Matrix Solutions: Find Linearly Independent Solutions

[UOAMCBURGER]

Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x₁+x₂+x₃ = 0
in the textbook it says which solutions are spanned by the vectors f₁ = (−1, 1, 0) ^T and f₂ = (−1, 0, 1) ^T

Homework Equations

The Attempt at a Solution

[/B]
I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x₁ = 1 (since x₁ is the only leading variable) and then clearly the system is only satisfied when (x₁, x₂, x₃) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?

 

[tnich]

Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x₁+x₂+x₃ = 0
in the textbook it says which solutions are spanned by the vectors f₁ = (−1, 1, 0) ^T and f₂ = (−1, 0, 1) ^T[/B]

Homework Equations

The Attempt at a Solution

I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x₁ = 1 (since x₁ is the only leading variable) and then clearly the system is only satisfied when (x₁, x₂, x₃) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?[/B]

What problem are you trying to solve? Finding the kernel of the given matrix?

 

[Ray Vickson]

Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x₁+x₂+x₃ = 0
in the textbook it says which solutions are spanned by the vectors f₁ = (−1, 1, 0) ^T and f₂ = (−1, 0, 1) ^T[/B]

Homework Equations

The Attempt at a Solution

I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x₁ = 1 (since x₁ is the only leading variable) and then clearly the system is only satisfied when (x₁, x₂, x₃) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?[/B]

(1) Turn off the bold font!
(2) The equation $x_1+x_2+x_3=0$ describes a 2-dimensional plane in 3 dimensions. You can express any vector in the plane as a linear combination of two linearly-independent vectors, so that for any point ${\bf p} = (x_1,x_2,x_3)$ in the plane we can write ${\bf p} = c_1 {\bf p}_1 + c_2 {\bf p}_2$. As we vary $(c_1,c_2)$ we sweep out the whole plane.

The solution is merely giving you two possible linearly-independent vectors ${\bf p}_1$ and ${\bf p}_2$ in the plane; there are infinitely many other possible choices.

 

[FactChecker]

Three variables and one linear constraint should tell you that there is a two-dimensional solution space. The solutions (1,-1,0) and (1,0,-1) are clearly two linearly independent solutions. The problem is "rigged" to make those solutions easy to spot. Any other solution, like (0,1,-1), can be obtained by a linear combination of the two: (0,1,-1) = -1(1,-1,0)+1(1,0,-1).
(Your statement that (1,-1,0) and (1,0,-1) are the only solutions is not correct. They form a basis for the entire space of solutions.)

 

[berkeman]

(1) Turn off the bold font!

Fixed.