- #1

- 349

- 1

## Homework Statement

## Homework Equations

## The Attempt at a Solution

2) No clue.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Highway
- Start date

- #1

- 349

- 1

2) No clue.

- #2

- 349

- 1

I am not sure if this is what I have to do. . .

2b) I am not familiar with basis other than S = {v

3) Apparently the question asks for the BASIS of the nullspace, and not the nullspace which I have found. . .

5) My professor told me that I need to be more clear on showing where those numbers and sets of vectors came from in my solution, but I am out of ideas. He mentioned that I would want to do something like: c

- #3

Deveno

Science Advisor

- 908

- 6

there are 2 closure conditions, and a third condition...any guess as to what these are?

as for finding a basis, can you think of linear combinations of a basis for M

do you

(3) you're close. suppose that every element of a subspace is a multiple of some vector v. prove that {v} is a basis for that subspace.

(5) if you keep careful track of which row-operations you performed, these will give you a way to construct a linear combination of the 3 vectors you started out with. what exactly did you do to row 2, that made it 0? express that as an equation involving u,v and w.

- #4

- 349

- 1

for (2), do you know how you check that a subset S of a vector space V is a subspace?

there are 2 closure conditions, and a third condition...any guess as to what these are?

It needs to be shown that they hold under addition, multiplication, and that the zero vector is contained in the set. . .

(I gotta think a bit about the rest of your post)

- #5

Deveno

Science Advisor

- 908

- 6

yes. so...how do you show that the sum of two skew-symmetric matrics is skew-symmetric?

- #6

- 349

- 1

yes. so...how do you show that the sum of two skew-symmetric matrics is skew-symmetric?

it's a theorem or definition in the book. . . you would just make up two arbitrary ones and show that it holds, right?

- #7

- 349

- 1

as for finding a basis, can you think of linear combinations of a basis for M_{nxn}that might be symmetric matrices?

do youknowof any bases for M_{nxn}(hint: think of a matrix as being n n-vectors laid "end-to-end"...what is an obvious basis for R^{n2}?)

I'm kinda confused about this, is this part of what we were already talking about in the other discussion posts?

(3) you're close. suppose that every element of a subspace is a multiple of some vector v. prove that {v} is a basis for that subspace.

I know what you are saying, but I'm having trouble figuring out how I would exactly do that. I know what you are saying to do, and how/why that works. . .

(5) if you keep careful track of which row-operations you performed, these will give you a way to construct a linear combination of the 3 vectors you started out with. what exactly did you do to row 2, that made it 0? express that as an equation involving u,v and w.

I have my 3 row operations listed, but I'm not sure how I would write that up. . . it seems like it would make more sense to re-do the reduction with column operations, so that everything would be in terms of u, v, w alone. . .

- #8

Deveno

Science Advisor

- 908

- 6

it's a theorem or definition in the book. . . you would just make up two arbitrary ones and show that it holds, right?

the key is you have to use the definition of skew-symmetric somehow. i do not know the definition of skew-symmetric you are using, the easiest one to use in my opinion is:

A

if this is true for A and B, what can you say about (A+B)

I'm kinda confused about this, is this part of what we were already talking about in the other discussion posts?

to show a set is a basis, you need to show 2 things:

a) the set is linearly independent

b) the set spans the subspace.

if we have a one-element set {v}, showing spanning should be easy, there's only one kind of linear combination: av (there's no other basis elements to add). for a one-element set, linear independence means showing av = 0 implies a = 0.

I know what you are saying, but I'm having trouble figuring out how I would exactly do that. I know what you are saying to do, and how/why that works. . .

I have my 3 row operations listed, but I'm not sure how I would write that up. . . it seems like it would make more sense to re-do the reduction with column operations, so that everything would be in terms of u, v, w alone. . .

yes, you could use column operations. both methods should give you an equation of the form:

(some stuff)u + (other stuff)v + (still more stuff)w = ?

remember, in any linear combination, you can always "collect like terms".

