# Math Help - Getting the image: involves LT composition and change of basis matrix.

1. ## Getting the image: involves LT composition and change of basis matrix.

Hi!

i have a doubt here:

Excercise:
Given:
$B_{1}=\{v_{1},v_{2},v_{3}\}$
$B_{2}=\{-v_{2}-v_{3},v_{1}+v_{2}+v_{3},v_{1}-v_{2}\}$
Both are basis from the V linear space.
and
$f:V \rightarrow V \ / \ M(f)_{B_{1}B_{2}}= \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right)$

-->Find $Image(fof)$

So, my plan was:

Get the change of basis matrix, this is easy because of the way the basis are expresed
$C_{B_{2}B_{1}} = \left(\begin{array}{ccc}0&1&1\\-1&1&-1\\-1&1&0\end{array}\right)$

Now i know that $C_{B_{2}B_{1}}.M(f)_{B_{1}B_{2}}=M(f)_{B_{1}B_{1}}$

$\left(\begin{array}{ccc}0&1&1\\-1&1&-1\\-1&1&0\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right) =
\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)$

Now i just do
$M(f)_{B_{1}B_{1}}M(f)_{B_{1}B_{1}}$
$\Rightarrow$ $\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) = \left(\begin{array}{ccc}-2&-11&10\\8&14&-16\\4&7&-8\end{array}\right)$

And those columns are the image of $(fof)$but expresed in $B_{1}$, so, image well expresed is
$\ Image = < -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} ; 10v_{1} -16v_{2} - 8v_{3} >$ but´s LD, so
$\ Image = \{ -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} \}$
-----------------------------------------

So, the problem is that in the book i found a proposition that says that

given linear functions $f:V \rightarrow W \$ and $g:W \rightarrow U$

$\Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}$

-----------------------------------------------------

So, that, what i have done is: $M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)$

and i know now that the image of the LT are the colums of the matrix, but those are expresed in $B_{2}$ so, the image is going to be formed by:

$\{ -4(-v_{2}-v_{3}) + 6(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}$
$\{ -7(-v_{2}-v_{3}) + 3(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}$
$\{ 8(-v_{2}-v_{3}) - 6(v_{1}+v_{2}+v_{3}) + 14(v_{1}-v_{2}) \}$

$\ Image = < -4v_{1} + 20v_{2} + 10v_{3} ; -7v_{1} + 20v_{2} + 10v_{3} ; -8v_{1} - 28v_{2} - 14v_{3} >$ but´s LD, so
$\ Image = \{ 10v_{1} + 20v_{2} + 1v_{3} ; 0v_{1} + 0v_{2} + 1v_{3} \}$

Unfortunately, both results are different!!
which is the correct one? where is the mistake?

2. Originally Posted by viko
Hi!

i have a doubt here:

Excercise:
Given:
$B_{1}=\{v_{1},v_{2},v_{3}\}$
$B_{2}=\{-v_{2}-v_{3},v_{1}+v_{2}+v_{3},v_{1}-v_{2}\}$
Both are basis from the V linear space.
and
$f:V \rightarrow V \ / \ M(f)_{B_{1}B_{2}}= \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right)$

-->Find $Image(fof)$

So, my plan was:

Get the change of basis matrix, this is easy because of the way the basis are expresed
$C_{B_{2}B_{1}} = \left(\begin{array}{ccc}0&0&1\\-1&1&-1\\-1&1&0\end{array}\right)$

Seems like you go by columns; fine, it's just the same...as long as you keep it being constant. Anyway, the second column's first entry (the matrix's 12 entry) must be 1, not zero.[/color]

Now i know that $C_{B_{2}B_{1}}.M(f)_{B_{1}B_{2}}=M(f)_{B_{1}B_{1}}$

$\left(\begin{array}{ccc}0&0&1\\-1&1&-1\\-1&1&0\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&1&-2\\-1&-3&3\end{array}\right) =$ $
\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)" alt="
\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right)" />

Now i just do
$M(f)_{B_{1}B_{1}}M(f)_{B_{1}B_{1}}$
$\Rightarrow$ $\left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) \left(\begin{array}{ccc}1&-2&1\\2&6&-6\\1&3&-3\end{array}\right) = \left(\begin{array}{ccc}-2&-11&10\\8&14&-16\\4&7&-8\end{array}\right)$

And those columns are the image of $(fof)$but expresed in $B_{1}$, so, image well expresed is
$\ Image = < -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} ; 10v_{1} -16v_{2} - 8v_{3} >$ but´s LD, so
$\ Image = \{ -2v_{1} + 8 v_{2} + 4v_{3} ; -11v_{1} + 14v_{2} + 7v_{3} \}$
-----------------------------------------

So, the problem is that in the book i found a proposition that says that

given linear functions $f:V \rightarrow W \$ and $g:W \rightarrow U$

$\Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}$

-----------------------------------------------------

So, that, what i have done is: $M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)$

[color=red]This is all you need: this is the correct answer.

Tonio

and i know now that the image of the LT are the colums of the matrix, but those are expresed in $B_{2}$ so, the image is going to be formed by:

$\{ -4(-v_{2}-v_{3}) + 6(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}$
$\{ -7(-v_{2}-v_{3}) + 3(v_{1}+v_{2}+v_{3}) - 10(v_{1}-v_{2}) \}$
$\{ 8(-v_{2}-v_{3}) - 6(v_{1}+v_{2}+v_{3}) + 14(v_{1}-v_{2}) \}$

$\ Image = < -4v_{1} + 20v_{2} + 10v_{3} ; -7v_{1} + 20v_{2} + 10v_{3} ; -8v_{1} - 28v_{2} - 14v_{3} >$ but´s LD, so
$\ Image = \{ 10v_{1} + 20v_{2} + 1v_{3} ; 0v_{1} + 0v_{2} + 1v_{3} \}$

Unfortunately, both results are different!!
which is the correct one? where is the mistake?
.

Note that the mistake in the change of basis matrix was just when I copied the matrix to computer, the other numbers have been got with the correct matrix

Anyway i still dont understand why are the results different if both procedures are correct! :S

4. Hi, i discovered the problem!

Originally Posted by myself
-----------------------------------------

So, the problem is that in the book i found a proposition that says that

given linear functions $f:V \rightarrow W \$ and $g:W \rightarrow U$

$\Rightarrow \ M(gof)_{B_{1}B_{3}} = M(g)_{B_{2}B_{3}}.M(f)_{B_{1}B_{2}}$

-----------------------------------------------------

So, that, what i have done is: $M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)$
$M(f)_{B_{1}B_{2}}.M(f)_{B_{1}B_{2}} = M(fof)_{B_{1}B_{2}} = \left(\begin{array}{ccc}-4&-7&8\\6&3&-6\\-10&-10&14\end{array}\right)$