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Ok so my question is how do you proceed when you need to find the image representation if the basis is a lower RX than your linear transformation.

For example: Find the image representation of the linear transformation f(x,y,z,t)=(x+y,z-x,t+y) fron $\Bbb R^4$ to $\Bbb R^3$ with respect to the following bases:

  • $\{(0,0,0,1),(0,1,0,1),(0,0,1,0,),(1,0,1,0)\}$ of $\Bbb R^4$
  • $\{(1,0,0),(0,1,0),(0,0,1)\}$ of $\Bbb R^3$

In the $\Bbb R^4$ basis all I have to do is find $f(x,y,z,t)$ using each of the vectors and I get the matrix $\{(0,0,1),(1,0,2),(0,1,0),(1,0,0)\}$

But I don´t know how to proceed with the $\Bbb R^3$ basis. Any help would be nice.

Thanks.

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    Once you find the image of each $\Bbb R^4$ basis vector, you then obtain the coordinate vectors with respect to the $\Bbb R^3$ basis. Since the $\Bbb R^3$ basis is the standard basis there is really no more work to be done. You need only find one "image representation", and the two bases you are given are both used for the same matrix (they are not used for two different matrices). – 2017-01-07

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I think you're misunderstanding. They want you to do it using that weird basis for R4 and the standard basis for R3, which is what you did. This isn't two separate problems. Any problem requires you specify the basis for each space.

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    Yes, you are right, thanks a lot. Now I can solve the problem. Thanks guys – 2017-01-07
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Given a linear transformation $T:\Bbb R^n\rightarrow\Bbb R^m$ and a choice of bases $\{e_1,...,e_n\}$ of $\Bbb R^n$ and $\{f_1,...,f_m\}$ of $\Bbb R^m$ is associated a matrix $A$ with $m$ rows and $n$ columns as follows. You write $$ A=\left(\begin{array}[cccc] {} a_{1,1}& a_{1,2}&\cdots& a_{1,n}\\ {} a_{2,1}& a_{2,2}&\cdots& a_{2,n}\\ {} ... & ... & ... & ... \\ {} a_{m,1}& a_{m,2}&\cdots& a_{m,n} \end{array}\right) $$ if $$ T(e_k)=\sum_{j=1}^ma_{j,k}f_j. $$ In words: the $k$-th column of $A$ lists the coordinates of $T(e_k)$ in terms of the basis $\{f_1,...,f_m\}$.

In your case you have $T(e_1)=f_3$, $T(e_2)=f_1+2f_3$, $T(e_3)=f_2$ and $T(e_4)=f_1$. Thus $$ A=\left(\begin{array}[cccc] {} 0& 1&0& 1\\ {} 0& 0&1& 0\\ {} 1 & 2 &0 & 0 \end{array}\right). $$