Find $\left[T\right]_\beta^\beta$ for NMH{823}

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SUMMARY

The discussion focuses on finding the transition matrix $\left[T\right]_\beta^\beta$ for the linear transformation T defined by T\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 2y- z \\ 2x- y+ z \\ x+ z\end{bmatrix}. The user calculates T\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} and derives the equations A + B + C = 0, 2B + C = 3, and A + B = 2 to find the coefficients A, B, and C. The resulting first column of the transition matrix is \begin{bmatrix}\frac{8}{5} \\ \frac{2}{5} \\ \frac{11}{5}\end{bmatrix}, demonstrating the correct approach to constructing the transition matrix.

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karush
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nmh{823}

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ok just want to see if I went the right direction with this before I launch into the rref
of which I would use a online calculator to do if I could find one.
$\left[T\right]_\beta^\beta$ should be on the right after rref
 
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Given that T\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 2y- z \\ 2x- y+ z \\ x+ z\end{bmatrix}

Then T\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}1+ 2(0)- 1 \\ 2(1)- 0+ 1 \\ 1+ 1\end{bmatrix}= \begin{bmatrix} 0 \\ 3\\ 2 \end{bmatrix}. That is very different from your "\begin{bmatrix}2 \\ 0 \\ 2 \end{bmatrix}". Am I misunderstanding something?

To write that in basis \beta you want to find numbers, A, B, C such that
A\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}+ B\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix}+ C\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}= \begin{bmatrix}A+ B+ C \\ 2B+ C \\A+ B \end{bmatrix}= \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix}.

So we have the three equations, A+ B+ C= 0, 2B+ C= 3, and A+ B= 2. From the second equation C= 3- 2B, and from the third, A= 2- B. Putting those into the first equation, A+ B+ C= 2- B+ B+ 3- 2B= 5- 2B= 0. 2B= 5 so B= 2/5. Then C= 3- 4/5= 11/5 and A= 2- 2/5= 8/5. The first column of the transition matrix is \begin{bmatrix}\frac{8}{5} \\ \frac{2}{5} \\ \frac{11}{5}\end{bmatrix}.
 
https://www.physicsforums.com/attachments/8893

how did they get these (red)
 
karush said:
how did they get these (red)
They applied the operator T to the vector using either the explicit form of the matrix T just above or by using the original definition.

[math]T \left [ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right ] = \left ( \begin{matrix} 5 & 0 & 1 \\ 3 & 2 & -3 \\ 5 & 0 & 0 \end{matrix} \right ) \left [ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right ] = \left [ \begin{matrix} 5(1) + 0(1) + 1(2) \\ 3(1) + 2(1) - 3(2) \\ 5(1) + 0(1) + 0(2) \end{matrix} \right ] = \left [ \begin{matrix} 7 \\ -1 \\ 5 \end{matrix} \right ] [/math]

-Dan
 
Looks obvious now
But i couldn't see that

Sometimes I wish they would show more steps rather that assume things but then ..;)

Just glad there is a good forum to call on
 

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