- #1

Benny

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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?

[tex]T:R^4 \to R^3 ,T\left( x \right) = Ax[/tex] where [tex]A = \left[ {\begin{array}{*{20}c}

1 & 2 & { - 1} & 1 \\

1 & 0 & 1 & 1 \\

2 & { - 4} & 6 & 2 \\

\end{array}} \right][/tex]

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?

[tex]T:R^4 \to R^3 ,T\left( x \right) = Ax[/tex] where [tex]A = \left[ {\begin{array}{*{20}c}

1 & 2 & { - 1} & 1 \\

1 & 0 & 1 & 1 \\

2 & { - 4} & 6 & 2 \\

\end{array}} \right][/tex]

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?

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