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I have a linear transformation: $$T_1:\Bbb R^2 \rightarrow \Bbb R^2, T_1(a_1,a_2) = (a_1+a_2, 2a_1+4a_2)$$ The following ordered bases of $\Bbb R^2$ are given as such: $ \beta = \{(1,2),(0,1)\}$ and $\gamma$ is the standard basis of $\Bbb R^2$.

I need to compute
$[T_1]_\beta^\gamma$


I'm doing this:
$T_1(1,2) = (3,6)$ // $\beta$, row 1
$T_1(0,1) = (1,4)$ // $\beta$, row 2
So $[T_1]_\beta=\{(3,6),(1,4)\}$ ... right?
... Now what? How do I "plug" that into $\gamma$ (or plug $\gamma$ into that?)

I keep thinking I understand it, then I do all the computations and theorems aren't being proven. Ultimately, this problem has the following parts; I need to compute a $[T_2]_\alpha^\beta$, then show that $[T_1T_2]_\alpha^\gamma = [T_1]_\beta^\gamma[T_2]_\alpha^\beta$. My book shows one example of a transformation with respect to a basis, but nothing about when bases change -- it has sums and proofs and whatnot, but I'm finding that more confusing than anything. I can't find any step-by-step example to understand what exactly it's saying.

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  • $\begingroup$ Do you have the correct answer in your book? $\endgroup$ – Bruno Reis Oct 1 '16 at 3:05
  • $\begingroup$ No I don't; it's not assigned from the book, just a downloaded .pdf page. $\endgroup$ – Asinine Oct 1 '16 at 3:09
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    $\begingroup$ Not sure if I understood the question correctly. Write $(3,6)=3(1,0)+6(0,1)$ and $(1,4)=1(1,0)+4(0,1)$. So, what you have written as $[T]_{\beta}$ is actually $[T]_{\beta}^{\gamma}$. $\endgroup$ – user276115 Oct 1 '16 at 3:13
  • $\begingroup$ In that case, is it just matrix multiplication, i.e. multiplying $[T]_\beta$ and $\gamma$? $\endgroup$ – Asinine Oct 1 '16 at 3:28
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    $\begingroup$ You might want to read: math.stackexchange.com/a/1799809/9464 and math.stackexchange.com/q/53525/9464 $\endgroup$ – Jack Oct 1 '16 at 21:12
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That's the way I would do it. Someone please correct me if I do something incorrect. I'm still studying Linear Algebra.

$[T1]_{b}^{\gamma}$ It's a linear transformation that get some vector in the standard basis ($\gamma$) for $R^2$, apply the transformation, and then return it in the basis $b$.

Starting with that, we can affirm that: $$ [T1]_{b}^{\gamma} = [I]_{b}^{\gamma}\cdot [T1]_{\gamma}^{\gamma} $$ $[I]_{b}^{\gamma}$ is a change of basis from $\gamma$ to $b$.

Now, all we need to do is get the matrix that represents $[I]_{b}^{\gamma}$ and multiply to the matrix of the transformation $[T1]_{\gamma}^{\gamma}$.

The column vectors for the matrix of transformation $[T1]_{\gamma}^{\gamma}$ is going to be the transformed vectors that compose the basis $\gamma$: $$ (1,0) \rightarrow (1+0,2\cdot 1 + 4\cdot 0) = (1,2)\\ (0,1) \rightarrow (0+1,2\cdot 0 + 4\cdot 1) = (0,4)\\ \left[\begin{matrix} 1 & 0 \\ 2 & 4 \end{matrix}\right] $$ Now for the matrix of change of basis $\gamma$ to $b$, we need to decompose $\gamma$ vectors in terms of $b$ vectors: $$ (1,0) = a(1,2)+b(0,1) \rightarrow a = 1 , b = -2\\ (0,1) = c(1,2)+d(0,1) \rightarrow c = 0 , d = 1\\ \left[\begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix}\right] $$

Now multiplying the matrices, we get:

$$ \left[\begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix}\right] \cdot \left[\begin{matrix} 1 & 0 \\ 2 & 4 \end{matrix}\right] = \left[\begin{matrix} 1 & 0 \\ 0 & 4 \end{matrix}\right] $$

Now, to check if everything is correct, let's pick a random vector from $R^2$ describe it in terms of $\gamma$: $$ V_{\gamma} = (a,b) $$ Then we apply the transformation $[T1]_{b}^{\gamma}$: $$ \left[\begin{matrix} 1 & 0 \\ 0 & 4 \end{matrix}\right]\cdot \left[\begin{matrix} a \\ b \end{matrix}\right] = \left[\begin{matrix} a \\ 4b \end{matrix}\right] $$ Now we just need to check if we apply $[T1]_{\gamma}^{\gamma}$ and then apply $[I]_{b}^{\gamma}$ is going to give us the same result: $$ \left[\begin{matrix} 1 & 2 \\ 0 & 4 \end{matrix}\right] \cdot \left[\begin{matrix} a \\ b \end{matrix}\right] = \left[\begin{matrix} a \\ 2a + 4b \end{matrix}\right]\\ \left[\begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix}\right] \cdot \left[\begin{matrix} a \\ 2a + 4b \end{matrix}\right] = \left[\begin{matrix} a \\ 4b \end{matrix}\right] $$

So it's indeed correct.

I really don't know if I was clear enough, of if it's 100% correct. I hope to have contributed a little with your learning proccess. Feel free to leave a comment if you have had any doubts on what I've done...

Let's share some knowledge :) Thanks!

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    $\begingroup$ Thank you very much! I'm going to have to study this a bit. $\endgroup$ – Asinine Oct 1 '16 at 21:43
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    $\begingroup$ Try to watch this my friend: youtube.com/watch?v=P2LTAUO1TdA Good luck with your studies... $\endgroup$ – Bruno Reis Oct 1 '16 at 22:56
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The next class after I posted this, my professor discussed (and gave) the "answer". He'd previously not brought it up when I went to ask him, I would assume because he didn't want to give it away so easily.

The original question read (in full: Linear Algebra 2.2.3, Friedberg): Let $T_1:\Bbb R^2 \rightarrow \Bbb R^2$ be defined by $T_1(a_1,a_2) = (a_1+a_2, 2a_1+4a_2)$. Let $\beta$ be the standard ordered basis for $\Bbb R^2$ and $\gamma = \{(1,1,0),(0,1,1),(2,2,3)\}$. Compute $[T]^\gamma _\beta$. If $\alpha = \{(1,2),(2,3)\}$, compute $[T]^\gamma _\alpha$. I was having issues with the last computation.

You put $\alpha$ into a matrix and set it equal to the computations for $\gamma$: ( $(1, 1, 2)$ and $(-1,0,1)$) then evaluate the matrix for those.

$$ \left[\begin{matrix} 1 & 0 & 2\\ 1 & 1 & 2\\ 0 & 1 & 3\\ \end{matrix}\right] = \left[\begin{matrix} 1\\1\\2 \end{matrix}\right]= \left[\begin{matrix} -1\\0\\1 \end{matrix}\right] $$

Solving for the RREF you get $$ \left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix}\right] = \left[\begin{matrix} -1/3\\0\\2/3 \end{matrix}\right]= \left[\begin{matrix} -1\\1\\0 \end{matrix}\right] $$

This seemed so simple, yet he never bothered mentioning this (at least in an obvious way) in class or in asking during office hours.

So hopefully this can provide some insight to others as well.

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