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Say I reduced a particular matrix to its reduced row echelon form and have this: $ \begin{bmatrix} 1 & 0 & 5 & 0\\ 0 & 1 & 3 & 0\\ 0 & 0 & 0 & -2 \end{bmatrix} $

And I let $\vec{b}= \begin{bmatrix} x\\ y\\ z \end{bmatrix}$ so that $Ak=b$ looks like this: $ \begin{bmatrix} 1 & 0 & 5 & 0\\ 0 & 1 & 3 & 0\\ 0 & 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} k_{1}\\ k_{2}\\ k_{3}\\ k_{4} \end{bmatrix}=\vec{b} $

Now, I can see that the column space of this matrix $A$ is likely to be a plane. But how do I find the equation of the plane, without using cross product? I could get to something like:

$-2k_{4}=z$

$k_{2}+3k_{3}=y$

$k{1}+5k_{3}=x$

But this is still far from the equation of the plane. I don't intend to use cross product because I am thinking if I could extend this to higher dimensions which cross product could be very cumbersome.

So, how do I derive the equation of the plane of the column space of matrix A from here?

Thanks for any help!

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    @xEnOn: I don't know if my answer will be clearer now, but I think you are confused about what the column space is or how it relates to the equations $A\mathbf{k}=\mathbf{b}$; the column space is not the set of solutions, it's the set of $\mathbf{b}$s for which the equation $A\mathbf{k}=\mathbf{b}$ has solutions. In your examples, that set is all of $\mathbb{R}^3$, not a plane.2011-06-29

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You seem to be confusing the solution set of $A\mathbf{x}=\mathbf{b}$ with the column space of $A$.

Remember: the system $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $\mathbf{b}$ is in the column space. But the solutions themselves (the $\mathbf{x}$ that satisfy the equation) are not in the column space generally. Here, they don't even "live" in the same space, as the column space of $A$ is a subspace of $\mathbb{R}^3$, but the solutions are vectors in $\mathbb{R}^4$.

Here, the column space of $A$ has dimension $3$: the column space includes the vectors $\left(\begin{array}{c}1 \\0\\0\end{array}\right),\quad \left(\begin{array}{c}0\\1\\0\end{array}\right),\quad\text{and}\quad \left(\begin{array}{r}0\\0\\-2\end{array}\right),$ which clearly span all of $\mathbb{R}^3$. So the column space is not a plane, it is all of $\mathbb{R}^3$.

I think your confusion lies in the following: the number of parameters (free variables/non pivot columns) determines the dimension of the solution space to $A\mathbf{x}=\mathbf{b}$ when it has a solution. And in order to decide whether $A\mathbf{x}=\mathbf{b}$ has a solution, you look at the column space. The system $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $\mathbf{b}$ lies in the columnspace of $A$.

Because here the column space is all of $\mathbb{R}^3$, the system has solutions for every possible choice of $\mathbf{b}\in\mathbb{R}^3$. The solution has dimension $\mathrm{nullity}(A)$, which by the Rank-Nullity Theorem is equal to the number of columns of $A$ minus the rank of $A$, i.e., the number of free variables/number of parameters/number of non pivot columns in $A$. Here, the rank is $3$, so the nullity is $1$. That is: the solution space has dimension $1$, and so will be a line in $\mathbb{R}^4$.

Then, to get the equation of that line, we have $\begin{align*} k_1 +5k_3 &= x\\ k_2 + 3k_3 &= y\\ k_4 &= -\frac{1}{2}z. \end{align*}$ Note that $x$, $y$, and $z$ are constants (they are the entries of your $\mathbf{b}$, and the coordinates in $\mathbb{R}^4$ (where the line "lives") are given by components called $k_1$, $k_2$, $k_3$, and $k_4$.

The system of lines in $\mathbb{R}^4$ that are solutions to the different systems are parametrized by your $x$, $y$, and $z$, but since they can be any point in $\mathbb{R}^3$, there is no relation among them; the collection of all $\mathbf{b}$s for which the system is consistent is all of $\mathbb{R}^3$, not a plane either.

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    @xEnOn: My pleasure, and you're very welcome.2011-06-29