View Full Version : What's a basis for the planar subspace (in R3) 2x - y + 2z = 0?

Sir Penguin

28-10-2005, 04:14:30

I think my brain's addled because I'm up past my bedtime. Is {(0,0,0), (-1,0,-1)} one? Does a basis vector have to be non-zero? How about {(2,-2,-1), (-1,0,-1)}? The plane passes through the origin, so it's just two solutions to the equation, right?

:cry:

SP

has to be linearly independent also

JM

the basis should be two lines.. which are linearly independent

Jon Miller

err.. also have to span the space as well

Jon Miller

Colon

28-10-2005, 04:45:01

I'm reading funny words.

(I think a basis vector (for R2) has to be non-zero, but both of yours are)

JM

(sorry if I am confusing, it has honestly been years since I have thought about these things)

Sir Penguin

28-10-2005, 16:29:20

Now that I can think, I'm sure I'm right. It's just two vectors from the origin to some point on the plane (x,y,z), s.t. the vectors are linearly independent. I'd never seen the problem of finding a basis for a plane expressed as an equation before, so it threw me.

Thanks for the assist, JM. :beer:

SP

Venom

28-10-2005, 16:37:21

It's always hilariously scary and confusing to see Noj Mmmmmmr talk nerdy science and math.

MOBIUS

28-10-2005, 16:57:15

Is this a foreign language? Looks a bit like Dutch, but it isn't unless it's not about beer...:cute:

That was my real passion once, I wonder why I haven't earnestly pursued it.

Now I piaf it.

Japher

28-10-2005, 17:14:12

passions smassions...

MOBIUS

28-10-2005, 17:19:30

Originally posted by MoSe

That was my real passion once, I wonder why I haven't earnestly pursued it.

Now I piaf it.

I thought we established that she didn't regret anything...:D

Or are you talking about Finches?

Provost Harrison

28-10-2005, 17:40:38

What's a basis for the planar subspace (in R3) 2x - y + 2z = 0?

Yes.

Lazarus and the Gimp

28-10-2005, 18:30:35

Originally posted by Sir Penguin

I think my brain's addled because I'm up past my bedtime. Is {(0,0,0), (-1,0,-1)} one? Does a basis vector have to be non-zero? How about {(2,-2,-1), (-1,0,-1)}? The plane passes through the origin, so it's just two solutions to the equation, right?

:cry:

SP

I can understand both the words and the numbers. It's the combinations of the two that lose me.

Japher

28-10-2005, 18:32:54

kind of like mr. g's posts

Spartak@CPH

28-10-2005, 18:38:16

except there is a hint of meaning to SP.

Japher

28-10-2005, 18:40:15

I wonder if the stuff JM said after makes any sense

Spartak@CPH

28-10-2005, 18:42:52

Nah!

Vincent

28-10-2005, 19:51:27

*dies*

Japher

28-10-2005, 19:53:33

*beats Vincent's body while laughing hysterically*

KrazyHorse@home

28-10-2005, 20:54:25

The normal vector to that plane is parallel to (2, -1, 2) (you get this from the coefficients in the equation describing the plane)

To find a basis for that plane, you first need to find a vector which is orthogonal to the plane's normal vector. Orthogonal vectors have 0 dot product, therefore, if our proposed basis vector is (x, y, z)

2x - y + 2z = 0

one solution to this is the vector (1, 2, 0)

another solution is the vector (1, 0, -1)

So this is a valid basis. It is not, however, an orthonormal basis. We can either orthonormalise the above basis via the Gram-Schmidt process or we can simply choose one of the basis vectors and take its cross product with the plane's normal:

(1, 0, -1) X (2, -1, 2) = (-1, -4, -1)

and then normalise both basis vectors, so an orthonormal basis for that plane is:

(1/sqrt(2))(1, 0, -1) and (1/sqrt(18))(-1, -4, -1)

To answer some more questions: a basis is by definition the smallest possible set of vectors which spans the space. Therefore, the set of basis vectors is linearly independent (if it wasn't then at least one of the basis vectors would be superfluous, as it could be written as a linear combination of the other basis vectors), so (0, 0, 0) can never be a basis vector (since it is never linearly independent)

Colon

28-10-2005, 21:01:51

Stop doing that you!

Dyl Ulenspiegel

28-10-2005, 21:18:54

Mit der Prinzipienlehre wird die Optimierung des Gewährleistungsgegenstands zum Prinzip und damit zur Grundrechtsnorm erhoben; beides bildet aber nur einen Bestandteil der Grundrechtsnorm, insbesondere wird hier die Beschränkung ausgeblendet. Dies mag seinen Grund darin haben, dass etwa für Alexy Schranken sich dadurch bestimmen, dass sie rechtmäßig „grundrechtliche prima facie-Positionen einschränken“. Diese prima facie-Position ist dann zB die (weit verstandene) Berufsfreiheit schlechthin. Dabei werden aber nicht die Grundrechte und nicht einmal die grundrechtlichen Gewährleistungen optimiert, sondern die isoliert betrachteten und im Ergebnis mit dem Grundrecht gleichgesetzten Gewährleistungsgegenstände. Nach der hier vertretenen Auffassung ist auch die „prima facie-Position“ nach wie vor nur der Schutz gegen ungerechtfertigte Beschränkungen.

Colon

28-10-2005, 21:24:46

That's much better. Except of the German of course.

Colon

28-10-2005, 21:25:28

It turns out that based on the evidence from the large German survey, the average degree of risk aversion is almost exactly equal to that suggested by large-scale surveys of US households. Using the estimate of the typical degree of risk aversion of Germans in our model of what real yields should be on safe government debt takes us to a figure well in excess of yields on inflation-proof bonds in the world today. We estimate that real yields might be expected to be in the region of 3% — but inflation-proof bonds issued by the French and UK governments have yields closer to 1.5% than the 3% figure; TIPS yields are around 2% at the longer end of the US real curve.

Venom

28-10-2005, 21:34:10

That could put a cokehead to sleep.

Sir Penguin

28-10-2005, 22:08:31

Originally posted by KrazyHorse@home

To answer some more questions: a basis is by definition the smallest possible set of vectors which spans the space. Therefore, the set of basis vectors is linearly independent (if it wasn't then at least one of the basis vectors would be superfluous, as it could be written as a linear combination of the other basis vectors), so (0, 0, 0) can never be a basis vector (since it is never linearly independent)

And linear combinations of 0 and x just form a straight line, not a plane. Right.

I was doing an easy projection to find the closest point in the plane to a point outside the plane, so the bullshit roots and stuff that come with an orthonormal basis were way too much trouble.

SP

Lurker the Second

28-10-2005, 23:47:18

:lol: Great thread! Until this, not a bad post in it.

Sir Penguin

29-10-2005, 02:12:36

I applaud you for resisting the urge to be sarcastic.

SP

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