How to find the null space and column space of a transpose

Transposing a matrix swaps the roles of its rows and columns. That changes the size of the input vectors and output vectors we use when finding its spaces. Let’s transpose a four-row matrix and work out both spaces.

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Write the transpose and track its dimensions

$M=\left[\begin{array}{rr}2& 0\\ -1& 4\\ 3& -2\\ 0& 7\end{array}\right]$

$M^T=\left[\begin{array}{rrrr}2& -1& 3& 0\\ 0& 4& -2& 7\end{array}\right]$

The transpose has two rows and four columns. Its null space lies in four-dimensional input space. Its column space lies in two-dimensional output space.

Solve the homogeneous system for the null space

$\begin{array}{l}2x_1-x_2+3x_3=0\\ 4x_2-2x_3+7x_4=0\end{array}$

Let $x_3=s$ and $x_4=t$. The second equation gives $x_2=\frac{1}{2}s-\frac{7}{4}t$. Substituting into the first gives $x_1=-\frac{5}{4}s-\frac{7}{8}t$.

Scaling the two parameter vectors separately gives the null-space basis $(-5,2,4,0)$ and $(-7,-14,0,8)$. There are two free variables, so this null space has dimension $2$.

Find the column space from the original columns of the transpose

The first two columns of $M^T$ are $(2,0)$ and $(-1,4)$. They are independent, so they form a basis for its column space. The column space is all of two-dimensional space and has dimension $2$.

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