• Got a question on how to use our new website? Check out our user guide here!

Vector space pls help (1 Viewer)

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
3,777
Gender
Male
HSC
N/A
If you have even heard of Calabi-Yau Manifold, then vector spaces should be a piece of cake.

By the way - can't see the question, so unable to help.
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
3,777
Gender
Male
HSC
N/A
My knowledge of vector space is rusty. For 3(a), I think you can use: (x-1) and (x-1)2 as a basis of W; dimension = 2
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
279
Location
Antartica
Gender
Male
HSC
N/A
My knowledge of vector space is rusty. For 3(a), I think you can use: (x-1) and (x-1)2 as a basis of W; dimension = 2
Still astonishing that you remember this stuff.

Unfortunately, your basis cannot produce x^2 - x - 2.
_____________________________________________



_____________________________________________

In linear algebra, a basis (for a vector space) is a set of vectors that are linearly independent and span the vector space.

If you unpack this, then you can interpret a basis as a set of the most fundamental building blocks for some world.
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,671
Gender
Male
HSC
2007
Unfortunately, your basis cannot produce x^2 - x - 2
That polynomial does not vanish at 1 and so is not in the span of either of your bases (nor does it need to be).

Your basis is equivalent to his since .

(Also you probably shouldn't use the double implication symbol as informally as in the above post, you should only use it between two propositions that imply each other. In one line, the RHS is an algebraic expression rather than a mathematical statement, and in the others you refer to a polynomial with previously undefined/unquantified constants a and b in it. These early linear algebra courses aren't hard mathematically, but they are meant to drill you in making (semi)formal logical arguments and chasing definitions.)
 
Last edited:

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
279
Location
Antartica
Gender
Male
HSC
N/A
That polynomial does not vanish at 1 and so is not in the span of either of your bases (nor does it need to be).

Your basis is equivalent to his since .

(Also you probably shouldn't use the double implication symbol as informally as in the above post, you should only use it between two propositions that imply each other. In one line, the RHS is an algebraic expression rather than a mathematical statement, and in the others you refer to a polynomial with previously undefined/unquantified constants a and b in it. These early linear algebra courses aren't hard mathematically, but they are meant to drill you in making (semi)formal logical arguments and chasing definitions.)
Woops, arithmetic error.

Thanks, I will take note of that in the future.
 

sida1049

Well-Known Member
Joined
Jun 18, 2013
Messages
824
Gender
Male
HSC
2015
pls help 3C I have no idea lol
What you need to do is to find the general form of a matrix that satisfies A=-A^T. That is, make each of the 9 entries a variable, and use that equation to make as many substitutions as you can to simplify A.

Once you've done that, you should have found the minimum number of variables you can use to describe a matrix satisfying such an equation. That number is the dimension of the vector space W.

Then, you need to decompose the matrix A as a linear combination of linearly independent 3x3 matrices. This shouldn't be too hard, as all you need to do is to look at entries with the same variable, and make it so A is equal to a sum of 3x3 matrices, with a variable as a coefficient in front of each matrix. Then each of those 3x3 matrices are your basis vectors.

I hope that makes sense.

For example, the matrix {{a,2b},{b,-a}} (first row is a and b, second row is b and a) has dimension 2, because the variables a and b can freely vary. I can decompose the matrix as a{{1,0},{0,-1}} + b{{0,2},{1,0}}. And so my basis vectors are the matrices {{1,0},{0,-1}} and {{0,2},{1,0}}.

Edit: corrected something and used a slightly better example.
 
Last edited:

hayabusaboston

Well-Known Member
Joined
Sep 26, 2011
Messages
2,371
Location
Calabi Yau Manifold
Gender
Male
HSC
2013
What you need to do is to find the general form of a matrix that satisfies A=-A^T. That is, make each of the 9 entries a variable, and use that equation to make as many substitutions as you can to simplify A.

Once you've done that, you should have found the minimum number of variables you can use to describe a matrix satisfying such an equation. That number is the dimension of the vector space W.

Then, you need to decompose the matrix A as a linear combination of linearly independent 3x3 matrices. This shouldn't be too hard, as all you need to do is to look at entries with the same variable, and make it so A is equal to a sum of 3x3 matrices, with a variable as a coefficient in front of each matrix. Then each of those 3x3 matrices are your basis vectors.

I hope that makes sense.

For example, the matrix {{a,2b},{b,-a}} (first row is a and b, second row is b and a) has dimension 2, because the variables a and b can freely vary. I can decompose the matrix as a{{1,0},{0,-1}} + b{{0,2},{1,0}}. And so my basis vectors are the matrices {{1,0},{0,-1}} and {{0,2},{1,0}}.

Edit: corrected something and used a slightly better example.
So I wrote out a 3x3 matrix abdefghi equals minus abcdefghi, so b+d=0, h+f=0 and g+c=0, so taking that into account I got basis as a long thing with dimension 6 right?

https://imgur.com/ai837hu

also is this right for b)

https://imgur.com/h1WnGLK
 

sida1049

Well-Known Member
Joined
Jun 18, 2013
Messages
824
Gender
Male
HSC
2015
Oh snap yay! lol I was blank for a while because totally forgot how to approach these things.

Linear algebra 80% final exam is in like 5 weeks, Imma start the heavy prep 2 weeks before i think.
You got this. If you haven't already, I recommend 3blue1brown's YouTube series on linear algebra to get some intuition for it.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top