- #9

- 349

- 1

- #10

Deveno

Science Advisor

- 908

- 6

think of a typical vector in R

one way, is to assign a point a set of coordinates: v = (x,y,z). now we've cut the infinity of points down to just something requiring 3 numbers: x, y and z (the 3 in R

let's relate this to a basis: we "split the coordinates" like this:

(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1).

so our "x's" are just x times the basis vector (1,0,0), and similarly with the y's and z's.

the set {(1,0,0),(0,1,0),(0,0,1)} has certain properties, which make it useful:

it is linearly independent (you can't "cancel y's" by adding in "x's" and "z's").

it spans all of R

now, although this particular basis is VERY convenient, it's not the only set of 3 vectors we could use. but we have to be careful, we can't just pick 3 vectors at random. for example:

{(1,1,0), (0,0,1), (1,1,1)} isn't any good: the last vector is just a sum of the first 2, so it's not adding any new information. so that set of 3 vectors fails on BOTH counts, it's not only linearly dependent:

1(1,1,0) + 1(0,0,1) + (-1)(1,1,1) = (0,0,0)

and it also fails to span: any linear combination of the 3 is:

a(1,1,0) + b(0,0,1) + c(1,1,1) = (a,a,0) + (0,0,b) + (c,c,c) = (a+c,a+c,b+c), so you can see that the first 2 coordinates are the same, thus, for example, (1,2,3) isn't in the span.

so a basis is SPECIAL, it is like the "instant" condensed form of a vector space (just add coordinates: that is, make linear combinations). we need to be extra careful when deciding if a set is a basis, because we're letting that basis represent the ENTIRE vector (possibly sub-)space.

we want linear independence so we don't have "extra vectors we don't need" (a basis is minimal among all spanning sets), and we want spanning so that we "don't miss anything" (a basis has a maximal span among all linearly independent sets).

bases are your friends, they let you simplify things. they are like a key to a code that lets you expand everything from that....well, basis (in the ordinary english meaning of the word).

- #11

Deveno

Science Advisor

- 908

- 6

this only shows closure of addition for the 2x2 case. you need to show it for the nxn case. since you can't write down arbitrarily large matrices (after all n might be a million), you'll need to work directly from the definition of a skew-symmetric matrix. what is this definition?

- #12

- 349

- 1

i really appreciate your help with all of this, and am sitting here trying to digest / process all of this information :D

- #13

- 349

- 1

is this the definition you are talking about?

where the first part is the skew-symmetric part and the second is the symmetric part?

- #14

- 349

- 1

is this the definition you are talking about?

where the first part is the skew-symmetric part and the second is the symmetric part?

Actually, using A

I have: (A+B)

So for the closure under multiplication, am I trying to show that kA is skew-symmetric or that (AB)

Thanks

- #15

Deveno

Science Advisor

- 908

- 6

- #16

- 349

- 1

does anyone know how i can show that a skew-symmetric matrix contains the zero vector?

as for #3, i wrote my answer to include the zero vectors for

for #5 i was able to write (0,0,0) as a linear combination of the three vectors that were given (in terms of

- #17

Deveno

Science Advisor

- 908

- 6

does anyone know how i can show that a skew-symmetric matrix contains the zero vector?

use the fact that 0

as for #3, i wrote my answer to include the zero vectors foruandvwithwbeing my vector solution from finding the nullspace sinceu,v,ware linearly independent and span...

never, never, never put a 0-vector in a basis. the 0-vector always makes ANY set linearly dependent.

for #5 i was able to write (0,0,0) as a linear combination of the three vectors that were given (in terms ofu,vandw).

good :)

- #18

- 349

- 1

use the fact that 0^{T}= 0....

never, never, never put a 0-vector in a basis. the 0-vector always makes ANY set linearly dependent.

good :)

awesome, thanks!!!!! :D

Share: