Transformation

**Simplicity** · June 18th 2008, 10:01 AM

If $T: \begin{pmatrix} x \\y \\z \end{pmatrix} ->\begin{pmatrix} x \\x+y \\x+z \end{pmatrix}$ then what does $T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}$ equal?

Can you show in steps as I find this topic confusing. Thanks in advance.

**bobak** · June 18th 2008, 10:10 AM

Still breathing after this morning huh ?

$T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix} = \begin{pmatrix} x+y \\x + 2y \\x +2y+z \end{pmatrix}$

Bobak

**Moo** · June 18th 2008, 10:47 AM

Hello !

Originally Posted by Air

If $T: \begin{pmatrix} x \\y \\z \end{pmatrix} ->\begin{pmatrix} x \\x+y \\x+z \end{pmatrix}$ then what does $T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}$ equal?

Can you show in steps as I find this topic confusing. Thanks in advance.

I'd suggest you to rewrite that

because using the same x & y & z stuff makes things complicated !

$T: \begin{pmatrix} m \\ n \\ p \end{pmatrix} \mapsto \begin{pmatrix} m \\ m+n \\ m+p \end{pmatrix}$

$T(\textbf{v})=T \begin{pmatrix} x+y \\ y \\ y+z \end{pmatrix}$

$x+y=m$
$y=n$
$y+z=p$

---> $m=x+y$
$m+n=(x+y)+y=x+2y$
$m+p=(x+y)+(y+z)=x+2y+z$

Got it ? =)

**Simplicity** · June 18th 2008, 12:11 PM

Originally Posted by Moo

$y=n$

How did you get $y=n$ ?

**Moo** · June 18th 2008, 12:13 PM

Originally Posted by Air

How did you get $y=n$ ?

Ow ^^

$ T: \begin{pmatrix} {\color{red}m} \\ {\color{blue}n} \\ {\color{green}p} \end{pmatrix} \mapsto \begin{pmatrix} m \\ m+n \\ m+p \end{pmatrix} $

$ T(\textbf{v})=T \begin{pmatrix} {\color{red}x+y} \\ {\color{blue}y} \\ {\color{green}y+z} \end{pmatrix} $

**bobak** · June 18th 2008, 12:46 PM

Can you find the 3 by 3 matrix that represents the transformation ?

Bobak

**Simplicity** · June 18th 2008, 01:11 PM

Originally Posted by bobak

Can you find the 3 by 3 matrix that represents the transformation ?

Bobak

**bobak** · June 18th 2008, 01:37 PM

As $T$ is a liner transformations there exists and matrix such that $A \textbf{v} = T( \textbf{v} )$ to find this matrix you need to find the images of the vectors $i, j$ and $k$ under $T$ . This all lies down the understanding how matrix multiplication works.

I do this row by row, you see that the $i$ vector is preserved under the transformation, so the first row is one $i$ combined linearly with no j's or k's

so the first row is $\left(\begin{array}{ccc}1 & 0 & 0 \\? & ? & ? \\? & ? & ?\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)$

for the next row you combining one $i$ with one $j$ and no k's

so we have $\left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\? & ? & ?\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)$

Then similarly for the next row your combining 1 i with 1 k.

giving $\left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)$

to check this is correct just multiple the two.

$\left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right) = \begin{pmatrix} x \\x+y \\x+z \end{pmatrix}$

so $T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}$ is given by $\left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right) \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}$

You following?

Bobak

**Simplicity** · June 18th 2008, 01:44 PM

Originally Posted by bobak

You following?

Bobak

Yes! Much better than the book too.

Thanks!

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