How Do You Find the Standard Matrix Representation for a Linear Transformation?

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Homework Help Overview

The discussion revolves around finding the standard matrix representation for a linear transformation defined by T1 and using it to compute T1(1,-3,0). The subject area is linear algebra, specifically focusing on linear transformations and matrix representations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the correct form of the standard matrix representation for the transformation T1, with some questioning the initial attempt and others providing alternative representations based on applying T1 to basis vectors.

Discussion Status

There is an ongoing exchange of ideas regarding the correct matrix representation, with some participants offering corrections and clarifications. Multiple interpretations of the transformation's representation are being discussed, and participants are actively engaging with each other's contributions.

Contextual Notes

There is mention of differing conventions that may affect the representation of the transformation, indicating that the discussion is influenced by varying textbook definitions or approaches.

DanielFaraday
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Homework Statement



Write the standard matrix representation for T1 and use it to find [T1(1,-3,0)]E.

Homework Equations



[tex] T_1\left(x_1,x_2,x_3\right)=\left(x_3,-x_1,x_3\right)[/tex]

The Attempt at a Solution



I just wanted to check to see if I am doing this right. Thanks in advance!

[tex] A=\left(<br /> \begin{array}{ccc}<br /> 0 & 0 & 1 \\<br /> -1 & 0 & 0 \\<br /> 0 & 0 & 1<br /> \end{array}<br /> \right)\[/tex]

[tex] \left[T_1(1,-3,0)\right]_E=A\left(<br /> \begin{array}{c}<br /> 1 \\<br /> -3 \\<br /> 0<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{ccc}<br /> 0 & 0 & 1 \\<br /> -1 & 0 & 0 \\<br /> 0 & 0 & 1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{c}<br /> 1 \\<br /> -3 \\<br /> 0<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 0 \\<br /> -1 \\<br /> 0<br /> \end{array}<br /> \right)[/tex]
 
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Your A is transposed from what it should be.
 
No, Officeshredder,
Applying T1 to each basis vector in turn gives the columns.

T1(1, 0, 0)= (0, -1, 0)
T1(0, 1, 0)= (0, 0, 0)
T1(0, 0, 1)= (1, 0, 1)

So T1 is represented by
[tex]\begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}[/tex]
exactly what DanielFaraday has.

And, of course, T1(1,-3,0)= (0,-1,0) as said.
 
Thank you both for your input!
 
Office shredder may be using a different convention than you and I:

[tex]T_1(1,-3,0)= \begin{bmatrix}1 & -3 & 0\end{bmatrix}\begin{bmatrix}1 & -1 & 0\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}[/tex]
 
Yes, these things often depend on the textbook. Thanks.
 
No, sorry, that was just a brain fart on my part
 

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