# Find a basis for the solution space of the homogeneous system of equations.

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• Oct 8th 2010, 04:12 PM
linalg123
Find a basis for the solution space of the homogeneous system of equations.
Find a basis for the solution space of the homogeneous system of equations.

2x1 + x2 + x3 + x4 = 0,
3x1 + x2 - x3 + 2x4 = 0,
x1 + x2 + 3x3 = 0,
x1 - x2 - 7x3 + 2x4 = 0

I tried putting it in row-echelon form which gave me

[ 2 1 1 1 ]
[ 0 -1 -5 1 ]
[ 0 0 0 0 ]
[ 0 0 0 0 ]

let x2=0 , x3=1 which gives [-3 0 1 5]
x4 =5 , x1= -3

let x3=0, x2=x4=1 which gives [-1 1 0 1]
x1=-1

and dim 2..

but i have a feeling this may be wrong?
Thanks in advance
• Oct 8th 2010, 04:38 PM
Ackbeet
Well, I agree with your row reduction. However, I'm not sure I follow your back substitution. I would probably substitute in for the higher-indexed variables first. For example, back substitution yields

$\displaystyle \displaystyle x_{2}=-5x_{3}+x_{4},$ and

$\displaystyle \displaystyle x_{1}=-\frac{1}{2}x_{2}-\frac{1}{2}x_{3}-\frac{1}{2}x_{4}=-\frac{1}{2}(-5x_{3}+x_{4})-\frac{1}{2}x_{3}-\frac{1}{2}x_{4}=2x_{3}-x_{4}.$

Generally, to find the basis required, you can't plug in specific values for variables. In the case of this underdetermined system, you're going to have to use parameters. That is, suppose $\displaystyle x_{3}=s$ and $\displaystyle x_{4}=t.$ Can you write the whole solution in terms of these two parameters? If so, what do you suppose the solution space looks like?

Can you continue from here?
• Oct 8th 2010, 04:52 PM
linalg123
Ok, i understand up to getting x2 and x1.

I don't know what you mean by this though.
Quote:

Can you write the whole solution in terms of these two parameters? If so, what do you suppose the solution space looks like?
does it mean that the basis is x3 and x4?
• Oct 8th 2010, 05:45 PM
Ackbeet
What you want to do is write the solution vector $\displaystyle \mathbf{x}$ as follows:

$\displaystyle \mathbf{x}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{ 4}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\d\end{bmat rix}+s\begin{bmatrix}e\\f\\g\\h\end{bmatrix}+t\beg in{bmatrix}i\\j\\k\\l\end{bmatrix},$

where all the letters $\displaystyle a-l$ represent numbers. Since your system is homogeneous, that will have implications for the numbers $\displaystyle a-d.$

Does this make sense?
• Oct 8th 2010, 06:25 PM
linalg123
i imagine it makes sense if you wrote it but i don't understand :(

if we have

x1= 2x3 - x4 and
x2= -5x3 + x4
x3 = x3
x4 = x4,

wouldn't a-d all be zero?
e=2
f=-5
g=1
h=0
i=-1
j=1
k=0
l=1?
• Oct 8th 2010, 06:30 PM
Ackbeet
Correct. Now you just need to interpret your results. Write out this:

$\displaystyle \mathbf{x}=s\begin{bmatrix}2\\-5\\1\\0\end{bmatrix} +t\begin{bmatrix}-1\\1\\0\\1\end{bmatrix}.$

So all solutions look like linear combinations of what?
• Oct 8th 2010, 06:43 PM
linalg123
the two vectors?
• Oct 8th 2010, 07:15 PM
Ackbeet
Which two vectors?
• Oct 8th 2010, 07:23 PM
linalg123
[2 -5 1 0] and [-1 1 0 1] ?
• Oct 8th 2010, 07:25 PM
Ackbeet
Right. So now check: are those two vectors linearly independent? You've pretty much shown already that they span the solution space (i.e., every solution can be written as in post # 9). If they span the solution space, and they're linearly independent, then by definition they are a basis for the solution space. Which is the problem you stated in the OP.

Make sense? Could you do another problem like this on your own now?
• Oct 8th 2010, 07:35 PM
linalg123
Awesome. Yes i think i could do another one of these now. Thanks so much, you're a great teacher
• Oct 8th 2010, 07:37 PM
Ackbeet
Thanks for the compliment. You're very welcome for the help! Have a good one.
• Oct 9th 2010, 03:30 AM
HallsofIvy
Quote:

Originally Posted by linalg123
Awesome. Yes i think i could do another one of these now. Thanks so much, you're a great teacher

Yes, he is. Now, if he could just convince the department chair and dean of that!