How Can a New Basis Change the Matrix Representation of a Linear Transformation?

[P3X-018]

I've given the linear transformation $L: \mathbb{R}^4 \rightarrow \mathbb{R}^3$, where

$$L(\mathbf x) = A\mathbf x$$

and

$$A = \left[ \begin{array}{cccc} 1&1&-4&-1 \\ -2&3&1&0 \\ 0&1&-2&2 \end{array} \right]$$

The first question of the problem is that, by using the standard basis $E$ for $\mathbb{R}^4$, I have to determine a new basis $F = [\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3]$ for $\mathbb{R}^3$ such that the matrix representing $L$ in $E$ and $F$ is the reduced row echelon form of the matrix $A$, which I
determined to be

$$U = \left[ \begin{array}{cccc} 1&0&0&-11 \\ 0&1&0&-6 \\ 0&0&1&-4 \end{array} \right]$$

I determined the basis by using, that if $B = (\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3)$, then

$$U = B^{-1}\left( L(\mathbf{e}_1), L(\mathbf{e}_2), L(\mathbf{e}_3), L(\mathbf{e}_4) \right) = B^{-1}A$$

So $B^{-1}$ must be "the product" of elementary operations performed on A, which can easily be found by

$$\left(A|I\right) \rightarrow \left(U|B^{-1}\right)$$

Since $B^{-1}\left(A|I\right) = \left(B^{-1}A|B^{-1}\right) = \left(U|B^{-1}\right)$. I found B as

$$B =\left[ \begin{array}{ccc} 1&1&-4 \\ -2&3&1 \\ 0&1&-2 \end{array} \right]$$

The next question is to explain why it is possible for an arbitrary $m\times n$ matrix A, to determine a new basis F for $\mathbb{R}^m$ (where $L: \mathbb{R}^n \rightarrow \mathbb{R}^m$), but still keep the standard basis for $\mathbb{R}^n$, so that the linear transformation corrosponding to A i the new basis is represented by the reduced row echelon form of A. Like in the last problem.

I'm sorry if the question is purely expressed, but that's how it's expressed in the problem.
The way I see it is that if we have the linear transformation L given by $L(\mathbf{x}) = A\mathbf{x}$, then by using the reduced row echelon form of A, let's call it U, the transformation can be expressed in new basis as

$$[L(\mathbf{x})]_F = U[\mathbf{x}]_E$$

But where E will be the standard basis.

I don't know exactly how to explain this. It doesn't look like it would be a sufficient explanation to use the steps in the 1. question. A hint to how I can explain it would really help.

 

[fopc]

Regarding your question about an "explanation", you might think of this problem in terms of equivalence relations on matrices. Here's some starting material.

1) Left Associate: (your case) Matrices A and B (both mxn) are left associate iff there exists a non-singular (mxm) Q such that B = (Q^-1)A. Multiplication by (Q^-1) corresponds to performing a sequence of elementary row operations. If A represents a linear transformation
T: U -> V relative to basis A in U and basis B in V, then matrix B represents T relative to A and a new basis in V. There is an accompanying theorem. It's really all you need in the way of explanation.

Here are some other equivalence relations you might explore.

2) Right Associate: You can imagine how this one is defined.

3) Equivalent: To be more consistent with the above naming convenvtion, some call it Associate, (e.g., E.Nering). Note that associate is definitely not a standard term for this relation, and I suspect left and right in 1) and 2) aren't either.

4) Similar: Perhaps the most interesting of the four.

 

[tacman]

One way to explain this is by considering the concept of change of basis. When we have a linear transformation from one vector space to another, the choice of basis for each vector space can affect how the transformation is represented. In this case, we have a linear transformation from \mathbb{R}^4 to \mathbb{R}^3 , and the standard basis for each of these spaces is defined as the set of unit vectors along each coordinate axis.

However, we can choose a different basis for \mathbb{R}^3 that is not aligned with the coordinate axes, but still spans the same space. This new basis, represented by the matrix B, will have a different set of unit vectors and will result in a different representation of the linear transformation L. By using the reduced row echelon form of A, we are essentially finding a new basis for \mathbb{R}^3 that is more convenient for representing the transformation L.

Furthermore, we can keep the standard basis for \mathbb{R}^4 because the transformation L is still being applied to vectors in this space. The new basis F only affects how the transformation is represented in \mathbb{R}^3 , but it does not change the input space for L. This is why we can determine a new basis for \mathbb{R}^3 while keeping the standard basis for \mathbb{R}^4 .