# Is is possible to find a basis for the column space of $A$,given reduced row echelon form of matrix $A$ and $A^T$,

Suppose $A$ is a $3$x$4$ matrix and the reduced row echelon form of $A$ is $\begin{pmatrix}1&0&0&1\\0&1&2&2&\\0&0&0&0\end{pmatrix}$

and the reduced row echelon form of $A^T$ is $\begin{pmatrix}1&0&2\\0&1&-1\\0&0&0\\0&0&0\end{pmatrix}$

Find a basis for $R(A)$, where $R(A)$ is the column space of $A$

I don't think this is possible, but in the answer key, it said that $R(A)$ = $R(A^T)^T$, which has basis $$\{(1, 0, 2)^T,(0, 1, -1)^T\}$$

How does this work?

• You mean $R(A^T)^\perp$. $S^T$ doesn't mean anything unless $S$ is a matrix (or vector). – Omnomnomnom Oct 26 '16 at 1:10
• @Omnomnomnom What do you mean by $S$? – user59036 Oct 26 '16 at 1:15
• Oh, excuse me. I think your book really means $$R(A) = R[(A^T)^T]$$ I was confused without the extra brackets. Now it makes sense. – Omnomnomnom Oct 26 '16 at 1:15
• Remember that row-reduction does not change the row-space of a matrix – Omnomnomnom Oct 26 '16 at 1:16
• – BCLC Oct 26 '16 at 4:03

## 2 Answers

TL;DR

Row space of $A^T$ = column space of $A$

When we want to find a basis for the row space of a matrix $A$, we could use the the rows of $A$ except that it is not always the case that the rows of $A$ are linearly independent.

1. So we have to eliminate rows which can be written as linearly combinations of other rows.

2. Now we perform EROs on $A$ until we reach row-echelon form (or reduced row-echelon form) to get row vectors that, like the original matrix $A$, span the row space of $A$.

3. This time however, the row vectors (apart from the zero row vectors) that we get are linearly independent.

4. Thus, we have a basis for the row space of $A$.

For the column space of $A$, the procedure is the same as above if we replace 'row' with 'column'. ECOs however are difficult because we are used to addition vertically as in the EROs.

So instead of performing ECOs on $A$, we perform EROs on $A^T$.

This gives us row vectors (apart from the zero row vectors) that are linearly independent and span the row space of $A^T$, which is equivalent to the column space of $A$.

• do you agree with the commenter? – Anonymous Oct 26 '16 at 4:11
• @Anonymous Not sure about the first comment but Omnomnomnom seems to be right about others – BCLC Oct 26 '16 at 4:17
• lol, I remember you, you commented once with that question in one of my solutions and I had no idea what you were actually trying to say. – Anonymous Oct 26 '16 at 4:23
• @Anonymous LOL – BCLC Oct 26 '16 at 5:13
• LOL yeah that one! – Anonymous Oct 26 '16 at 5:16

The non-zero rows of of the row echelon form of $A^T$ give you a basis of the column space of $A$ if you tranpose them.

The row space of a matrix is preserved as perform elementary row operation. As you perform elementary row operations to the transpose of the matrix, you are actually performing column operations to the original matrix while preserving the column space.

It is known that the non-zero rows of the row echelon forms are linearly independent and hence form a basis to the row space.