Basis Transformation Matrix

Question

I have a problem related to transformation matrix. The problem is like that;

Given basis $B$: $b_1 = \begin{pmatrix}1&2\end{pmatrix}^T$ and $b_2 =\begin{pmatrix}2&1\end{pmatrix}^T$ and $A$: $a_1=\begin{pmatrix}1&2\end{pmatrix}^T$, $a_2=\begin{pmatrix}2&7\end{pmatrix}^T$. Find the transformation matrix $T$ from the basis $B$ into basis $A$?

I would appreciate if anyone help me with this.

Answer 1

The solution by provided by @NDewolf. Let's fill out the details.

---

Vectors will be colored according to the basis membership, and named based upon color: $$ \color{blue}{\mathbf{B}\ (standard)}, \qquad \color{red}{\mathbf{R}\ (a)}, \qquad \color{green}{\mathbf{G}\ (b)}. $$

---

$$ \mathbf{T}_{\color{red}{R}\to \color{blue}{B}}= \color{black}{\left[ \begin{array}{r|r} 1 & 2 \\ 2 & 7 \\ \end{array} \right]} $$

The matrix $\mathbf{T}_{\color{red}{B}\to \color{blue}{A}}$ is an operator which maps a $\color{blue}{blue}$ vector to a $\color{red}{red}$ vector.

The inverse matrix is a map which connects vectors in the $\color{blue}{standard}$ basis to vectors in the $\color{red}{\mathbf{R}}$ basis: $$ \mathbf{T}^{-1}_{\color{blue}{B}\to \color{red}{R}} = % \color{blue}{ \frac{1}{3} \left( \begin{array}{rr} 7 & -2 \\ -2 & 1 \\ \end{array} \right)} $$

---

For the green basis $\color{green}{\mathbf{G}}$, the maps are $$ \mathbf{G}_{\color{green}{G}\to \color{blue}{A}}= \color{black}{\left[ \begin{array}{r|r} 1 & 2 \\ 2 & 1 \\ \end{array} \right]}, \qquad % \mathbf{G}^{-1}_{\color{blue}{A}\to \color{green}{G}}= \color{black}{\left[ \begin{array}{rr} -1 & 2 \\ 2 & -1 \\ \end{array} \right]} $$

Transition from red $\color{red}{\mathbf{R}\ (a)}$ to green $\color{green}{\mathbf{G}\ (b)}$

As so succinctly expressed by @Bernard, connect all bases through the hub of the blue basis, $\color{blue}{standard}$. Start with a vector in the $\color{red}{\mathbf{R}}$ basis, map that to a vector in the $\color{blue}{standard}$ basis, then map that to a vector in the $\color{green}{\mathbf{G}}$ basis: $$ \color{red}{ \left[ \begin{array}{c} x_{1} \\ y_{1} \\ \end{array} \right]} % \quad \Longrightarrow \quad % \color{blue}{ \left[ \begin{array}{c} x_{2} \\ y_{2} \\ \end{array} \right]} % \quad \Longrightarrow \quad % \color{green}{ \left[ \begin{array}{c} x_{3} \\ y_{3} \\ \end{array} \right]} % $$

The solution is $$ \begin{align} T_{\color{red}{R}\to\color{green}{G}} &= \color{green}{G^{-1}} \color{red}{R} \\ & = \left[ \begin{array}{rr} -1 & 2 \\ 2 & -1 \\ \end{array} \right] % \left[ \begin{array}{cc} 1 & 2 \\ 2 & 7 \\ \end{array} \right] \\ % &= \left[ \begin{array}{cr} 1 & 4 \\ 0 & -1 \\ \end{array} \right] \end{align} $$

Answer 2

The definition of the transformation matrix is $A=TB$ where $A$ is a vector in the new basis and $B$ the same vector in the old basis. So by taking $A$ and $B$ as the basis vectors, you can see that finding $T$ comes down to finding an expansion of the new basis in terms of the old basis.

Answer 3

The change of basis matrix from basis $\mathscr B$ to basis $\mathscr A$ is the matrix of the identity map from $(\mathbf R^2,\mathscr A)$ to $(\mathbf R^2,\mathscr B)$.

Let $T_B$ be the matrix with column vectors $b_1$ and $b_2$: it's the change of basis matrix from the canonical basis to basis $\mathscr B$. Similarly the matrix $T_A$ with colum,n vectors $a_1ˆ$ and $a_2$ is the change of basis matrix from the canonical basis to basis $\mathscr A$. The commutative diagram: $$\begin{array}{cl}(\mathbf R^2,\textit{Can})&\stackrel{=}{\longleftarrow}(\mathbf R^2,\mathscr B)\\\uparrow\, =&\nearrow\,=\\(\mathbf R^2,\mathscr A)\end{array}$$ shows at once that $$T=T_B^{-1}T_A.$$