Question regarding linear transformation

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Hi

I have a linear transformation T which maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex] a A is the standard matrix for the linear transformation.

I'm suppose to determain that T maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]

I was told by my professor about the following theorem.

If [tex]T:\mathbb{R}^n \rightarrow \mathbb{R}^m[/tex] is a linear transformation and A is the standard matrix for T. Then

a/ T maps [tex]\mathbb{R}^n \rightarrow \mathbb{R}^m [/tex] if and only if A [tex] \mathrm{span} \{ \mathbb{R}^m \}[/tex]

b/ T is one-to-one if and only if columns of A are linearly independent.

If I then apply (a) from to my problem:

[tex]A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ][/tex]

A being the standard matrix of the linear transformation.

A can also be written:

[tex]\left[ \begin{array}{cc} 1 \\2 \end{array} \right ] \mathrm{x} + \left[ \begin{array}{cc} 0 \\1 \end{array} \right ] \mathrm{y} + \left[ \begin{array}{cc} 4 \\6 \end{array} \right ] \mathrm{z} = \left[ \begin{array}{cc} 0 \\0 \end{array} \right ][/tex]

This is equal to:


[tex]\begin{equation}\nonumber
x + 4z = 0
\end{equation}[/tex]

[tex]\begin{equation}\nonumber
2x + y + 6z = 0
\end{equation}[/tex]

x in equation 1 can be written as [tex]x = -4z[/tex]

If I insert that x into equation 2 then I get

[tex]-2z + y = 0[/tex]

But what do I then do to prove point (a) in the theorem ???

Sincerley and many thanks in advance

Fred
 

Answers and Replies

  • #2
HallsofIvy
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Actually this: "I have a linear transformation T which maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]
a A is the standard matrix for the linear transformation.

I'm suppose to determain that T maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]" doesn't quite make sense! I assume you mean simply that T is from[tex]\mathbb{R}^3 [/tex]to [tex]\mathbb{R}^2[/tex] and you want to prove this is an "onto" mapping.
Okay, you are given A= [tex]A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ][/tex]

I don't understand why you then set Ax= 0: that would prove that the columns are linearly independent which is your professor's (b), not (a) (and this is clearly not one to one). To show that A is "onto", you need to show that for all y in [tex]\mathbb{R}^2[/tex],there exist x in [tex]\mathbb{R}^3[/tex] so that Ax= y. The simplest way to do that is to row-reduce A. If the second row does not reduce to all 0s, it's true.
 

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