Let $S$ be the standard basis for $\mathbb{R}^5$. Let $B = (b_1, b_2, b_3, b_4, b_5)$ be the ordered basis with:
$b_1 = (2, 1, 1, -2, -2)$;
$b_2 = (0, -2, 4, 5, -4)$;
$b_3 = (1, -4, 5, 5, -4)$;
$b_4 = (5, -4, 2, 3, 1)$;
$b_5 = (4, 1, 2, -3, -2)$;
I'm having difficulty wrapping my head around the concepts.
How would you write the transition matrix $P_{S,B}=\alpha_{B,S}(\text{id}_{\mathbb{R}^5})$?
Would $P_{S,B}$ simply be the vector $[ (b_1) (b_2) (b_3) (b_4) (b_5) ]$ and $P_{B,S}$ the inverse of that?
Given a linear transformation $$T : \mathbb{R}^5 \to \mathbb{R}^5:T(x_1, x_2, x_3, x_4, x_5)=(x_1 - 2(x_2), x_2 + x_3, x_3 + x_4, x_4, x_5)$$
how would you then calculate the matrix representation $[T]B$ of $T$ with respect to $B$?
Thanks a lot!!
Writing $T$ in matrix form: $$ T x = \left( \begin{matrix} 1 & -2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix} \right) $$ The coordinate tansformation $B$ from standard to $b_i$ base should give $B b_i = e_i$ or $B (b_1, \ldots, b_5) = I$ this gives $B = (b_1, \ldots, b_5)^{-1}$.
So $T$ regarding the base $b_i$ has the matrix $$ B T B^{-1} $$