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My question is as follows:

Find a basis for the row space and a basis for the column space by first reducing the matrix to row echelon form:

$$ A =\left[\begin{array}{rrr} 5 & 9 & 3 \\ 3 & -5 & -6 \\ 1 & 5 & 3 \\ \end{array}\right] $$

Row operations:

$$(1/5) R_1 \rightarrow R_1$$ $$(-3) R_1 + R_2 \rightarrow R_2$$ $$(-1) R_1 + R_3 \rightarrow R_3$$ $$(-5/52) R_2 \rightarrow R_2$$ $$(-16/5) R_2 + R_3\rightarrow R_3$$

Gives us the matrix $A$ in row echelon form:

$$ A =\left[\begin{array}{rrr} 1 & 9/5 & 3/5 \\ 0 & 1 & 3/4 \\ 0 & 0 & 0 \\ \end{array}\right] $$

Now, we find the column space:

$$C_1 =\left[\begin{array}{rrr} 1 \\ 0 \\ \end{array}\right] \in \mathbb{R^2} $$

$$C_2 =\left[\begin{array}{rrr} 9/5 \\ 1 \\ \end{array}\right] \in \mathbb{R^2} $$

$$C_3 =\left[\begin{array}{rrr} 3/5 \\ 3/4 \\ \end{array}\right] \in \mathbb{R^2} $$

Now, we find the row space:

$$r_1 =\left[\begin{array}{rrr} 1 & 9/5 & 3/5 \\ \end{array}\right] \in \mathbb{R^3} $$

$$r_2 =\left[\begin{array}{rrr} 0 & 1 & 3/4 \\ \end{array}\right] \in \mathbb{R^3} $$

Column spaces $$ C_1, C_2, C_3 $$ and my row spaces $$ r_1, r_2 $$

Is this all wrong? How is this done correctly?

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  • $\begingroup$ Doing the row operations showed you that you can take linear combinations and delete the last row which means the first two rows in the original matrix span the rows. And now looking at A transpose and putting that in row echelon will show you that the first two columns span. $\endgroup$ – Mr.Fry May 12 '15 at 22:28
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http://www.math.jhu.edu/~jmb/note/rowcol.pdf

This is much better and apply it to your example first and when you get something comment.

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Note that $$ \DeclareMathOperator{rref}{rref}\rref \begin{bmatrix} 5 & 9 & 3 \\ 3 & -5 & -6 \\ 1 & 5 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -3/4\\ 0 & 1 & 3/4\\ 0 & 0 & 0 \end{bmatrix} $$ This tells us that $\DeclareMathOperator{rank}{rank}\rank(A)=2$ so any basis for $\DeclareMathOperator{Row}{Row}\Row(A)$ and $\DeclareMathOperator{Col}{Col}\Col(A)$ consists of two vectors.

For $\Col(A)$ we need only select two linearly independent columns of $A$. A nice trick is to select the two columns of $A$ that correspond to the pivot columns of $\rref(A)$. This gives \begin{align*} \begin{bmatrix} 5\\3\\1 \end{bmatrix}&& \begin{bmatrix} 9\\-5\\5 \end{bmatrix} \end{align*} as a basis for $\Col(A)$.

To find a basis for $\Row(A)$ we can choose the first two columns. This gives \begin{align*} \begin{bmatrix} 5&9&3 \end{bmatrix}&& \begin{bmatrix} 3&-5&-6 \end{bmatrix} \end{align*} as a basis for $\Row(A)$.

To find a basis for $\Row(A)$ in general we can compute $\rref(A^\top)$ and select the rows of $A$ that correspond to the pivot columns of $\rref(A^\top)$.

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