# Without elimination, find bases for the four subspaces for this matrix. [GStrang P191 3.6.6]

Without elimination, find bases for the four subspaces for $A = \begin{bmatrix} 0 & 3 & 3 & 3 \\ & \color{green}{\mathbf{0}} & &\\ 0 & 1 & 0 & 1 \end{bmatrix}$.

Without elimination, I don't apprehend how to do so. Though with elimination, I can:

$REF(A) = \begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ & \mathbf{0} & &\\ \end{bmatrix}$. There are $2$ pivots, one each in column $2$ and $3$, so $rank(A) = 2$.
Basis for $C(A) = \{\text{ pivot column 2 of A, pivot column 3 of A }\}$
Basis for $C(A^T) = \{\text{ the two nonzero rows of$A$or$REF(A)$}\}$
Basis for $Null(A) = \{ \, (\color{#B8860B}1, 0, 0, \color{#B8860B}0)^T, (\color{#B8860B}0, -1, 0, \color{#B8860B}1 \}$ where the bronze denote the special solutions.
Since the 1st and 3rd rows of $\mathbf{A \neq 0}$, thus by inspection and by column-row multiplication, $\mathbf{y^TA = 0 \implies y^T} = (0, \color{green}{\text{ any scalar }}, 0)$ = basis $N(A^T)$. $\quad \square$

The row space is generated by two nonzero vectors. It is easy to see these are linearly independent, so they form a basis for the row space.

Similarly, the column space is generated by three vectors. Two are identical, and the other is independent of the duplicated vector, so we obtain from these a basis of two vectors for the column space.

For the null-space of $A$, we note first that it must be 2-dimensional by the rank-nullity theorem. Since the first column is null, the vector $(1,0,0,0)$ is in the null space. To find a second vector in the nullspace, write out $Ax = 0$ explicitly.

$x_1\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} + x_2\begin{bmatrix} 3 \\ 0 \\ 1 \\ \end{bmatrix} + x_3\begin{bmatrix} 3 \\ 0 \\ 0 \\ \end{bmatrix} + x_4\begin{bmatrix} 3 \\ 0 \\ 1 \\ \end{bmatrix} = 0$.

The 2nd column equals the 4th, so subtracting the 4th from the 2nd gives zero. This is the same as $Ax$ when $x = (0,1,0,-1)$. Both vectors are linearly independent, so they form a basis.

By the rank-nullity theorem again, we see that the null space of $A^T$ has dimension $1$, so all we need to do is find one non-zero vector in this space. Since the second column of $A^T$ is null, $(0,1,0)$ works.

• Thank you! I upvoted. I edited your answer to dilate on the origin of the second vector in the nullspace. Does it match with your experimentation? Is this an easier way? Please let me know. If it's right, then I'll accept. – Greek - Area 51 Proposal Dec 2 '13 at 7:48
• @LePressentiment Yes, that is right. – Potato Dec 2 '13 at 13:06