Finding standard matrix and proving questions

Question

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?

Answer 1

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.

Answer 2

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?