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

linear algebramatrices

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?

Best Answer

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 $A$.

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