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This is a question that has stumped me. It asks "Without computing A, find bases for the four fundamental subspaces." The $LU$ equation is as follows:

$$\begin{bmatrix} 1 & 0 & 0 \\ 6 & 1 & 0 \\ 9 & 8 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$

I understand that $$N(A) = \begin{bmatrix} 0 \\ 2 \\ -1 \\ 1 \end{bmatrix} $$ and that there is no $N(A^T)$ because there are no zero rows in $U$, but I do not know what to do for the column/row spaces without computing $A$. I could compute columns 1, 2, and 3 because it's not "$A$" but I know that's not what they want.

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up vote 0 down vote accepted

The row space of $A=LU$ is the same as the rowspace of $U$ since $L$ is not singular. $L$ represents only a change of basis on the rows of $U$. The column space is the full span, since the null space is nothing.

Also, I think this is the null space: $$N(A) = \begin{bmatrix} 0 \\ 1 \\ -2 \\ 1 \end{bmatrix}$$

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Yes, you are correct. I copied my answer down incorrectly. I completely forgot about how there were three pivots and the vectors were in R^3. Thank you for pointing that out! – Goyatuzo Oct 4 '12 at 18:30

If we are given two matrices $A$ and $B$, then the columns of $AB$ are linear combinations of the columns of $A$ and the rows of $AB$ are the linear combinations of the rows of $B$. This follows immediately from block multiplication $$AB = \begin{pmatrix} A\mathbf{b_1} & \cdots & A\mathbf{b_n}\end{pmatrix} = \begin{pmatrix} \mathbf{a_1}^\mathrm{T}B \\ \vdots \\ \mathbf{a_m}^\mathrm{T}B\end{pmatrix}$$ where $a_i$ and $b_i$ denote the row/column vectors of $A$ and $B$ respectively. From this, what can you conclude?

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