Does a 2×3 matrix span ℝ² using the given theorem?

[Mathman23]

Hi

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

I'm suppose to determain that T maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$

I was told by my professor about the following theorem.

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

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

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:

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

A being the standard matrix of the linear transformation.

A can also be written:

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

This is equal to:


$$\begin{equation}\nonumber<br /> x + 4z = 0<br /> \end{equation}$$

$$\begin{equation}\nonumber<br /> 2x + y + 6z = 0<br /> \end{equation}$$

x in equation 1 can be written as $$x = -4z$$

If I insert that x into equation 2 then I get

$$-2z + y = 0$$

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

Sincerley and many thanks in advance

Fred

 

[HallsofIvy]

Actually this: "I have a linear transformation T which maps $$\mathbb{R}^3 \rightarrow \mathbb{R}^2$$
a A is the standard matrix for the linear transformation.

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

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 $$\mathbb{R}^2$$,there exist x in $$\mathbb{R}^3$$ 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.

 

[M98Ranger]

Hi Fred,

To prove point (a) in the theorem, you need to show that the span of the columns of A is equal to \mathbb{R}^m. In other words, you need to show that any vector in \mathbb{R}^m can be written as a linear combination of the columns of A.

In your case, since A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ], the span of the columns of A will be all possible linear combinations of the three columns. This means that any vector in \mathbb{R}^2 can be written as a linear combination of the columns of A. For example, the vector \left[ \begin{array}{cc} 2 \\ 3 \end{array} \right ] can be written as 2\left[ \begin{array}{cc} 1 \\ 2 \end{array} \right ] + 3\left[ \begin{array}{cc} 0 \\ 1 \end{array} \right ] + 0\left[ \begin{array}{cc} 4 \\ 6 \end{array} \right ]. This shows that the span of the columns of A is equal to \mathbb{R}^2, and therefore, T maps \mathbb{R}^3 \rightarrow \mathbb{R}^2.

I hope this helps. Let me know if you have any other questions.