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?

• (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 '14 at 18:34

(1) Yes.

(2) Any three of them.

(3) Exactly.

(4) Because the rank of the matrix is three: the rank of the matrix is the dimension of the span of its column vectors.

• Thank you, but i'm still confused about the rank, is the rank the # of pivots of the # of non-zero rows? also, for question 4, if the rank of the matrix is equal to the # of rows in the row reduced echelon form, it's onto? – user125535 Feb 14 '14 at 18:40
• In row-echelon form, these end up being the same thing. – gregkow Feb 14 '14 at 18:43
• Read it out loud: “The rank of a matrix is the number of non-vanishing rows in its echelon form. That rank is the dimension of the span of the column vectors of the matrix.” Read it again out loud. Now consider your fourth question again. – Michael Hoppe Feb 14 '14 at 18:54