0
$\begingroup$

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.

  • 1
    @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.2011-05-30

1 Answers 1

2

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.

  • 1
    @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.2011-05-30