~The Council Of Youngers of BYC! (COY)~

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OH NO. Uh nuh, uh uh.. What fit of madness could have possessed the owners of that house to have possibly done something so maniacal?!?!?
But seriously, who in their right mind would carpet a kitchen? Ya might as well carpet the garage. Or a chicken coop some days. lol :P
 
Oooooo!

lol, it just makes kitchen messes that much harder to clean. Especially if I am in there. :P
 
Hello Adam. Hello Ooj. Wazzup, how are mah homiez? (lol, I can't believe I just typed that...)

Random facts you can [re]familiarise yourselves with if you wish... (or just ignore it because it's for my notes. I forgot my book.)



An nxn matrix has exactly n eigenvalues.

A matrix with an eigenvalue of 0 is not invertible.

An eigenspace is a subspace of dimension at most equal to the multiplicity of its corresponding eigenvalue.


Eigenvectors corresponding to distinct eigenvalues are linearly independent. Proof by induction.

True for one eigenvector: eigenvectors are not 0, so the equation cv = 0 has only the trivial solution and v is linearly independent. (What was the point of that...)

Assume it's true for k eigenvectors.

Now consider the set of eigenvectors up to v(k+1). To prove the set is linearly independent we show that all the constants in c(1)v(1) + .... +c(k)v(k) + c(k+1)v(k+1) = 0 are zero. (EQ 1)

Multiply this equation by A. Each term becomes c(k)A.v(k), which is also c(k)L(k)v(k) because eigenvectors are awesome like that. (EQ 2)

Multiply the original equation by L(k+1). (EQ 3)

EQ 2 - EQ 3 removes the final term involving v(k+1). This means we only have eigenvectors 1 through k left in the equation. Those are linearly independent so their coefficients must be 0. As none of the eigenvectors are the same, L(n)-L(k+1) can't ever be zero. That means all the constants have to be 0. If c(1) through c(k) are zero, EQ 1 is reduced to c(k+1)v(k+1) = 0. v(k+1) is not zero because it's an eigenvector.

Therefore c(k+1) is zero and the equation has only the trivial solution and the set is linearly independent.

*inserts induction conslusion sentence*




I've forgotten to bring my notebook to uni so I have to memorise key facts from the maths lecture to write down later. I scheduled this hour for watching the lecture recording and there's no way I'm doing it again. :P

I will keep this page loaded on my phone to write it up later. (much better than the other options, a) typing on my phone, b) depleting my printing quota and half a tree, c) having to move to go get my USB and then having to load that onto my ancient laptop again, which is always a battle to turn on, d) watching it again tomorrow. :P)
 
Last edited:
Fierlin1182 said:
Hello Adam. Hello Ooj. Wazzup, how are mah homiez? (lol, I can't believe I just typed that...)

Random facts you can [re]familiarise yourselves with if you wish... (or just ignore it because it's for my notes. I forgot my book.)



An nxn matrix has exactly n eigenvalues.

A matrix with an eigenvalue of 0 is not invertible.

An eigenspace is a subspace of dimension at most equal to the multiplicity of its corresponding eigenvalue.


Eigenvectors corresponding to distinct eigenvalues are linearly independent. Proof by induction.

True for one eigenvector: eigenvectors are not 0, so the equation cv = 0 has only the trivial solution and v is linearly independent. (What was the point of that...)

Assume it's true for k eigenvectors.

Now consider the set of eigenvectors up to v(k+1). To prove the set is linearly independent we show that all the constants in c(1)v(1) + .... +c(k)v(k) + c(k+1)v(k+1) = 0 are zero. (EQ 1)

Multiply this equation by A. Each term becomes c(k)A.v(k), which is also c(k)L(k)v(k) because eigenvectors are awesome like that. (EQ 2)

Multiply the original equation by L(k+1). (EQ 3)

EQ 2 - EQ 3 removes the final term involving v(k+1). This means we only have eigenvectors 1 through k left in the equation. Those are linearly independent so their coefficients must be 0. As none of the eigenvectors are the same, L(n)-L(k+1) can't ever be zero. That means all the constants have to be 0. If c(1) through c(k) are zero, EQ 1 is reduced to c(k+1)v(k+1) = 0. v(k+1) is not zero because it's an eigenvector.

Therefore c(k+1) is zero and the equation has only the trivial solution and the set is linearly independent.

*inserts induction conslusion sentence*




I've forgotten to bring my notebook to uni so I have to memorise key facts from the maths lecture to write down later. I scheduled this hour for watching the lecture recording and there's no way I'm doing it again.
tongue.png


I will keep this page loaded on my phone to write it up later. (much better than the other options, a) typing on my phone, b) depleting my printing quota and half a tree, c) having to move to go get my USB and then having to load that onto my ancient laptop again, which is always a battle to turn on, d) watching it again tomorrow.
tongue.png
)
Click to expand...

BYC has so many uses.
gig.gif
 

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