Transformation Matrix T in Terms of β1, β2 with Row Reduction Explained

Click For Summary
The transformation matrix T is expressed in terms of the vectors α1, α2, and α3 as Tα1 = (1,−3), Tα2 = (2,1), and Tα3 = (1,0). Row reduction is discussed in the context of calculating the matrix of T relative to the basis B and the standard basis of \mathbb{R}^2. It is noted that since the basis vectors β1 and β2 correspond to the standard basis vectors e1 and e2, simply swapping the rows suffices instead of performing row reduction. However, if the relationship between B and the standard basis were more complex, row reduction might be necessary. The discussion emphasizes the importance of understanding the basis transformations in linear algebra.
jolly_math
Messages
51
Reaction score
5
Homework Statement
Let T be the linear transformation from R3 into R2 defined by T(x1, x2, x3) = (x1 + x2, 2x3 - x1).
B={α1,α2,α3} and B′ ={β1,β2}, where α1 = (1,0,−1), α2 = (1,1,1), α3 = (1,0,0), β1 = (0,1), β2 = (1,0). What is the matrix of T relative to the pair B, B′?
Relevant Equations
transformation
T(α1), T(α2), T(α3) written in terms of β1, β2:
Tα1 =(1,−3)
Tα2 =(2,1)
Tα3 =(1,0).
Then there is row reduction:
1667976594734.png

Therefore, the matrix of T relative to the pair B, B' is
1667976607260.png

I don't understand why the row reduction takes place? Also, how do these steps relate to ## B = S^{-1}AS ##? Thank you.
 
Last edited:
Physics news on Phys.org
The solution has calculated the matrix of T relative to the basis B and the standard basis of \mathbb{R}^2, \{ e_1 = (1,0), e_2 = (0,1) \}. But \beta_1 = e_2 and \beta_2 = e_1 so the rows of that matrix need to be swapped to give the matrix with respect to B and B', But you really don't need row reduction to do that; you can just swap the rows.

If B' depended on the standard basis in a less straightforward manner, then the row reduction may have been necessary to get to the required matrix.
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...