Say I have a $2\times 3$ matrix with $\operatorname{rank}(A)=1$ and $\operatorname{nullity}(A)=2$, i.e. only one pivot point and two free variables in the solution set of this system because we have interrelationships between the two rows in this matrix. Also, we know that
$$\operatorname{rank}(A)+\operatorname{nullity}(A)=n$$
where $n$ is the number of columns. Now, I transposed this matrix and we know that $\operatorname{rank}(A)=\operatorname{rank}(A^T)$ (linear dependence still present), and when we transpose a matrix the number of columns is the number of rows of the original matrix, and vice versa; i.e. now the matrix is $3\times 2$ and we already know the rank. And with that we can determine the nullity as follows: $$\begin{align*}
\operatorname{rank}(A^T)+\operatorname{nullity}(A^T)&=2\\
1+\operatorname{nullity}(A^T)&=2\\
\operatorname{nullity}(A^T)&=1
\end{align*}$$ Can this be true? Because I'm getting the feel that the transpose should have $\operatorname{nullity}(A^T)=1$ because the number of columns is $2$ now. I googled and all I could find was regarding rank but not so much nullity; the course-book is not very clear (at least for me) when it comes to this, although from the exercises I've done so far it should be the case unless there are some exceptions that I am unaware of.
[Math] Is it possible for a matrix to have nullity different from its transpose
linear algebra
Best Answer
Just consider
$$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$$
The nullspace of $A$ is
$$N(A) = \langle (0,1,0)^T, (0,0,1)^T \rangle$$
which has dimension $2$.
Now the transpose of $A$ is
$$A^T = \begin{pmatrix} 1 & 0\\ 0 & 0\\ 0 & 0\end{pmatrix}$$
The nullspace of $A^T$ is
$$N(A^T) = \langle (0,1)^T \rangle$$
which has dimension $1$.