $A$ = 
$$ \left[
\begin{array}{cccc}
  a_{11}&a_{12}&\cdots&a_{1n}\\
  a_{21}&a_{22}&\cdots&a_{2n}\\
  \vdots&\vdots&\ddots&\vdots\\
  a_{n1}&a_{n2}&\cdots&a_{nn}
\end{array}
\right] $$, standard matrix = $(A-wI)$. However, I am not quite sure how to proceed from here to parts b and c. Do I prove part b by applying matrix multiplication manually, or is there another way to do it?
 A: Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
A: For b) 
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A \cdot A -l A \cdot I -w I \cdot  A + w l  I \cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A \cdot I = I \cdot A$.
 I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)\dots(A-w_iI) \dots(A-w_kI))v$$ $$=\left( (A-w_1I)(A-w_2I) \dots(A-w_kI)(A-w_iI) \right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
