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I am watching the linear algebra course of MIT presented by Gilbert Strang and in Lecture 13 he solves the equation $Ax = b$. This is not the usual "find $x$", instead $x$ and $b$ are given and we need to find the matrix $A$ The link is at https://www.youtube.com/watch?v=l88D4r74gtM&t=841s at about minute 14:00. We know:

$Ax = b$

$b=\begin{bmatrix}2\\4\\2\end{bmatrix}$

$x=\begin{bmatrix}2\\0\\0\end{bmatrix}$ + $c\begin{bmatrix}1\\1\\0\end{bmatrix}$ + $d\begin{bmatrix}0\\0\\1\end{bmatrix}$

$\dim N(A)=2$

$A$ is a $3\times 3$ matrix.

But you can see all of that in the video, it looks much clearer on a blackboard.

I am having problems understanding what did professor Gilbert do to get the answer $A=\begin{bmatrix}1&&-1&&0\\2&&-2&&0\\1&&-1&&0\end{bmatrix}$

I just don't seem to understand what happened, at all. How did he come up with the first column? Then how about the other 2? How did he use the nullspace? Why did it work? How did he use $x$ and $b$ ?

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  • $\begingroup$ Is it all clear now? $\endgroup$ – user Oct 27 '18 at 20:25
  • $\begingroup$ @gimusi Yep, got it. Thanks a lot. $\endgroup$ – user1502 Oct 28 '18 at 11:46
  • $\begingroup$ Well done! The course by Gilbert Strang is one of the best introductory courses for linear algebra and applications! Enjoy it! Bye $\endgroup$ – user Oct 28 '18 at 11:47
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Recall that

$$A\begin{bmatrix}a\\b\\c\end{bmatrix}=a\cdot col_1(A)+b\cdot col_2(A)+c\cdot col_3(A)$$

then we know that

$$A\begin{bmatrix}2\\0\\0\end{bmatrix}=\begin{bmatrix}2\\4\\2\end{bmatrix} \implies col_1(A)=\begin{bmatrix}1\\2\\1\end{bmatrix}$$

morover from

$$A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} \implies col_2(A)=-col_1(A)=\begin{bmatrix}-1\\-2\\-1\end{bmatrix}$$

Can you conclude for $col_3(A)$?

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It's again a linear system, with unknowns living in a vector space, precisely the $3\times 1$ column vectors. Write $A=[a_1\ a_2\ a_3]$; then you know that $$ \begin{cases} 2a_1 = b & (c=0,\;d=0) \\[4px] 3a_1+a_2=b & (c=1,\;d=0) \\[4px] 2a_1+a_3=b & (c=0,\;d=1) \end{cases} $$ This immediately entails that $a_3=0$, $a_1=\frac{1}{2}b$ and $$ a_2=b-3a_1=-\frac{1}{2}b $$ You could even do (formal) Gaussian elimination! \begin{align} \left[\begin{array}{ccc|c} 2 & 0 & 0 & b \\ 3 & 1 & 0 & b \\ 2 & 0 & 1 & b \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 0 & 0 & b/2 \\ 3 & 1 & 0 & b \\ 2 & 0 & 1 & b \end{array}\right] && R_1\gets \tfrac{1}{2}R_1 \\[4px]&\to \left[\begin{array}{ccc|c} 1 & 0 & 0 & b/2 \\ 0 & 1 & 0 & -b/2 \\ 0 & 0 & 1 & 0 \end{array}\right] && \begin{aligned} R_2&\gets R_2-3R_1\\R_3&\gets R_3-2R_1\end{aligned} \end{align}

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  • $\begingroup$ That's nice even if I think that G. Strang is using a more direct argument based on the intepretation of $Ax$ as a linear combination of the columns of $A$. $\endgroup$ – user Oct 26 '18 at 23:17

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