Find standard matrix of linear transformation satisfying conditions

Click For Summary
SUMMARY

The discussion focuses on finding the standard matrix for the linear transformation T: R^3 → R^3, given specific transformations of vectors. The vectors T([1 2 2]) = [1 0 -1], T([-1 -4 -5]) = [0 1 1], and T([1 5 7]) = [0 2 0] are used to construct an augmented matrix. The solution involves row reducing this matrix to express the standard basis vectors as linear combinations of the given vectors, ultimately leading to the determination of T's action on the standard basis vectors.

PREREQUISITES
  • Understanding of linear transformations in R^3
  • Familiarity with augmented matrices and row reduction techniques
  • Knowledge of linear combinations and scalar multiplication
  • Basic concepts of matrix representation of linear transformations
NEXT STEPS
  • Study the process of row reducing augmented matrices in linear algebra
  • Learn about the properties of linear transformations and their matrix representations
  • Explore the concept of basis vectors and their role in vector spaces
  • Investigate the application of linear combinations in solving linear equations
USEFUL FOR

Students studying linear algebra, educators teaching matrix theory, and anyone interested in understanding linear transformations in vector spaces.

oracle104
Messages
1
Reaction score
0

Homework Statement



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.

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.
 
Last edited:
Physics news on Phys.org
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.
 
Last edited:

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K