- #1

Mathman23

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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