Hi:
I see an example about nullspace and orthogonality, the example is following:

$$Ax=\begin{bmatrix} 1 & 3 &4\\ 5 & 2& 7 \end{bmatrix} \times \left[ \begin{array}{c} 1 \\ 1\\-1 \end{array} \right]=\begin{bmatrix} 0\\0\end{bmatrix}$$

The conclusion says the nullspace of $A^T$ is only the zero vector(orthogonal to every vector). I don't know why the columns of A and nullspace of $A^T$ are orthogonal spaces.
I know nullspace is the solution of Ax=0; but in this theorem, why columns of A is related
to nullsapce of $A^T$.
Thanks.

Last edited:

Notice the null space of A^T is orthogonal to (1,5), (3,2) and (4,7). So it is orthogonal to the space spanned by the vectors, which is the column space of A.

HallsofIvy
$$A^T= \begin{bmatrix}1 & 5 \\ 3 & 2 \\ 4 & 7\end{bmatrix}$$
so that the condition that a vector be in the null space of $A^T$ is
$$\begin{bmatrix}1 & 5 \\ 3 & 2 \\ 4 & 7\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix}x+ 5y \\ 3x+ 2y \\ 4x+ 7y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$