# Vector space pls help (1 Viewer)

#### hayabusaboston

##### Well-Known Member
https://imgur.com/a/J5ZRs

2 c) and all of three have me completely lost pls advise what to do

Also is 2 b) a subspace? I said it was after some working

#### hayabusaboston

##### Well-Known Member
Imgur.com/a/4za4z

#### Drongoski

##### Well-Known Member
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.

#### hayabusaboston

##### Well-Known Member
help pls someone :'(

Heeeeeeelep

aw

#### He-Mann

##### Vexed?
2 b) is a subspace.

What do you need to prove for 2 c) to be a subspace?

#### hayabusaboston

##### Well-Known Member
2 b) is a subspace.

What do you need to prove for 2 c) to be a subspace?
I got 2) but I cant get 3) now :/

#### Drongoski

##### Well-Known Member
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?
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.
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\bg_white \begin{align*}p(1) = 0 &\iff (x-1)(ax + b) = 0 \quad [\text{where }a,b \in \mathbb{C}]\\ &\iff ax(x-1)-b(x-1) = 0 \\ &\iff \mathrm{span}\{x-1, x(x-1)\}\end{align*} \\Thus, a basis for W is \{x-1, x(x-1)\} which means \mathrm{dim}(W) = 2.

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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.

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#### seanieg89

##### Well-Known Member
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 $\bg_white (x-1)^2=x(x-1)-(x-1)$.

(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.)

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#### He-Mann

##### Vexed?
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 $\bg_white (x-1)^2=x(x-1)-(x-1)$.

(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.

#### hayabusaboston

##### Well-Known Member
Ohh thankyou goyss! <3

Im trying to figure out 3c atm heh

#### hayabusaboston

##### Well-Known Member
pls help 3C I have no idea lol

#### sida1049

##### Well-Known Member
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.

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#### hayabusaboston

##### Well-Known Member
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

#### hayabusaboston

##### Well-Known Member
Yep, and yep! You got this.
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.

#### sida1049

##### Well-Known Member
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.