So i'm working through a LA textbook, and i'm at the end of chapter 3 (span, basis, dimensions).
I thought i was doing ok, i thought i was getting the answers right, and then i came to this question.
I completely do not understand how the answer is obtained.
From elimination I come out to
1 3 4
0 -5 -5
0 0 0
and i don't know what to do now to get that answer.
Also when it asks for a basis, does it mean column or row?
Let's try to understand what is being asked. First let's write the problem in matrix notation:
[eqn]
\left[ \begin{array}{ccc} 1 & 3 & 4 \\ 2 & 1 & 3 \\ -1 & 2 & 1 \\ \end{array} \right]
\left[ \begin{array}{c} x \\ y \\ z \\ \end{array} \right]
=
\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right]
[/eqn]
You are tasked to find the solution set for this system of equations. That means you want to find every [x;y;z] (column) vector that satisfies the equation.
It is a basic fact of LA that the solution set to a linear system either has 0 solutions, 1 solution, or infinite solutions. If you did not know this, we can go over the proof.
So the question is asking you to find a basis for this solution set of (column) vectors, so necessarily the basis will be composed of column vectors. Capiche?
To solve systems of equations, we often use elimination, as you attempted. Unfortunately you didn't correctly calculate the Reduced Row Echelon form of the matrix (don't sweat it, it's really easy to make an arithmetic mistake). The matrix you should arrive at is:
[eqn]
\left[ \begin{array}{ccc} 1 & 3 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right]
[/eqn]
Now that the matrix is in RRE form, it should be immediately obvious that there is only one free parameter (if not, review elimination and free parameters). That tells us that the dimension of the solution set is 1, or in other words a line. To calculate a basis for this line, we just have to find one non-zero vector on the line; any other solution is simply a scaled copy of the basis vector.
For arbitrary problems, you finish the full blown Gauss-Jordan Elimination algorithm to get an answer. In this case, it's not hard to see that only the first equation matters from the RRE form, and it's easy to see by inspection that one solution to this system is [-1;-1;1]. Granted, any non-zero scalar multiple of this vector is also a valid basis vector, so something like [2;2;-2] is also a valid answer.
Make sense?
Wow, you are phenomenal, thanks so much! This all makes a lot more sense to me now.
The problem i am still having now is that i cant get the RRE form :(
I'm not sure what i'm doing wrong.
1 3 4
2 1 3
-1 2 1
to
1 3 4
0 -5 -5
0 5 5
to
1 3 4
0 -1 -1
0 1 1
to
1 3 4
0 1 1
0 0 0
I thought i understood elimination and gauss-jordan elimination but i guess not :(
They're asking for the nullspace.
>0x -5y -5z =0
This tells you z=-y
>1x + 3y + 4z = 0
>1x + 3y - 4y = 0
This tells you x=y
So all solutions are of the form (x,x,-x) for any x. This is a 1 dimension space.
Ha, I fucked up. It's been a while since I last did elimination myself...
You did the Gaussian elimination correctly to get:
[eqn]
\left[\begin{array}{ccc}1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\\end{array}\right]
[/eqn]
Next step is back substitution (which when combined with Gaussian elimination is Gauss-Jordan elimination) to get
[eqn]
\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\\end{array}\right]
[/eqn]
At this stage you have two leading variables (often called pivots) in columns 1 and 2, and a free variable in column 3. I implicitly misidentified them in my previous post, but everything else is ok. You can see this by writing out the full system of equations
[eqn]
\begin{array}{cccc}
x & & & +& z = 0 \\
& & y &+& z = 0 \\
\end{array}
[/eqn]
that if you pick an arbitrary z, then x and y are solved for in terms of z.
z is the free parameter that gives rise to that 1D line of solutions.
ohhh ok!
seriously thanks again, this was so helpful :D
i find the way the text presents things is different from how i learn or something.
its all symbols and few words, and when it defines things with words, they just use more symbols, so it's hard to get a grasp of things.
i really appreciate you taking the time to explain it all to me :D
cheers
Differnet guy here:
How do you deduct that the solution is a line? If the only equation left is 1*x+3*y+4*z=0 doesnt that mean that its a plane?
I would have said that both the second and third column are free parameters but I do assume that you are correct.
It's the solution where two planes intersect (in a line). The two planes being x + z = 0, and y + z = 0
I as ponderig about how the set of convergent series forms a vector space, but what would the basis of such a space be?
Ok thanks for clearing that up.
If the RREF Matrix would have actually been
>1 3 4
>0 0 0
>0 0 0
the solution would have been a plane though, right?