Change of basis for linear transformation - Linear algebra

Question

So I'm having a lot of difficulties with change of basis. Watched tons of tutorials on youtube but they only seem to confuse me more.

Let $T: \mathbb{R^2} \to \mathbb{R^2}$ be defined by $T(a,b) = (a + 2b, 3a - b)$. Let $\mathcal{B} = \{(1,1),(1,0)\}$ and $\mathcal{C} = > \{(4,7),(4,8)\}$. Find $[T]_\mathcal{B}$ and $[T]_\mathcal{C}$ and show that $[T]_\mathcal{C} = Q^{-1} \cdot [T]_\mathcal{B}\cdot Q$ for some invertible matrix $Q$.

So I've done some thinking, and the matrix $Q$ might be the matrix that goes from basis $c$ to $b$? and then if I invert that one I get matrix $Q^{-1}$? I just have 0 idea where to start to problem from, like what's the first matrix that I have to work with?

Do I start with the basic basis in $R^2$ that is $\{(1,0),(0,1)\}$ and apply the transformation to it?

Answer 1

Take a look here

How do I express ordered bases for polynomials as a matrices? Linear Algebra.

In your case you know the matrix for the canonical basis:

$$T_A=\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$$

And two other basis:

$$\mathcal{B} = \{(1,1),(1,0)\}$$ and

$$\mathcal{C} = \{(4,7),(4,8)\}$$

and have to find the matrices $T_{B}$ and $T_{C}$ with respect these two basis.

You can follow the previous example to do it.

Let's see for example how to find out $T_B$.

Let be $w$ any vector given in the canonical basis and indicate with: $v_1=(1,1)$ and $v_2=(1,0)$ the two vectors of the basis $\mathcal{B}$, also given with respect to the canonical basis.

We are loking for the coefficient $x_1$ and $x_2$ such that:

$$w=x_1\cdot v_1+x_2\cdot v_2$$

or in matrix form:

$$w=V\cdot x$$

$$V=\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$$

Note that $V$ contains as column the vectors of the new basis with respect to the canonical basis. It is important to note that $V$ represent the matrix of change of basis from $\mathcal{B}$ to the canonical.

Thus, the components of any vector $w$ with respect to the new basis are given by:

$$x=V^{-1}\cdot w$$

$$V^{-1}=\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$$

Thus $V^{-1}$ represent the matrix of change of basis from the canonical to the new basis $\mathcal{B}$ .

The mapping of $w$ with respect to the canonical basis is given by:

$$u=T_A\cdot w$$

Since we are looking for the matrix $T_B$ that represent $T$ with respect to the new basis $\mathcal{B}$, let's plug in the change of basis:

$u=V \cdot y$

$w=V \cdot x$

to obtain:

$$u=T_A\cdot w \iff V \cdot y=T_A\cdot V \cdot x \iff y=V^{-1}T_A\cdot V \cdot x=T_B\cdot x$$

Thus the matrix :

$$T_B=V^{-1}\cdot T_A\cdot V$$

represent the transformation with respect to the new basis $\mathcal{B}$.

For $T_C$ you can proceed in the same manner finding: $$T_C=W^{-1}\cdot T_A\cdot W$$ Now since $$T_B=V^{-1} \cdot T_A\cdot V \implies T_A=V \cdot T_B\cdot V^{-1}$$ you conclude $$T_C=W^{-1} \cdot T_A\cdot W=W^{-1} \cdot V \cdot T_B\cdot V^{-1}\cdot W$$