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Given that $A = LU$, where $L =$ \begin{bmatrix}1&0&0\\2&1&0\\3&0&1\end{bmatrix} and $U =$ \begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix}

I need to find the following without extensive calculation to show insight into the topic:

• the rowspace of $A$.

• the columnspace of $A$.

• the nullspace of $A$.

• the nullspace of $A^T$.

I know that the rowspace is just the non-zero rows from U. The columnspace is the columns of $L$ that correspond to the pivot columns of $U$. The nullspace is the nullspace of $U$. But I can't find an efficient way to calculate a basis for $\text{null}(A^T)$. I know that $\text{null}(A^T)$ is the orthogonal projection of $\text{col}(A)$, but I still can't manage to find a short, easy solution based on some properties without calculating $A^T$ and doing it that way.

If anyone knows some properties to solve this, please let me know :)

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  • $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented Aug 15 at 15:25
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    $\begingroup$ Hint: $A^T = U^T L^T$. $\endgroup$ Commented Aug 17 at 0:35

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Note that $A^T = U^TL^T$. $$U^T = \begin{bmatrix} 1 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}$$ $$L^T = \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ By inspection we know that the vectors that get sent to $\vec{0}$ under $U^T$ are $\begin{bmatrix}0,0,t\end{bmatrix}^T$ for $t\in\mathbb{R}$, so we just need to find the vectors that get sent to $\begin{bmatrix}0,0,t\end{bmatrix}^T$ under $L^T$. I went ahead and simply did $$\begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ t \end{bmatrix}$$ The resulting system of equations has the solution $$\begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} -3t\\ 0\\ t \end{bmatrix}$$ for $t\in\mathbb{R}$. This solves the problem.

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