How to find a column space and decide whether Ax = b has a solution

A matrix-vector product is a combination of the columns of the matrix. That means we can solve Ax = b exactly when those columns can combine to make b. Let’s find a column-space basis and connect it with the null space.

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Read the pivot positions from a reduced matrix

$A=\left[\begin{array}{rrrr}2& 1& 3& 1\\ 4& -2& 8& 4\\ 5& 6& -2& -3\end{array}\right]$

Row reduction gives the following matrix.

$\left[\begin{array}{rrrr}1& 0& 0& -\frac{3}{31}\\ 0& 1& 0& -\frac{8}{31}\\ 0& 0& 1& \frac{15}{31}\end{array}\right]$

The pivots are in columns 1, 2, and 3. Use those positions to select the corresponding columns from the original matrix A.

Choose the basis from the original columns

A column-space basis is $(2,4,5)$, $(1,-2,6)$, and $(3,8,-2)$. These are three independent vectors in three-dimensional space, so the column space is all of three-dimensional space. For this matrix, $Ax=b$ has a solution for every three-component vector $b$.

Use the free column to find the null-space dimension

There are four variable columns and three pivots. The fourth variable is free. Letting $x_4=31t$ in the reduced system gives

$x=t(3,8,-15,31)$

So the null space has dimension 1 and the column space has dimension 3. The dimensions add to the number of columns, 4. More generally, a matrix with n columns and r pivots has null-space dimension n minus r.

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