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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:




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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Actually, since the rank of $A$ is clearly $3$, and the columns all live in $\mathbb{R}^3$, the column space of $A$ must be all of $\mathbb{R}^3$, not a plane at all. Why do you think "the column space of $A$ is likely to be a plane"? –  Arturo Magidin Jun 29 '11 at 19:20
I am thinking that it could be a plane because the last 2 columns are non-pivot columns. Since only the first 2 columns are pivot columns, which means this matrix only offers 2 dimensions in the space, which could possibly be plane, is this right? –  xenon Jun 29 '11 at 19:22
@xEnOn: No, it's not right. You are confusing solutions with columnspace. The solutions to $A\mathbf{x}=\mathbf{0}$ form a $2$-dimensional subspace of $\mathbb{R}^4$, but the column space of $A$ is the range of $A$. The column space of $A$ clearly contains $(1,0,0)^T$, $(0,1,0)^T$ (the first and second column), and also $(0,0,1)^T = \frac{1}{5}(2,3,5)^T - \frac{2}{5}(1,0,0)^T - \frac{3}{5}(0,1,0)^T$ (which is a linear combination of the first three columns); so the column space, which "lives" in $\mathbb{R}^3$, is all of $\mathbb{R}^3$. –  Arturo Magidin Jun 29 '11 at 19:25
ohh..actually I got what you meant and edited my question before realising you had a new reply. I updated my question a little to the equation. Sorry for the confusion. Now, is this a reduced row echelon form that could have a plane for its column space? Or is it still not? –  xenon Jun 29 '11 at 19:28
@xEnOn: No, your column space is still of dimension $3$, and in any case that's not what you are trying to do, I think. You are trying to figure out the equations that describe the solution set; the dimension of the solution set is the number of columns minus the dimension of the column space, so in this case the solutions form an affine variety of dimension $1$, not $2$. That is, the solutions form a line, not a plane. And they are not in the column space, the column space is still dimension $3$ sitting in $\mathbb{R}^3$, so all of $\mathbb{R}^3$. –  Arturo Magidin Jun 29 '11 at 19:37
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1 Answer

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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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Thanks so much! Just to clarify a little more, the column space is the column combinations of $A$ and therefore the column space is all the values of b that $Ax$ can attain, right? As for the solution space, we are referring to all the possible values of $x$ in $Ax=b$ when give an specific $b$? –  xenon Jun 29 '11 at 20:16
@xEnOn: The column space of an $n\times m$ matrix ($n$ rows, $m$ columns) is the subspace of $\mathbb{R}^n$ that is spanned by the columns of $A$; yes, the column space of $A$ is the range of the map that sends $\mathbf{x}\in\mathbb{R}^m$ to $A\mathbf{x}\in \mathbb{R}^n$, i.e., "all values that $Ax$ can attain). The solution space/set (it's not a subspace unless $b=0$) is indeed the set of all solutions: all $x\in\mathbb{R}^m$ such that $Ax=b$, where $b$ has been given and is fixed. –  Arturo Magidin Jun 29 '11 at 20:20
Thank you so much Arturo Magidin! I think I finally cleared this confusion that I have had for some time. Thank you so much for your help! :) –  xenon Jun 29 '11 at 20:27
@xEnOn: My pleasure, and you're very welcome. –  Arturo Magidin Jun 29 '11 at 20:28
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