Basis for set of solutions for linear equation

[TheFallen018]

[solved] Basis for set of solutions for linear equation

Hi,

I have this problem I was working through, but I'm not sure that I've approached it from the right way. The problem consists of 3 parts, which build off of each other. I'm pretty confident about the first two parts, but no so much for the third. Here's the problem

Consider the linear equation $x+y-2z=0$
A). Find all solutions of this equation in vector form:
B). Give two vectors that span the set of solutions:
C). Prove that those two vectors are linearly independent and hence give a basis for the set of solutions

The first part, I basically pointed out that y and z are free variables, and expressed them as such
$y=t$
$z=s$
$x=-t+2s$
Solutions given by $(-t+2s,t,s)$

Then the second part, I just split up that vector into two separate vectors:
$t(-1,1,0)+s(2,0,1)$

For the third part, I used a basic proof that two vectors in ${R}^{n}$ are linearly depended iff one vector is a scalar multiple of the other. Therefore they are linearly independent.

Here's where I'm really not sure about much at all. I basically used the initial system, and turned the two vectors in part 2 into linear equations.
$x+y-2z=0$
$-x+y=0$
$2x+z=0$
I then used these three systems to make a 3x3 matrix which I row reduced. It came out as a matrix with a unique solution, so I thought the basis would be $(1,0,0),(0,1,0),(0,0,1)$, or maybe the initial equations expressed as vectors $(1,1,-2),(-1,1,0),(2,0,1)$.

Though, I'm really not sure here, as these span ${R}^{3}$ and seem a little too broad for what we are after. Any help would be great. Thanks :)

 

[I like Serena]

Hi,

I have this problem I was working through, but I'm not sure that I've approached it from the right way. The problem consists of 3 parts, which build off of each other. I'm pretty confident about the first two parts, but no so much for the third. Here's the problem

Consider the linear equation $x+y-2z=0$
A). Find all solutions of this equation in vector form:
B). Give two vectors that span the set of solutions:
C). Prove that those two vectors are linearly independent and hence give a basis for the set of solutions

The first part, I basically pointed out that y and z are free variables, and expressed them as such
$y=t$
$z=s$
$x=-t+2s$
Solutions given by $(-t+2s,t,s)$

Then the second part, I just split up that vector into two separate vectors:
$t(-1,1,0)+s(2,0,1)$

For the third part, I used a basic proof that two vectors in ${R}^{n}$ are linearly depended iff one vector is a scalar multiple of the other. Therefore they are linearly independent.

Hi fallen angel,

It's all correct so far.

Here's where I'm really not sure about much at all. I basically used the initial system, and turned the two vectors in part 2 into linear equations.
$x+y-2z=0$
$-x+y=0$
$2x+z=0$
I then used these three systems to make a 3x3 matrix which I row reduced. It came out as a matrix with a unique solution, so I thought the basis would be $(1,0,0),(0,1,0),(0,0,1)$, or maybe the initial equations expressed as vectors $(1,1,-2),(-1,1,0),(2,0,1)$.

Though, I'm really not sure here, as these span ${R}^{3}$ and seem a little too broad for what we are after. Any help would be great. Thanks :)

The equation $x+y-2z=0$ represents a plane. Its normal vector is $(1,1,-2)$.
And you've found that $(-1,1,0)$ and $(2,0,1)$ are 2 independent vectors in that plane.
Note that they are both perpendicular to $(1,1,-2)$.

However, when we turn $(-1,1,0)$ into the equation $-x+y=0$, we're defining a new plane that is perpendicular to $(-1,1,0)$.
In other words $(-1,1,0)$ is not in that new plane.

Similarly the 3rd equation is yet another plane.
These 3 planes intersect each other uniquely in the origin.

 

[TheFallen018]

Hi fallen angel,

It's all correct so far.
The equation $x+y-2z=0$ represents a plane. Its normal vector is $(1,1,-2)$.
And you've found that $(-1,1,0)$ and $(2,0,1)$ are 2 independent vectors in that plane.
Note that they are both perpendicular to $(1,1,-2)$.

However, when we turn $(-1,1,0)$ into the equation $-x+y=0$, we're defining a new plane that is perpendicular to $(-1,1,0)$.
In other words $(-1,1,0)$ is not in that new plane.

Similarly the 3rd equation is yet another plane.
These 3 planes intersect each other uniquely in the origin.

Hmm, ok, so I wasn't on the right path there. It seems after doing some research, that a basis is basically a linearly independent spanning set. Since in question 2 I showed that these vectors, ${(-1,1,0),(2,0,1)}$ span the set of solutions, and since I showed in the first part of question 3 that they are linearly independent, is it right to say that ${(-1,1,0),(2,0,1)}$ form a basis for the set of solutions? Thanks

 

[I like Serena]

Hmm, ok, so I wasn't on the right path there. It seems after doing some research, that a basis is basically a linearly independent spanning set. Since in question 2 I showed that these vectors, ${(-1,1,0),(2,0,1)}$ span the set of solutions, and since I showed in the first part of question 3 that they are linearly independent, is it right to say that ${(-1,1,0),(2,0,1)}$ form a basis for the set of solutions? Thanks

Yep. (Nod)

 

[TheFallen018]

Yep. (Nod)

You're a legend. Thanks! :)

 

[topsquark]

You're a legend. Thanks! :)

Oh don't tell him that! Now he's going to get the idea that he's good at what he does and that he's a better Mathematician than I am. And that's just crazy talk. (I've got the medication to know what that is.) (Nod)

-Dan

 

[I like Serena]

Oh don't tell him that! Now he's going to get the idea that he's good at what he does and that he's a better Mathematician than I am. And that's just crazy talk. (I've got the medication to know what that is.) (Nod)

-Dan

It's too late.
I AM LEGENDARY! (Poolparty)