MHB Basis of a vector space - apparently simple problem

[Math Amateur]

I am revising vector spaces and have got stuck on a problem that looks simple ... but ... no progress ...

Can anyone help me get started on the following problem ..

Determine the basis of the following subset of $$ \mathbb{R}^3 $$:

... the plane $$ 3x - 2y + 5z = 0 $$

From memory (I studied vector spaces a long time ago!) I suspect that the solution would involve expressing the problem (but how?) in the form $$ Av = b $$ and then forming the matrix $$(A|b)$$ and then reduce to reduced row echelon form ...

... but then, in this case, it seems that $$ Av = b $$ is as follows:

$$ Av = (3 \ -2 \ \ 5) (x \ y \ z)^T = 0 $$

... ... so we are dealing with reducing the matrix $$(3 \ -2 \ \ 5 \ 0)$$ ... ?

This does not seem right ... must have made a mistake in formulating the problem ...

Can someone please clarify this situation for me ...

Help would be appreciated

Peter

 

[Nono713]

There is nothing wrong here, you are on the right track. Finding a basis of the plane amounts to finding a basis of the nullspace of the following matrix (since that nullspace is the plane by definition):
$$\left [ \begin{matrix} 3 &-2 &5 \end{matrix} \right ]$$
So you row-reduce it as usual (don't be put off by the fact that it only has one row - apply the usual rules):
$$\left [ \begin{matrix} 1 &\frac{-2}{3} &\frac{5}{3} \end{matrix} \right ]$$
The matrix has rank $1$, and so its nullspace has dimension $2$ (as expected, it's a plane) and so it has two free variables, and is of the form:
$$N = \left [ \begin{matrix} \frac{2}{3} y + \frac{-5}{3} z \\ y \\ z \end{matrix} \right ]$$
And so:
$$N = \left [ \begin{matrix} \frac{2}{3} y \\ y \\ 0 \end{matrix} \right ] + \left [ \begin{matrix} \frac{-5}{3} z \\ 0 \\ z \end{matrix} \right ]$$
Which can be rewritten as:
$$N = y \left [ \begin{matrix} \frac{2}{3} \\ 1 \\ 0 \end{matrix} \right ] + z \left [ \begin{matrix} \frac{-5}{3} \\ 0 \\ 1 \end{matrix} \right ]$$
And so we conclude that a basis of the nullspace of the matrix (and so, a basis of the plane) is:
$$\left \{ \left [ \begin{matrix} \frac{2}{3} \\ 1 \\ 0 \end{matrix} \right ], \left [ \begin{matrix} \frac{-5}{3} \\ 0 \\ 1 \end{matrix} \right ] \right \}$$
​

 

[Math Amateur]

There is nothing wrong here, you are on the right track. Finding a basis of the plane amounts to finding a basis of the nullspace of the following matrix (since that nullspace is the plane by definition):
$$\left [ \begin{matrix} 3 &-2 &5 \end{matrix} \right ]$$
So you row-reduce it as usual (don't be put off by the fact that it only has one row - apply the usual rules):
$$\left [ \begin{matrix} 1 &\frac{-2}{3} &\frac{5}{3} \end{matrix} \right ]$$
The matrix has rank $1$, and so its nullspace has dimension $2$ (as expected, it's a plane) and so it has one free variable, and is of the form:
$$N = \left [ \begin{matrix} \frac{2}{3} y + \frac{-5}{3} z \\ y \\ z \end{matrix} \right ]$$
And so:
$$N = \left [ \begin{matrix} \frac{2}{3} y \\ y \\ 0 \end{matrix} \right ] + \left [ \begin{matrix} \frac{-5}{3} z \\ 0 \\ z \end{matrix} \right ]$$
Which can be rewritten as:
$$N = y \left [ \begin{matrix} \frac{2}{3} \\ 1 \\ 0 \end{matrix} \right ] + z \left [ \begin{matrix} \frac{-5}{3} \\ 0 \\ 1 \end{matrix} \right ]$$
And so we conclude that a basis of the nullspace of the matrix (and so, a basis of the plane) is:
$$\left \{ \left [ \begin{matrix} \frac{2}{3} \\ 1 \\ 0 \end{matrix} \right ], \left [ \begin{matrix} \frac{-5}{3} \\ 0 \\ 1 \end{matrix} \right ] \right \}$$
​

Thanks Bacterius ... really appreciate your help...

But ... can you explain how you arrived at the matrix N and how it enters the process?

Peter

 

[Nono713]

Thanks Bacterius ... really appreciate your help...

But ... can you explain how you arrived at the matrix N and how it enters the process?

Peter

N isn't a matrix, it's a vector. Actually it's the parametric form of the nullspace of the matrix $\left [ \begin{matrix} 3 &-2 &5 \end{matrix} \right ]$. You obtain it the usual way, by rewriting dependent variables in terms of free variables (here we have $1x + (-2/3)y + (5/3)z = 0$, and $x$ depends on $y$ and $z$, so $x = (2/3)y - (5/3)z$) and then separating the variables to obtain a basis for this nullspace. I thought I'd give it a name, $N$ seemed suitable :D

 

[Deveno]

In "high-school" one would solve this like so:

$3x - 2y + 5z = 0$

$3x - 2y = -5z$

We have 1 equation, 3 unknowns. Picking any two values for $x,y$ determines $z$, picking any less than two values, isn't any help:

Suppose $x$ (for example), was known to be the constant $a$. This gives:

$-2y = -5z - 3a$ or equivalently:

$2y = 5z + 3a$. Now $\frac{3}{2}a$ is just another constant, let's call it $b$ for now. Dividing by 2 gives:

$y = \frac{5}{2}z + b$.

This is a line in the $yz$-plane, and as its slope is not 0, it takes on ALL values for $y$.

We could do this for each variable in turn, the results aren't all that illuminating, but they do tell us we need to know at least TWO variables to find the third. There is nothing restricting our choice (hence the name "free variables" or "free parameters").

On the other hand, if we know $x,y$ then $z = \frac{2}{5}y - \frac{3}{5}x$.

In other words, all of our vectors $(x,y,z)$ that satisfy this are of the form:

$\left(x,y,\frac{2}{5}y - \frac{3}{5}x\right) = x\left(1,0,-\frac{3}{5}\right) + y\left(0,1,\frac{2}{5}\right)$.

If one wants to avoid fractions, one can put then in the form:

$(5x,5y,2y-3x) = x(5,0,-3) + y(0,5,2)$, which means our plane is spanned by the vectors (5,0,-3) and (0,5,2).

Now, YOU need to show that this IS a linearly independent set (that (5,0,-3) and (0,5,2) do not lie on the SAME line through the origin).

As a sanity check note that if $(x,y,z) = (5,0,-3)$ that:

$3x -2y + 5z = 15 - 0 + -15 = 0$, and similarly for the other vector.