I have no idea how to solve this and do not even understand the question. Excuse my lack of formatting and assume that that when a number is after a letter like $e_2$ the number is subscript. $V_2$ is a linear space with basis $e = (e_1,e_2)$ and the systems of vectors $a=(a_1,a_2)$ and $b=(b_1,b_2)$ where $a_1 = e_1+ 2e_2$, $a_2 = e_1 + 3e_2$, $b_1=-e_1+e_2$,$b_2 = -e_1+ 2e_2$. Let $f$ be a linear transformation on $V_2$.

1) If $f(e)=b$ find the matrix $A$ of $f$ in basis $e$ and $f(v)$ where $v=3e_1-2e_2$.

2) If $f(b)=a$ find the matrix $M$ of $f$ in base $e$

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    $\begingroup$ I thnk the function f is supposed to be applied to the 2x2 matricies of e and b $\endgroup$ – waxo99 Dec 13 '19 at 19:48
  • $\begingroup$ Never mind. I haven’t had my coffee yet this morning. It looks like you’re meant to read $f(e)=b$ as $f(e_1)=b_1$ and $f(e_2)=b_2$, and similarly for the second question. Use linearity to compute $f(v)$ and use the fact that the columns of a transformation matrix are the images of the basis vectors to find the matrices. $\endgroup$ – amd Dec 13 '19 at 19:48
  • $\begingroup$ If I could get a detailed answer I would appreciate it but I understand if you do not have the time. $\endgroup$ – waxo99 Dec 13 '19 at 19:55

Question 1:

As amd says in his comment, we are meant to read $f(e) = b$ as $f(e_j) = b_j$ for all indices $j$. With that in mind, we can compute the matrix of $f$ with respect to $e$ as follows.

To find the first column of the matrix, note that $$ f(e_1) = b_1 = -e_1 + e_2 = (-1)\cdot e_1 + (1) \cdot e_2. $$ It follows that $(-1,1)$ is the coordinate vector of $f(e_1)$ relative to $e$, and is thus the first column of our matrix. In other words, we have $$ A = \pmatrix{-1&?\\1&?}. $$ Similarly, find that the second column is $(-1,2)$, so that the matrix is given by $$ A = \pmatrix{-1&-1\\1&2}. $$ From there, we could compute $f(v)$ in two ways. One option is to use the linearity of $f$ as follows: $$ f(v) = f(3e_1 - 2e_2) = 3f(e_1) - 2f(e_2) = 3b_1 - 2b_2 \\ = 3 (-e_1 + e_2) - 2(-e_1 + 2e_2) \\ = -3e_1 + 3e_2 + 2e_1 - 4e_2 \\ = -e_1 - e_2 $$ The other option is to compute the transformation matrix that we just computed. Since $v = 3e_1 - 2e_2$, the coordinate vector of $V$ relative to $e$ is $(3,-2)$. Thus, the coordinate vector of $f(v)$ is given by $$ A [v]_e = \pmatrix{-1&-1\\1&2} \pmatrix{3\\-2} = \pmatrix{-1\\-1}. $$ This is the coordinate vector of $f(v)$, which is to say that $f(v) = -e_1 - e_2$, confirming the earlier result.

Question 2:

This one is a little bit trickier. Remember that to find the matrix of $f$ with respect to $e$, we need to figure out what $f(e_1)$ and $f(e_2)$ are. We were given this information directly in the last problem, but now we have to get that information using what we know about $f(b_1)$ and $f(b_2)$.

We can use the following procedure:

  • Find numbers $x_1$ and $x_2$ such that $x_1 b_1 + x_2 b_2 = e_1$. This is technically a system of equations, but it's easy to solve by "guessing" an answer.
  • Note that $f(e_1) = f(x_1b_1 + x_2 b_2) = x_1 f(b_1) + x_2 f(b_2)$. The vector of coefficients is the coordinate vector of $f(e_1)$ relative to $e$, and is our first column of $M$.
  • Do the same thing to find the second column, which is the vector of coefficients of $f(e_2)$.

Regarding the matrix-multiplication formula: suppose that $P = [f]_{b \to e}$, i.e. $P$ is the matrix of $f$ relative to the bases $b$ and $e$. For your problem, we would have $$ P = \pmatrix{1&1\\2&3}. $$ Suppose that $B = [I]_{b \to e}$, i.e. $B$ is the change of basis matrix from $b$ to $e$. For your problem, we would have $$ B = \pmatrix{-1&-1\\1&2}. $$ Then the matrix $M$ of $f$ relative to the basis $e$ would be given by $$ M = [f]_{b \to e}[I]_{b \to e}^{-1} = PB^{-1} = \pmatrix{1&1\\2&3}\pmatrix{-1&-1\\1&2}^{-1} = \pmatrix{-1&0\\-1&1}. $$

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  • $\begingroup$ What about matrix A $\endgroup$ – waxo99 Dec 13 '19 at 20:29
  • $\begingroup$ I had them mixed up, see my latest edit $\endgroup$ – Ben Grossmann Dec 13 '19 at 20:30
  • $\begingroup$ What about matrix M then $\endgroup$ – waxo99 Dec 13 '19 at 20:31
  • $\begingroup$ See my latest edit $\endgroup$ – Ben Grossmann Dec 13 '19 at 20:33
  • $\begingroup$ Can I get a detailed answer for the second part and can I ask if in general we have f(a) and F is the matrix of the function F and A is the matrix of a system of equations does this equal F times A or A times F ? $\endgroup$ – waxo99 Dec 13 '19 at 20:51

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