0
$\begingroup$

Let $S \left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}\right) = $ $\begin{pmatrix} x_1 & -2x_2 & x_3 & x_4\\ 2x_1 & - 4x_2 & -3x_3 & -3x_4\\ x_1 & -2x_2 &-4x_3 & -3x_4 \end{pmatrix} $

Determine if this linear transformation is surjective or injective. I already determined that it is not injective because the $\dim(Ker(S)) \not= 0$ therefore it is not injective. And I know a transformation is surjective iff $\dim(Range(S)) = $ dimesion of the codmain. However in a question like this the codomain is not given so its kind of throwing me off. Any help would be great!

$\endgroup$
  • $\begingroup$ If nothing is given, the codomain is typically just the biggest thing that makes sense; in this case that would be $\mathbb{R}^3$ (or $K^3$ if you're working in a more general setup). But it's a good question; one could have simply taken the codomain to be $\mathrm{Range}(S)$ -- then surjectivity would have been a lot easier to (dis)prove. $\endgroup$ – fuglede Apr 23 '14 at 11:37
  • $\begingroup$ To me, it looks like the codomain is the set of $3\times4$ matrices. Note that it doesn't say that $S$ is represented by the matrix; it says that the matrix is the value of $S$. Of course this could be a transcription error by the OP.... $\endgroup$ – Greg Martin Apr 23 '14 at 12:29
  • $\begingroup$ @GregMartin Judging by the fact that OP says it's not injective I think the result is meant to be a vector with three coordinates, otherwise the map is injective. I guess he wanted the operands to be equally spaced out. then again, it's missing some plus signs in this case so who knows. $\endgroup$ – GPerez May 23 '15 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.