0
$\begingroup$

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 $e$

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $$ M = [f]_{b \to e}[I]_{b \to e}^{-1} = PB^{-1} = \pmatrix{1&1\\2&3}\pmatrix{-1&-1\\1&2}^{-1} = \pmatrix{-1&0\\-1&1}. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.