# basic linear algebra

i have the row reduced form of a matrix:

$\begin{bmatrix}1&0&0&3&4\\0&1&0&1&2\\0&0&1&0&2\end{bmatrix}$

and let $T: R^{5}\to R^{3}$ be the linear transformation defined by $T(x) = Ax$ for all $x$ in $R^5$. Show that $T$ is onto but not 1-1.

1. is the rank of the matrix the number of pivot columns or the number of non-zero rows?
both evaluate to 3.

2. which columns are linearly independent? is it the columns with the leading variables? in this case, would the linearly independent columns be $C_1$, $C_2$, and $C_3$? and so $C_4$, and $C_5$ span $\{C_1,C_2,C_3\}$ ?

3. Is it not one-to-one if there are free variables? in this case, there are free variables $x_4$ and $x_5$, so that means it's not 1-1?.

4. How do i know when it's onto?

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(1) Yes. (2) Any three of them. (3) Exactly. (4) Because the rank of the matrix is three: the rank of the matrix is teh –  Michael Hoppe Feb 14 at 18:34