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I tried making the column vectors $(1,1,1)$ and $(1,2,1)$ into a matrix, but because it's $3\times2$, it's noninvertible. Does this mean that I cannot recover the linear transformation without a third vector? Do the vectors that define the transformation $(1,1,1)$ and $(1,2,1)$ have to form a basis for the subspace they're being transformed from for the transformation to be defined? Sorry if I'm mixing things up or wording things in a confusing manner.

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You're looking for a matrix $ \begin{pmatrix} a&b&c\\ e&f&g\\ h&i&j \end{pmatrix}$ such that $$\begin{pmatrix} a&b&c\\ e&f&g\\ h&i&j \end{pmatrix} \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} = \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix} $$ and $$\begin{pmatrix} a&b&c\\ e&f&g\\ h&i&j \end{pmatrix} \begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix} = \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} .$$ If you do the multiplications you get six equations $$\begin{cases} a+b+c=1 \\ e+f+g=2 \\ h+i+j=3 \\ a+2b+c=1 \\ e+2f+g= 1 \\ h+2i+j=1 \end{cases}.$$ You can easily find (by subtraction [e.g. $a+b+c=1$ and $a+2b+c=1$]) that $b=0$, $f=-1$, $i=-2$. Then for example choose $a=1, c=0, e=3,g=0, h=5,j=0$ and you can conclude that $$ \begin{pmatrix} 1&0&0\\ 3&-1&0\\ 5&-2&0 \end{pmatrix}$$ is a solution to your problem.

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  • $\begingroup$ Thank you! The way my professor taught us to find these was nowhere near as clear as this. $\endgroup$ – Wayfinder Mar 11 '16 at 0:40

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