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$f$ is a linear map from $\mathbb R^3$ to $\mathbb R^2$, such that: $$ f(1,0,0) = (1,2)\\ f(0,1,0) = (2,1)\\ f(0,1,1) = (0,0)\\ $$ a) Find an explicit expression for $f(x,y,z)$.

b) Calculate the matrix of the transformation $f$ in the basis $B,D$ and $C,D$, where: $$ B = \left\{ (1,0,0),(0,1,0),(0,0,1) \right\}\\ C = \left\{ (1,1,0),(0,1,1),(1,0,0) \right\}\\ D = \left\{ (1,1),(1,0) \right\} $$

I was able to get one solution for a, but then I couldn't solve letter b. Here's my solution for a: $$ f(x,y,z) = (x+2y-2z,2x+y-z) $$ Here's what I've done for b:

I've calculated what $f$ of each vector in basis B is, to get the following matrix:$$ \begin{pmatrix} 1 & 2 & -2\\ 2 & 1 & 1 \end{pmatrix} $$ and same for C:$$ \begin{pmatrix} 3 & 0 & 1\\ 3 & 0 & 2 \end{pmatrix} $$ I know that these matrices maps any vector in $\mathbb R^3$ to a vector in $ \mathbb R^2$ with the constraint that I've found in letter a, but they are related with a basis from $ \mathbb R^3$, in these cases, $B$ and $C$. Now what should I do to get the result of these transformations in terms of the basis $D$? Am I in the way?!

Thanks!

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Good work so far. To complete the problem, the key is to understand the meaning of the matrix of a linear map. Consider a linear map $T:V \rightarrow W$. For simplicity, assume the basis for $V$ is $\{\alpha_{1},\alpha_{2},\alpha_{3}\}$ and the basis for $W$ is $\{\beta_{1},\beta_{2}\}$. The first step, as you did above, is to consider the action of $T$ on the basis vectors in $V$. Say we did that and we have $T(\alpha_1), T(\alpha_2), T(\alpha_3)$. Now, we need to express each of these vectors that lie in $W$ as a linear combination of the basis in $W$ i.e $\{\beta_{1},\beta_{2}\}$ \begin{align} T(\alpha_1) &= a_{11} \beta_{1} + a_{12} \beta_{2} \\ T(\alpha_2) &= a_{21} \beta_{1} + a_{22} \beta_{2} \\ T(\alpha_3) &= a_{31} \beta_{1} + a_{32} \beta_{2} \end{align} where the $a_{ij}$ are some coefficients you find. As an example, the coordinate representation of $T(\alpha_1)$ is simply $$ T(\alpha_1)_{\beta}= \begin{pmatrix} a_{11}\\ a_{12} \end{pmatrix} $$ where the subscript $\beta$ denotes the coordinate represetation with respect to the basis $\beta$. The matrix is nothing but the coordinate representation of the mapped vectors i.e $$ \begin{pmatrix} \vdots & \vdots & \vdots \\ T(\alpha_1)_{\beta} & T(\alpha_2)_{\beta} & T(\alpha_3)_{\beta}\\ \vdots & \vdots & \vdots \\ \end{pmatrix} $$ As you can see, its representation depends on the basis we choose. The operator is one and one but based on the basis you choose, you have different matrix representations. The matrix of the linear map is then $$ \begin{pmatrix} a_{11} & a_{21} & a_{31}\\ a_{12} & a_{22} & a_{32} \end{pmatrix} $$ So for your problem, all that remains to be done is to compute the coordinate representation of the mapped vectors interms of the basis in $\mathbb R^{2}$.

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  • $\begingroup$ Perfect explanation, but that's exactly what I've done so far... I know that what remains is "to compute the coordinate representation of the mapped vectors interms of the basis in $\mathbb R^{2}$". But how can I represent them? I need to get my matrix of the linear map, and then do what?! I'm lost in that part... $\endgroup$ – Bruno Reis Oct 13 '16 at 2:09
  • $\begingroup$ Ok. You have already computed the action of f on each vector in the basis $B$. Let's just take the first one. $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. We can write it as linear combination of the basis vectors in $D$ as follows $$\begin{pmatrix} 1 \\ 2 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix}-1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$. It then follows that the coordinate representation of $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ is simply $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Do the same for others. Is that clear? $\endgroup$ – felasfa Oct 13 '16 at 2:23
  • $\begingroup$ To complete the above, the matrix representation of f with respect to the basis $B$, $D$ is given by $$\begin{pmatrix} 2 & 1 & 1 \\ -1 & 1 & -3\end{pmatrix}$$ Do you see why? Could you apply this for $C$ and $D$? $\endgroup$ – felasfa Oct 13 '16 at 2:29
  • $\begingroup$ it's the application of the linear map, and then, change of basis? so a matrix multiplication in that order? $\endgroup$ – Bruno Reis Oct 13 '16 at 2:34
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    $\begingroup$ Exactly, that is right. $\endgroup$ – felasfa Oct 13 '16 at 2:42
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Note that: \begin{pmatrix} 1 & 2 & -2\\ 2 & 1 & 1 \end{pmatrix} is the matrix $|f|_{BE}$ where B is the given basis and E is the standard basis for $\mathbb R^2$. Now recall that for two given bases, we have the respective change of basis matrices. Having this in mind, what you can use to get $|f|_{BD}$ and $|f|_{CD}$ is simply use matrix multiplication like so:

$$|f|_{BD}=C_{ED}|f|_{BE}C_{DE}=C_{DE}^{-1}|f|_{BE}C_{DE}$$ where $C_{ED}$ and $C_{DE}$ are the change of basis matrices for E to D and D to E respectively. The same idea can be used to calculate $|f|_{CD}$

I hope this helps you!

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