I have
$$A = \begin{pmatrix} 1 & 2 \\ 2 & 2 \\ 2 & 1 \end{pmatrix}$$
I have found that $range(A) = \left\{ \begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}, \begin{pmatrix} 2\\ 2\\ 1 \end{pmatrix} \right\} $
and that $null(A^T) = \left\{ \begin{pmatrix} 1\\ -\frac{3}{2}\\ 1 \end{pmatrix} \right\} $
I have been asked to state the rank and nullity theorem for $A$ and $A^T$. I know the rank and nullity theorem as rank($A$) + nullity($A$) = $n$ where A is an mxn matrix.
I know that nullity($A$) is the number of vectors present in the null space of matrix A, but I'm not too sure how to go on from there.
Can someone help? The solution states dim(null($A^T$)) + dim(range($A$)) = 3
Yet I'm not sure how they arrived there
Since the columns of $A$ are linearly independent, the rank of $A$ is simply $2$. Since $A$ has only two columns the Rank-Nullity theorem for $A$ simply says that the nullity of $A$ is $0$, rank($A$) + nullity($A$) = n = 2. Since the rank of $A$ is two, then the dimension of the rowspace is two and in particular this means that the rows are linearly dependent. Thus $A^T$ has two linearly independent columns and so its rank is also $2$. But as it has three columns, the nullity must be $1$ to satisfy the Rank-Nullity theorem.
By simple Gaussian elimination, we get the following Gauss-Jordan forms for $A$ and $A^T$: \begin{align*}GJ(A) = \begin{pmatrix} 1&0\\ 0&1\\ 0&0 \end{pmatrix}, \end{align*} and \begin{align*} GJ(A) = \begin{pmatrix} 1&0&0\\ 0&1&0\\ \end{pmatrix}. \end{align*} From here it is easy to see the ranks and nullities.