Finding a matrix whose null-space equals a specific subspace

Given the column space $$W=\mathcal{R}(A)$$, $$A= \begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 1 & 2 & 0\\ 1 & 1 & 1\\ \end{pmatrix}$$, I want to find a matrix $$B$$ such that the null space $$\mathcal{N}(B)=W$$.

I found that $$\mathcal{R}(A)$$ is spanned by two vectors only: $$(1,2,0,1),(0,-1,1,0)$$, but I don't know how to find the number of rows that $$B$$ must have. Also, by orthogonal complement decomposition theorem $$dim \mathcal{R}(B^T)=2$$.

• I would let $$b_3=(1,2,0,1)^T \\ b_4=(0,-1,1,0)^T$$ Extend $\{b_3,b_4\}$ to a basis $\{b_1,b_2,b_3,b_4\}$ of $\mathbb{R}^4$. Take $D=\text{diag}(1,1,0,0)$ and let $P$ be a matrix whole columns are $b_1,b_2,b_3,b_4$. Can you show that $B=PDP^{-1}$ works?
– user721971
Apr 7, 2023 at 2:35

You've shown that $$B^T$$ is of rank 2. It is known that a matrix and its transpose have the same rank, so $$B$$ must have a 2-d column space.
However, the number of rows in $$B$$ is not fixed by the given information. A simple way to see this is the fact that once you find a suitable $$B$$, you can tack on a couple of $$(0,0,0,0)$$'s to the bottom, and it won't change anything. In other words, we know that the vectors outputted by $$B$$ must lie on a 2-d plane, but we have full freedom in deciding the dimension of the space that this output plane is embedded in. If you want to be parsimonious and have $$B$$ be of full rank, it should have 2 rows for this reason (and let the output plane be the standard x-y plane).
Now how do we actually find $$B$$? This is somewhat routine. Let $$B$$ be $$\begin{pmatrix}a&b&c&d\\e&f&g&h\end{pmatrix}$$. We know that $$B\cdot(1,2,0,1)$$ and $$B\cdot(0,-1,1,0)$$ both equal $$\begin{pmatrix}0\\0\end{pmatrix}$$. This gives you 4 equations. Now find a basis of $$W^\perp$$, say $$v_1$$ and $$v_2$$, and act on them with $$B$$. This should give you two linearly independent vectors in 2-d space. For simplicity, $$B\cdot v_1=\begin{pmatrix}1\\0\end{pmatrix}$$ and $$B\cdot v_2=\begin{pmatrix}0\\1\end{pmatrix}$$. This should give you another 4 equations. WIth 8 equations and 8 unknowns, you can find $$B$$.
• I don't understood why taking a basis of $W^{\perp}$ is necessary. Apr 7, 2023 at 2:54
• A matrix is characterized by what it does. You know what $B$ does to stuff in $W$, now you want to look at what it does to stuff not in $W$ to be able to know everything about it Apr 7, 2023 at 2:57
• So taking $W^{\perp}=\mathcal{N}(A^T)$ must be equal to $\mathcal{R}(B^T)$, giving additional information about $B$? I think I'm fine so. Apr 7, 2023 at 3:02