Find standard matrix of linear transformation satisfying conditions


by oracle104
Tags: conditions, linear, matrix, satisfying, standard, transformation
oracle104
oracle104 is offline
#1
May16-11, 02:22 PM
P: 1
1. The problem statement, all variables and given/known data

Find the standard matrix for the linear transformation T: R^3-->R^3 satisfying:
T([1 2 2]) = [1 0 -1], T([-1 -4 -5]) = [0 1 1], T([1 5 7]) = [0 2 0]

All of the vectors are columns not rows, I couldn't type them correctly as columns.

3. The attempt at a solution
I tried constructing a matrix using the vectors being applied to T and row reducing it. I cannot figure out where to go from there. I assume I need to find T of the standard basic vectors in some way. I believe I can figure it out if I can get a step in the right direction.
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kru_
kru_ is offline
#2
May16-11, 06:17 PM
P: 85
Since [tex]T[/tex] is linear, we know that [tex]T\left(\vec{u} + \vec{v}\right) = T\left(\vec{u}\right) + T\left(\vec{v}\right)[/tex].
Since [tex]T[/tex] is linear, we know that [tex]T\left(c \vec{v}\right) = c T\left(\vec{v}\right)[/tex], for any real scalar c.

You can find the standard vectors as linear combinations of the given vectors by constructing an augmented matrix and row reducing, as you did.

For example:

[tex]\begin{pmatrix}1&& -1&& 1&& 1&&\\2&& -4&& 5&& 0&&\\ 2&& -5&& 7&& 0&&\end{pmatrix} -> \begin{pmatrix}1&& 0&& 0&& 3&&\\0&& 1&& 0&& 4&&\\ 0&& 0&& 1&& 2&&\end{pmatrix}[/tex]

So we can write [tex]\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}[/tex] as a linear combination of [tex](3)\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} + (4)\begin{bmatrix} -1 \\ -4 \\ -5 \end{bmatrix} + (2)\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}[/tex]

Now we know that:

[tex]T\left(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\right) = T\left( (3)\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} + (4)\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} + (2)\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix}\right) = (3)T\left( \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \right) + (2)T\left(\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} \right) + (4)T\left(\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix} \right)[/tex]

So, what is:

[tex](3)T\left( \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \right) + (4)T\left(\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} \right) + (2)T\left(\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix} \right)[/tex]

The values for the other standard vectors can be found with a similar process.


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