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## Main Question or Discussion Point

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

[tex] L(\mathbf x) = A\mathbf x [/tex]

and

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

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

determined to be

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

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

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

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

[tex] \left(A|I\right) \rightarrow \left(U|B^{-1}\right) [/tex]

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

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

The next question is to explain why it is possible for an arbitrary [itex] m\times n [/itex] matrix A, to determine a new basis F for [itex] \mathbb{R}^m [/itex] (where [itex] L: \mathbb{R}^n \rightarrow \mathbb{R}^m [/itex]), but still keep the standard basis for [itex] \mathbb{R}^n [/itex], 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 [itex] L(\mathbf{x}) = A\mathbf{x} [/itex], then by using the reduced row echelon form of A, let's call it U, the transformation can be expressed in new basis as

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

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 explaination to use the steps in the 1. question. A hint to how I can explain it would really help.

[tex] L(\mathbf x) = A\mathbf x [/tex]

and

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

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

determined to be

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

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

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

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

[tex] \left(A|I\right) \rightarrow \left(U|B^{-1}\right) [/tex]

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

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

The next question is to explain why it is possible for an arbitrary [itex] m\times n [/itex] matrix A, to determine a new basis F for [itex] \mathbb{R}^m [/itex] (where [itex] L: \mathbb{R}^n \rightarrow \mathbb{R}^m [/itex]), but still keep the standard basis for [itex] \mathbb{R}^n [/itex], 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 [itex] L(\mathbf{x}) = A\mathbf{x} [/itex], then by using the reduced row echelon form of A, let's call it U, the transformation can be expressed in new basis as

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

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 explaination to use the steps in the 1. question. A hint to how I can explain it would really help.

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