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

\begin{align} A = \begin{bmatrix} 1&3&0\\2&1&1\\0&-1&2\\-1&0&1 \end{bmatrix} \end{align}
\begin{align} \vec{R}_1 = (1,3,0) \\ \vec{R}_2 = (2,1,1) \\ \vec{R}_3 = (0,-1,2) \\ \vec{R}_4 = (-1,0,1) \\ \end{align}

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

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

\begin{align} A = \begin{bmatrix} 2&-4&-2&3\\6&-9&5&8\\2&-7&-3&9\\4&-2&-2&-1\\-6&3&3&4 \end{bmatrix} \implies \begin{bmatrix} 2&-4&-2&3\\0&3&1&-1\\0&0&0&5\\0&0&0&0\\0&0&0&0 \end{bmatrix} \end{align}

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 $\begin{Bmatrix} \begin{bmatrix} 2\\6\\2\\4\\-6 \end{bmatrix} & \begin{bmatrix} -4\\-9\\-7\\-2\\3 \end{bmatrix} & \begin{bmatrix} 3\\8\\9\\-1\\4 \end{bmatrix} \end{Bmatrix}$


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

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