Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
1  
@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
show 2 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.