Column Space of a matrix: best way to find it

[Trenters4325]

Is the best way to find the column space of a matrix to take the transverse, row reduce it, and then take the row vectors as your independant basis?

 

[galactus]

If a matrix is in row-echelon form, then the column vectors with the leading 1's of the row vectors form a column space.

For instance, let's choose a matrix we'll say in in REF.

\(\displaystyle \L\\\begin{bmatrix}1&-2&5&0&3\\0&1&3&0&0\\0&0&0&1&0\\0&0&0&0&0\end{bmatrix}\)

Let's call the row vectors r1, r2. r3.

The row vectors are \(\displaystyle \L\\r_{1}=\begin{bmatrix}1&-2&5&0&3\end{bamtrix}\).

\(\displaystyle \L\\r_{2}=\begin{bmatrix}0&1&3&0&0\end{matrix}\)

\(\displaystyle \L\\r_{3}=\begin{bmatrix}0&0&0&1&0\end{bmatrix}\)

These are a basis for the row space.

Now, the column space is, as we said, made up of the row vectors with the leading 1's. Let's call them c1, c2, c3.

\(\displaystyle \L\\c_{1}=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\)

\(\displaystyle \L\\c_{2}=\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}\)

\(\displaystyle \L\\c_{3}=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\)


See?.

 

[Trenters4325]

What if that matrix was reduced from one in which there were non-zero entries in the bottom row? Wouldn't that basis be wrong because, with linear combinations of the original columns, you could get vectors having a fourth dimension (a non-zero value in the bottom slot)? Are you claiming that that is a basis for the column space of just that matrix or of every matrix that row reduces to that matrix?

 

[Trenters4325]

OK, I understand. I was right -- that is only a basis for the matrix in row reduced echelon form. You do not need to respond again.