I'm given the following here.
Let $\mathcal D =\{(1,1),(1,-1)\}$ be a basis for $\mathbb{R}^2$ and let $\mathcal B$ and $\mathcal C$ be the bases of $\mathbb{R}^2$ satisfying $P_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix} 1 & 1\\ 5 & 2 \end{bmatrix}$ and $P_{\mathcal{D}\leftarrow\mathcal{C}}=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}$
a) If $\mathcal{B}=\{\vec{b_1},\vec{b_2}\}$ find $\vec{b_1}$ and $\vec{b_2}$.
b) Find $P_{\mathcal{B}\leftarrow\mathcal{D}}$ and use your answer to express $3(1,1)-4(1,-1)$ as a linear combination of $\vec{b_1}$ and $\vec{b_2}$.
Here's what I attempted:
a)
$P_{\mathcal{D}\leftarrow\mathcal{B}}=P_{\mathcal{D}\leftarrow\mathcal{C}}P_{\mathcal{C}\leftarrow\mathcal{B}}$
$P_{\mathcal{D}\leftarrow\mathcal{B}}=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}\begin{bmatrix} 1 & 1\\ 5 & 2 \end{bmatrix}$
$P_{\mathcal{D}\leftarrow\mathcal{B}}=\begin{bmatrix} 16 & 7\\ 22 & 10 \end{bmatrix}$
$P_{\mathcal{B}\leftarrow\mathcal{D}}=P^{-1}_{\mathcal{D}\leftarrow\mathcal{B}}$
$P_{\mathcal{B}\leftarrow\mathcal{D}}=\begin{bmatrix} \frac{10}{6} & \frac{-7}{6}\\ \frac{-11}{3} & \frac{8}{3} \end{bmatrix}$
So this implies:
$\mathcal{B}_1=\begin{bmatrix} \frac{10}{6} & \frac{-7}{6}\\ \frac{-11}{3} & \frac{8}{3} \end{bmatrix}\begin{bmatrix}1 \\ 1\\ \end{bmatrix}=\begin{bmatrix}\frac{1}{2} \\ -1\\ \end{bmatrix}$
$\mathcal{B}_2=\begin{bmatrix} \frac{10}{6} & \frac{-7}{6}\\ \frac{-11}{3} & \frac{8}{3} \end{bmatrix}\begin{bmatrix}1 \\ -1\\ \end{bmatrix}=\begin{bmatrix}\frac{17}{6} \\ \frac{-19}{3}\\ \end{bmatrix}$
Is this the correct approach for part a) ?
Also, I am not sure how to approach part b… Can someone lend me a hand on that?
Best Answer
You're close, but not quite for (a). Remember that a transition matrix has as columns the representations of the old basis vectors in the new basis. Let $\mathcal{E}$ be the standard basis, so then $[\vec{d_1}]_\mathcal{E}=\begin{bmatrix} 1 \\ 1\end{bmatrix}_\mathcal{E}$ and $[\vec{d_2}]_\mathcal{E}=\begin{bmatrix} 1 \\ -1\end{bmatrix}_\mathcal{E}$. So, for example, we can construct another transition matrix $\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{D}}=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$. (It's a surprise tool that will help us later.) So to find the representation of $\vec{b_1}$ and $\vec{b_2}$ in the standard basis, i.e. $[\vec{b_1}]_\mathcal{E}$ and $[\vec{b_2}]_\mathcal{E}$, we will need a transition matrix $\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{B}}$. We can read the columns of this matrix to get our $\vec{b_1}$, $\vec{b_2}$ in the standard basis. As you've already demonstrated you know how to do, we have $\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{B}}=\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{D}}\mathcal{P}_{\mathcal{D}\leftarrow\mathcal{C}}\mathcal{P}_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix} 38 & 16 \\ -6 & -3\end{bmatrix}$. We can read off the columns to get $[\vec{b_1}]_\mathcal{E}=\begin{bmatrix} 38 \\ -6\end{bmatrix}_\mathcal{E}$ and $[\vec{b_2}]_\mathcal{E}=\begin{bmatrix} 16 \\ -3\end{bmatrix}_\mathcal{E}$.
With regards to part (b), the matrix $\mathcal{P}_{\mathcal{B}\leftarrow\mathcal{D}}$ takes as input the representation of a vector over the basis $\mathcal{D}$ and spits out the representation of the same vector in the basis $\mathcal{B}$. You are correct as you listed in your solution to part (a) that $\mathcal{P}_{\mathcal{B}\leftarrow\mathcal{D}}=\left(\mathcal{P}_{\mathcal{D}\leftarrow\mathcal{B}}\right)^{-1}$ (it just so happened that was unnecessary for part (a), but we do need it now). You calculated that matrix correctly. So our vector that we want to find the representation of over the basis $\mathcal{B}$ (i.e. as a linear combination of $\vec{b_1}$ and $\vec{b_2}$) is, in the standard basis, $3\begin{bmatrix} 1 \\ 1\end{bmatrix}_\mathcal{E}-4\begin{bmatrix} 1 \\ -1\end{bmatrix}_\mathcal{E}$. But what is this vector over the basis $\mathcal{D}$? Well, since $[\vec{d_1}]_\mathcal{E}=\begin{bmatrix} 1 \\ 1\end{bmatrix}_\mathcal{E}$ and $[\vec{d_2}]_\mathcal{E}=\begin{bmatrix} 1 \\ -1\end{bmatrix}_\mathcal{E}$, we have that $3\begin{bmatrix} 1 \\ 1\end{bmatrix}_\mathcal{E}-4\begin{bmatrix} 1 \\ -1\end{bmatrix}_\mathcal{E}=3\vec{d_1}-4\vec{d_2}$, or in other words, that over the basis $\mathcal{D}$ it has representation $\begin{bmatrix} 3 \\ -4 \end{bmatrix}_\mathcal{D}$. This vector we can send through $\mathcal{P}_{\mathcal{B}\leftarrow\mathcal{D}}$ to yield $\mathcal{P}_{\mathcal{B}\leftarrow\mathcal{D}}\begin{bmatrix} 3 \\ -4 \end{bmatrix}_\mathcal{D}=\begin{bmatrix} \frac{29}{3} \\ -\frac{65}{3} \end{bmatrix}_\mathcal{B}$
To check that this is correct, we can send the representation of this vector over $\mathcal{B}$ back through the transition matrix we made in part (a), that of $\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{B}}$, and see if it matches our representation over $\mathcal{E}$, and indeed $\mathcal{P}_{\mathcal{E}\leftarrow\mathcal{B}} \begin{bmatrix} \frac{29}{3} \\ -\frac{65}{3} \end{bmatrix}_\mathcal{B} = \begin{bmatrix} -1 \\ 7 \end{bmatrix}_\mathcal{E}$, which as you can verify is in fact equal to our original vector.