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Linear transformations help?

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I'm totally lost and could use some help (I'm new here). Vectors I've already tried that are incorrect: <-3,-3,4> <-3,-6,28> <-3,1>,<-3,4>,<4,3> <-6,8>

Advice? THANK YOU

Edit: I looked through every single example given in class notes as well as the appropriate sections of the textbook and I have yet to find a transformation of the form M(nxm) -> Rn. All of the ones I've found are Rn->Rm or Rm->Rm

Re: Linear transformations help?

The vectors you give for a basis of the range of E can't possibly be right because they are in $\displaystyle R^3$. The operator E (multiplication by the matrix A with 2 rows and 3 columns) is from $\displaystyle R^6$ to $\displaystyle R^2$. The range is a subspace of $\displaystyle R^2$.

The "complicating point is that we are here thinking of the matrix, A, as the variable, and the vector v, as fixed: $\displaystyle v= \begin{pmatrix}1 \\ 2 \\ 7\end{pmatrix}$. Okay, since A is "variable", lets just write it as $\displaystyle \begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}$.

The simplest way to do this is to use the "rank-nullity" relation: if L is a linear transformation from U to V, then the "rank" of L (dimension of its range in V) plus the "nullity" of L (dimension of its null-space or "kernel" in U) is equal to the dimension of U.

Here, "U" is the set of all 2 by 3 real matrices and so has dimension 2(3)= 6. The null-space is the set of all matrices, M, such that Mv= 0.

Taking M as $\displaystyle \begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}$, we have $\displaystyle E(M)= Mv=\begin{pmatrix}a & b & c \\ d & e & f\end{pmatrix}$$\displaystyle \begin{pmatrix}1 \\ 2\\ 7\end{pmatrix}$$\displaystyle = \begin{pmatrix}a+ 2b+ 7c \\ d+ 2e+ 7f\end{pmatrix}$. If M is in the null-space then we must have $\displaystyle \begin{pmatrix}a+ 2b+ 7c \\ d+ 2e+ 7f\end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix}$. That is, we must have a+ 2b+ 7c= 0 and d+ 2e+7f= 0. From those two equations, we get a= -2b- 7c and d= -2d- 7f. That is, b, c, d, and f can be any numbers whatsoever and we can calculate a and d from them. That tells us that the null-space of E is a **four** dimensional subspace of the 6 dimensional space of all such matrices. The range of E, then, is a 6-4= 2 dimensional subspace of $\displaystyle R^2$. But $\displaystyle R^2$ is itself 2 dimensional! That means that E maps that set of matrices to **all** of [itex]R^2[/tex] and we can take as basis vectors of the range, any basis vectors for [itex]R^2[/itex], in particular, the "standard basis vectors", $\displaystyle \begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $\displaystyle \begin{pmatrix}0 \\ 1\end{pmatrix}$.

Re: Linear transformations help?

Thank you SO so much!

It's crazy how much better you are at explaining than my prof...

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Re: Linear transformations help?

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I'm probably going to have several questions, I'll just put them all in here to not clog up this forum. From what I can tell T takes the negative coefficient of v1 and uses it for w2 and it takes the absolute value of v2's coefficient and uses it on w1. Thanks for any clarifications...