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Let $f_k \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ a linear map defined by $$ f_k \begin{pmatrix} 1 \\ 2 \\ k \end{pmatrix}= \begin{pmatrix} 2+k \\ 3 \\ 0 \end{pmatrix}, \quad f_k \begin{pmatrix} 2 \\ k+1 \\ -1 \end{pmatrix}= \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}, \quad f_k \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix}= \begin{pmatrix} 1 \\ k \\ 2 \end{pmatrix}, $$ For $k=-1$ how can i find the matrix associated to $f_k$ with respect the standard basis in the domain and standard basis in codomain?

Added by Arturo: To put this question in context, see this previous question.

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@Katy: Please don't post simply by quoting the problem in the imperative. It reads as if you were giving orders. Could you perhaps put it in a quote box, and then add some words as to what you have managed to accomplish or why/where you are confused? In your most recent question it wasn't until after another answer was posted that you indicated in comments that you already knew the material contained in the answer and was looking for something else; that belongs in the question, because it guides potential repliers as to what you already know, and what it is you want to hear. – Arturo Magidin May 29 '11 at 22:16
Given what has been answered in the previous question, this should now be very easy: the "basis in the domain" is the basis you get by pluggin in $k$ to the three vectors listed in the definition (the ones you apply $f_k$ to); you know what the answers are; you should be able to write down the matrix. What is confusing you? – Arturo Magidin May 29 '11 at 22:18
excuse me. you are right – Katy23 May 29 '11 at 22:19
can you tell me the first steps of solution? – Katy23 May 29 '11 at 22:22
@Yuval: It's not really a duplicate, as in this case we simply have a specific linear transformation and we are asked for the matrix that represents it (relative to some bases), while the other question asked something different about a family of functions of which this is one member. It might more accurately be said that this question is "too localized" in the absence of the previous one, or that the only reason for looking at this is in light of the previous question, rathere than to say it is a duplicate of the previous one. – Arturo Magidin May 30 '11 at 1:50
up vote 2 down vote accepted

Plug in $k=-1$. That gives you a basis for $\mathbb{R}^3$, as discussed in your previous question, namely $$\left(\begin{array}{r}1\\2\\-1\end{array}\right),\qquad\left(\begin{array}{r}2\\0\\-1\end{array}\right),\qquad\left(\begin{array}{r}-3\\1\\5\end{array}\right).$$ That's the "basis in the domain".

You know what the image of the basis vectors under $f_{-1}$ is: you are told what they are.

So you have a basis $\beta$, and the value of the linear transformation at the vectors of $\beta$. How do you find the matrix of $f_{-1}$ from $\mathbb{R}^3$ with basis $\beta$ to $\mathbb{R}^3$ with the standard basis? The first column is the image of the first vector of $\beta$ written in terms of the standard basis. The second column of the matrix is...

Edit. I see now that the question asks for the matrix of $f_{-1}$ relative to the standard bases, which means you need to find $f_{-1}(e_1)$, $f_{-1}(e_2)$, and $f_{-1}(e_3)$. How to do that?

Well, since $(1,2,-1)$, $(2,0,-1)$, and A$(-3,1,5)$ are a basis, we can write $e_1$, $e_2$, and $e_3$ as linear combinations of them. For example, $$e_1 = -\frac{1}{15}\left(\begin{array}{r}1\\2\\-1\end{array}\right) + \frac{11}{15}\left(\begin{array}{r}2\\0\\-1\end{array}\right) + \frac{2}{15}\left(\begin{array}{r}-3\\1\\5\end{array}\right)$$ so that means that $$f_{-1}(e_1) = -\frac{1}{15}f_{-1}\left(\begin{array}{r}1\\2\\-1\end{array}\right) + \frac{11}{15}f_{-1}\left(\begin{array}{r}2\\0\\-1\end{array}\right)+ \frac{2}{15}f_{-1}\left(\begin{array}{r}-3\\1\\5\end{array}\right).$$ Continue this way to get the matrix.

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The first column of the matrix is the image through $f_{-1}$ of $e_1=(1,0,0)$. How i can evaluate this? – Katy23 May 30 '11 at 9:20
Let $B=(e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1))$ i must find $[f_{-1}]_B^B$. Or not :)? – Katy23 May 30 '11 at 9:22
@Katy23: I see that now. You need to write each of $e_1$, $e_2$, and $e_3$ in terms of $(1,2,-1)$, $(2,0,-1)$, and $(-3,1,5)$ (which is possible, since the latter 3 are a basis), and then use linearity of $f_{-1}$ to get the values. – Arturo Magidin May 30 '11 at 12:41

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