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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.

3 Answers3

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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 $M(T)$.

Arthur
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  • Thanks for your input. Does this happen because $B_2=((1,−1),(1,1))$ is already expressed in terms of $B_1$? –  Jun 17 '18 at 20:14
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    @JoshNg Yes and no. Yes, the vectors $(1,-1)$ and $(1,1)$ are expressed in $B_1$, and at some point you will need to translate into $B_2$. However, it is in $B_1$ you know the expression for $T$. So you have to start with $B_2$ expressed in $B_1$ and then apply $T$ to those if you are ever going to get anywhere to begin with. $B_2$ expressed in $B_2$ is the same as $B_1$ expressed in $B_1$, and this is the same for any basis (that's kindof the point of a basis). – Arthur Jun 18 '18 at 05:48
  • Thanks. Using your suggestion, I've found the correct expression of $M(T)$ in $B_2$ is $M(T) = \begin{bmatrix} -3 & 1 \ 2 & 2\ \end{bmatrix}$, which gives $M(T).v = \begin{bmatrix} 4 \ 4\ \end{bmatrix}$ (with $v = (-1/2,5/2)$), which is really $\begin{bmatrix} 8 \ 0\ \end{bmatrix}$ in terms of $B_1$. The remaining issue is that this still does not agree with the value I get when computing $T(-1/2,5/2)$ directly. Do I need to also translate $T(x,y) = (x+2y,3x-2y)$ into $B_2$? –  Jun 18 '18 at 10:54
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    @JoshNg Yes, the description of $T$ as $T(x,y) = (x+2y,3x-2y)$ relies heaviliy on the fact that we're in basis $B_1$, since we're using the actual coordinates directly. In basis $B_2$, the corresponding description of $T$ is $T(x, y) = (-3x+y, 2x+2y)$. Geometrically, what we mean when we say $T(x,y) = (x+2y,3x-2y)$ is that if $b_1, b_2$ are our chosen basis vectors, then $T(x\cdot b_1 + y\cdot b_2) = (x+2y)b_1 + (3x-2y)b_2$. If you're using different vectors than $b_1$ and $b_2$ to span your space, then the resulting coefficients will (almost certainly) be different. – Arthur Jun 18 '18 at 11:00
  • Could you please elaborate the method? –  Jun 18 '18 at 11:10
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    @JoshNg The method for what? The expression of $T(x, y)$ in basis $B_2$? There I'm just transcribing from the matrix representation of $T$ in basis $B_2$, the same way (or rather opposite way) of how you found the $B_1$-matrix representation of $B$ in the first place. The second half of my comment is just translating the actual geometric meaning of coordinates with respect to a basis. – Arthur Jun 18 '18 at 11:13
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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: $$P^{-1}=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix}$$.

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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).

Aravind
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