1
$\begingroup$

Problem: Let $L: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map with \begin{align*} [L]_{\alpha}^{\beta} = \begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix} \end{align*} as the matrix representation with respect to the standard bases $\alpha$ for $\mathbb{R}^2$ and $\beta$ for $\mathbb{R}^3$. Now find a basis $\mathcal{V}$ for $\mathbb{R}^2$ and a basis $\mathcal{W}$ for $\mathbb{R}^3$ such that \begin{align*} [L]_{\mathcal{V}}^{\mathcal{W}} = \begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix} \end{align*} where each $O$ represents a block matrix with all zeroes and/or does not appear.

Attempt at solution: I'm not sure if I understand what's being asked here. Since the matrix $[L]_{\alpha}^{\beta}$ is with respect to the standard bases, we have $L(1,0) = (2,4,6)$ and $L(0,1) = (3,6,9)$. From this I determined the general formula of $L$ as $L(x,y) = (2x + 3y, 4x + 6y, 6x + 9y)$.

Now, I assume the condition $\begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix}$ means we want a matrix representation of the form $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$. Let $\mathcal{V} = \left\{v_1, v_2\right\}$ and $\mathcal{W} = \left\{w_1, w_2, w_3\right\}$ be the other two bases we seek. So we want \begin{align*} L(v_1) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 1 w_1 + 0 w_2 + 0 w_3 \qquad \text{and} \qquad L(v_2) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 0 w_1 + 1 w_2 + 0 w_3 \end{align*} Now I'm stuck, and I don't know how to find a concrete example of $\mathcal{V}$ and $\mathcal{W}$ that would fit with the explicit formula for $L$ I found earlier.

$\endgroup$
  • 1
    $\begingroup$ actually I think what you want is not possible, since your matrix is of rang 1...so you won't be able to get to \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} but you can get to \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} $\endgroup$ – user190080 Jul 8 '15 at 21:07
  • $\begingroup$ as a small guideline: first try to find a basis in $\mathbb{R}^2$ such that the matrix looks like \begin{pmatrix} 2 & 0 \\ 4 & 0 \\ 6 & 0 \end{pmatrix} and then modify the basis in $\mathbb{R}^3$ $\endgroup$ – user190080 Jul 8 '15 at 21:49
  • $\begingroup$ Can you post what you have in mind as an answer please? I tried what you said but I can't do it still. $\endgroup$ – Kamil Jul 8 '15 at 21:54
  • $\begingroup$ Ok, done :) I skipped the computation but I am sure you can carry them out on your own, I hope the idea of choosing the basis' is getting through? Of course you can find other pairings of basis, but this seems to me the most straight forward way $\endgroup$ – user190080 Jul 8 '15 at 22:18
1
$\begingroup$

so as I mentioned in the comment, we first choose a proper basis $B$ in $\mathbb{R}^2$, we take $$ B=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1.5 \\ 1 \end{pmatrix}\right\} $$ and first keep the standard basis $A$ in $\mathbb{R}^3$, therefore our matrix looks like $$ L^{B}_A=\begin{pmatrix} 2 & 0 \\ 4 & 0 \\ 6 & 0 \end{pmatrix} $$ Next thing we do is to find the proper basis $C$ in $\mathbb{R}^3$, this is rather obvious and we take $$ C=\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\} $$ and therefore our linear maps L looks with respect to the basis $B$ and $C$ $$ L^B_C=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} $$ and that's it. By the way, the choice of $ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ is arbitrary, you just need to fill up $\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}\right\}$ to a basis of $\mathbb{R}^3$. The complete line of computation looks like $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}= \begin{pmatrix} 2 & 0 & 0\\ 4 & 1 & 0\\ 6 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix} \begin{pmatrix} 1 & -1.5 \\ 0 & 1 \end{pmatrix} $$ bests

$\endgroup$
  • $\begingroup$ Are you aware of \left and \right? They make braces / brackets / parentheses automatically adjust to their content, rather than hand-tuning them with \big etc. $\endgroup$ – joriki Jul 9 '15 at 4:16
  • 1
    $\begingroup$ @joriki I do now! Thanks, it indeed is much more convenient and looks so much better - crazy, I've never heard of that before... $\endgroup$ – user190080 Jul 9 '15 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.