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I'm trying to understand why is the following statement is true.

$\newcommand\V[3]{\begin{bmatrix}#1\\ #2\\ #3\end{bmatrix}}$ If $T\V100 = \V{a_1}{a_2}{a_3}$, $T\V110 = \V{b_1}{b_2}{b_3}$, and $T\V101 = \V{c_1}{c_2}{c_3}$, then $A = \begin{bmatrix}a_1 & b_1-a_1 & c_1-a_1 \\ a_2 & b_2-a_2 & c_2-a_2 \\ a_3 & b_3-a_3 & c_3-a_3\end{bmatrix}$.

Can someone help me please?

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  • $\begingroup$ It's not true. Did you mean $T$ instead of $A$? $\endgroup$ – user14972 Oct 29 '12 at 8:07
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    $\begingroup$ @Hurkyl I think the statement is fine (assuming $A$ is the standard matrix of $T$). $\endgroup$ – EuYu Oct 29 '12 at 8:08
  • $\begingroup$ @EuYu Yeah but why? $\endgroup$ – SyndicatorBBB Oct 29 '12 at 8:09
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Remember the key fact that if $A$ is a matrix, then $A\mathbf{e_i}$ extracts the $i$th column of $A$, where $\mathbf{e_i}$ is the $i$th standard basis vector.

If we let $A$ denote the standard matrix of the mapping $T$, then $A\mathbf{e_i} = T(\mathbf{e_i})$. What can you then say about the columns of $A$?

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  • $\begingroup$ They span ImT.. $\endgroup$ – SyndicatorBBB Oct 29 '12 at 8:13
  • $\begingroup$ You can say a lot more than that. $\endgroup$ – EuYu Oct 29 '12 at 8:19
  • $\begingroup$ Yeah they are also linear independence. $\endgroup$ – SyndicatorBBB Oct 29 '12 at 8:27
  • $\begingroup$ @Guy You can't conclude that they're linearly independent. The $i$th column of $A$ is precisely the image of $\mathbf{e_i}$ under $T$. There is nothing constraining the images of the standard basis vectors to be linearly independent. $\endgroup$ – EuYu Oct 29 '12 at 8:28
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    $\begingroup$ I was only looking for the fact that the columns of $A$ are the images of the standard basis vectors under $T$. After all, this is the fact that allows you to find the standard matrix given $T$. So if you can find what $T(\mathbf{e_i})$ is for each $i$, then you automatically have your $A$. $\endgroup$ – EuYu Oct 29 '12 at 8:32
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Because matrix $A$ is with respect to the canonical basis $\{(1,0,0),(0,1,0),(0,0,1)\}$. Notice to get $T(0,1,0)$, you need $T(1,1,0) - T(1,0,0) = (b_1,b_2,b_3) - (a_1,a_2,a_3) = (b_1 - a_1, b_2 - a_2, a_3 - b_3) = T(0,1,0)$. And this is your second column. Similarly, you obtain $T(0,0,1)$, which will be your third column.

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  • $\begingroup$ Very nicely put ! $\endgroup$ – SyndicatorBBB Oct 29 '12 at 8:10
  • $\begingroup$ I accept both answers....Thank you! $\endgroup$ – SyndicatorBBB Oct 29 '12 at 8:39

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