Change of Basis: finding $A$ for $T(x)=Ax$

Question

So my basis is the column vectors of \begin{pmatrix}a&0&0\\ \:b&a&0\\ \:c&b&a\end{pmatrix}

I was taught the relationship $\text{A = Q$^{-1}$[T]$_\text B$Q}$ and I think $Q^{-1}$ should be equal to my basis and then $Q$ will be the inverse of that. How do I find $[\text T]_\text B$ if $T(x)=Ax$?

Answer 1

To answer the question in full (as discussed in the comments):

Consider $A \in \mathrm{Mat}_{3 \times 3}(\Bbb R)$ and a vector $v \in \Bbb R^3$ such that $A^3 v = 0$ but $A^2 v \neq 0$.

1. Show that the vectors $A^2v$, $Av$, and $v$ form a basis of $\Bbb R^3$ Hint: It is sufficient to show they are linearly independent. Set up the proper equation, and multiply by $A^2$.

2. Find the matrix of the transformation $T(x) =Ax$ with respect to the basis $\{A^2v,Av,v\}$.

To answer question 1 (part a), recall the definition of linear independence. If $c_1 A^2v + c_2 Av + c_3 v = 0$, then it is supposed to hold that $c_1,c_2,c_3$ are all necessarily zero. So, suppose that $c_1,c_2,c_3$ are such that $c_1 A^2v + c_2 Av + c_3 v = 0$. Note that $$ c_1 A^2v + c_2 Av + c_3 v = 0 \implies \\ A(c_1 A^2v + c_2 Av + c_3 v) = A(0) \implies\\ c_2A^2 v + c_3 Av = 0 \implies\\ A(c_2A^2 v + c_3 Av) = A(0) \implies\\ c_3A^2 v = 0 $$ Now, by considering the first, third, and fifth lines above, conclude that $c_1,c_2,c_3$ are all necessarily zero.

To answer question 2 (part b), we can use the following procedure to find $[T]_{\mathcal B}$. If $\mathcal B = \{v_1,v_2,v_3\}$ and $M = [T]_{\mathcal B}$, then the elements of the $j$th column of $M$ are the numbers $m_{1j},m_{2j},m_{3j}$ for which $$ T(v_j) = m_{1j}v_1 + m_{2j}v_2 + m_{3j}v_3. $$ So for example: to find the second column of $[T]_{\mathcal B}$ for our $T$ and $\mathcal B = \{A^2 v, Av, v\}$, note that $$ T(Av) = A(Av) = A^2 v = 1 \cdot A^2 v + 0 \cdot Av + 0 \cdot v. $$ So, the second column of our matrix is $(1,0,0)$. In other words, we have $$ [T]_{\mathcal B} = \pmatrix{?&1&?\\?&0&?\\?&0&?}. $$