Finding a Basis for Im(T): Elementary Linear Algebra

[Benny]

Hi, I'm having trouble with some questions. I did elementary linear algebra a few months ago but seeing as I've forgotten most of it I'm effectively new to this. Anyway can someone help find a basis for Im(T) the following transformation?

$$T:R^4 \to R^3 ,T\left( x \right) = Ax$$ where $$A = \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} & 1 \\ 1 & 0 & 1 & 1 \\ 2 & { - 4} & 6 & 2 \\ \end{array}} \right]$$

I tried applying T to each of the four vectors in the standard basis for R^4 and apply T to each in turn I got: (1,1,2), (2,0,-4), (-1,1,-4), (1,1,2). The basis for Im(T) is supposed to be {(1,1,2),(2,0,-4)} so I've done something wrong since my work shows that my answer should also have (-1,1,-4) in the basis.

Can someone explain what is required to find a basis for Im(T)?

Edit: I re-checked my textbook. Im(T) is just the column space of A isn't it?

 

[George Jones]

I don't think you did the multiplication correctly - I get (-1,1,6), not (-1,1,-4).

Show: 1) (1,1,2), (2,0,-4) are linearly independent; 2) (-1,1,6) is a linear combination of (1,1,2) and (2,0,-4).

Regards,
George

 

[HallsofIvy]

And it's not so much "multiplication" as it is just copying the columns correctly!

 

[Benny]

Thanks for the help guys. The vectors I obtained was from taking each vector from the standard basis for R^4 and applying the transformation to each of those vectors. I found that I could ge tthe correct answer by taking the columns, reducing and then taking the relevant columns. This isn't the same method that I used to get my original answer(which had 3 vectors in it).

 

[HallsofIvy]

They are the same method! George Jones' point was that you multiplied wrong: Your matrix times the basis vector (0, 0, 1, 0) is (-1, 1, 6), not (-1, 1, -4) as you give. My point was that since you are just multiplying by 1 and 0, The result of multiplying each basis vector by the matrix is just the corresponding column. Since (0, 0, 1, 0) has the one in the third place, multiplying it by this matrix is just the third column of the matrix, (-1, 1, 6).

You know that those 4 vectors can't all be independent since this is only 3 dimensional space. It turns out that only two are- the image is a two dimensional subspace of R³. Another way to find a basis, since, as you say, the image is just the "column space", is to take the transpose of the matrix and "row reduce" it.

(Taking the transpose and row reducing is simpler for most people than "column reducing"!)

 

[Benny]

Hmm...I'll have to check that(the method not the calculation since I've just spotted my error). I'm more used to using the method of taking the matrix A, reducing and then taking the columns which correspond to the pivots.