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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 $dim \mathcal{R}(B^T)=2$.

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    $\begingroup$ 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? $\endgroup$
    – user721971
    Apr 7, 2023 at 2:35

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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 $B$.

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  • $\begingroup$ I don't understood why taking a basis of $W^{\perp}$ is necessary. $\endgroup$
    – piero
    Apr 7, 2023 at 2:54
  • $\begingroup$ 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 $\endgroup$ Apr 7, 2023 at 2:57
  • $\begingroup$ 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. $\endgroup$
    – piero
    Apr 7, 2023 at 3:02

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