## Row Space

Each row of matrix A can be defined with a vector in $\Re^n$

The set of linear combinations of the row vectors is the row space denoted as row(A)

So if

(1)But be warned, Row Operations effect the row dependency of a matrix!

However,

If B is the row reduction of A, then row(A) = row(B)

The last two rows of the reduced A matrix are therefore our dependent rows. Thus, when we write our row(A), we leave those out.

row(A) = span$\begin{Bmatrix} (2&-4&-2&3)\\(0&3&1&-1)\\(0&0&0&5) \end{Bmatrix}$

BUT DON"T USE THE REDUCED COLUMNS FOR THE COLUMN SPACE. This is because in our reduced matrix A there are no non-zero values in the fourth and fifth entries in our vectors. This makes it sound like we cannot make a vector that exists in $\Re^4 \text{ or } \Re^5$. We should be able to though, and if we use the original vectors of A, we are able to do so. Do note, however, as we don't have a pivot in the third column, that is our free variable and we remove it from our column space.

Our Column space is therefore

Col(A) = span $\begin{Bmatrix} \begin{bmatrix} 2\\6\\2\\4\\-6 \end{bmatrix} & \begin{bmatrix} -4\\-9\\-7\\-2\\3 \end{bmatrix} & \begin{bmatrix} 3\\8\\9\\-1\\4 \end{bmatrix} \end{Bmatrix}$

## Rank

The rank of a matrix A is the dimension of the column space of A