Why doesn't this example of basis change work?

Question

I'm learning about the change of basis in linear algebra, and trying to come up with an example to understand it. But somehow my example below doesn't make sense.

Let $B_1 = ((1,0),(0,1))$ be the standard basis in $R^2$, and $v = (2,3)$ be a vector.

Let $B_2$ be another basis of $R^2$, $B_2 = ((1,-1),(1,1))$. My understanding is that, in this system, the same $v$ above has the coordinate $v = (-1/2, 5/2)$.

Let $T: R^2 \rightarrow R^2$, $T(x,y) = (x+2y, 3x - 2y)$.

With $B_1$, $M(T) = \begin{bmatrix} 1 & 2 \\ 3 & -2\\ \end{bmatrix}$, and $T(2,3) = M(T).v = \begin{bmatrix} 1 & 2 \\ 3 & -2\\ \end{bmatrix} .\begin{bmatrix} 2 \\ 3\\ \end{bmatrix} =\begin{bmatrix} 8\\ 0\\ \end{bmatrix}$

With $B_2$, $M(T) = \begin{bmatrix} -1 & 3 \\ 5 & 1\\ \end{bmatrix}$, but $T(-1/2,5/2) = (9/2,-13/2)$, and $M(T).v = \begin{bmatrix} -1 & 3 \\ 5 & 1\\ \end{bmatrix} .\begin{bmatrix} -1/2 \\ 5/2\\ \end{bmatrix} =\begin{bmatrix} 8\\ 0\\ \end{bmatrix}$

I don't understand why in the calculations using $B_2$, $T(-1/2,5/2)$ is not equal to $M(T).v$, and why $M(T).v=\begin{bmatrix} 8\\ 0\\ \end{bmatrix}$. I was expecting that $M(T).v$ would give me something different, because we're using another basis. This "something" could then be "converted" back to $\begin{bmatrix} 8\\ 0\\ \end{bmatrix}$ in the standard basis.

Answer 1

As far as I can see, you found $M(T)$ in basis $B_2$ by letting the columns be the images $T(b_1),T(b_2)$ of the basis vectors $b_i$ of $B_2$... as expressed in basis $B_1$. You need to express the vectors $T(b_1),T(b_2)$ in basis $B_2$, and use that as the columns of $M(T)$.

Answer 2

Let $P=\begin{pmatrix}1&1\\-1 &1\end{pmatrix}$. This changes basis from $B_2$ to $B_1$.

Notice $$P^{-1}\begin{pmatrix}1&2\\3&-2\end{pmatrix}P=[T]_{B_2}$$.

I noticed that you forgot to multiply by $P^{-1}$. You need:

$$P^{-1}\begin{pmatrix}-1&3\\5&1\end{pmatrix}$$

I leave this calculation to you...

Note: $$P^{-1}=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix}$$.

Answer 3

What you want is to figure out how $A$ will look like in the new basis. The short answer is that it will be $B^{-1}AB$, where $B=\begin{bmatrix} 1 & 1 \\ -1 & 1\\ \end{bmatrix}$.

Explanation: Let $v=(2,3)^T$ and $x=(-1/2,5/2)^T$.

The vector $x$ of the new basis is the same as $v$ in the old basis, so $v=Bx$. Thus $x=B^{-1}v$. That is: for converting from the new basis $B_2$ to the standard basis $B_1$,you multiply by $B$, and for converting from standard to new basis, you must multiply by $B^{-1}$.

Now, what happens to $A$ in the new basis? Let $u$ be a vector in the standard basis, so that $w=B^{-1}u$ is the corresponding vector in the new basis.

Which matrix $X$ should be applied to $w$ that corresponds to multiplying $u$ by $A$? To figure this out, let's do the following:

1. Move $w$ to the standard basis by applying $B$. [We get $u$]

2. Now apply $A$; [We get $Au$]

3. Now apply $B^{-1}$. [We get the vector in the new basis corresponding to $Au$.]

Thus, the matrix $X$ is $B^{-1}AB$.

Instead, what you've done is to use the matrix $AB$, and then multiplied it with $B^{-1}v$ (the vector in the new basis) and this naturally gives $ABB^{-1}v=Av$, the vector in the standard basis (rather than in the new basis).