Basis for Im(T)

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?

Last edited:

Staff Emeritus
Gold Member
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

Homework Helper
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).