T_01

For relational structures A and B, what is the difference between an embedding of A in B and a partial isomorphism between A and B?

My professor is a luminary on this subject

That is a very funny idea

I have to prepare a presentation of 90 min about modular representations and the cde triangle until monday and I dont have a clue of almost anything the book is talking about. Any handy survival tips?

Yep, okay. Thanks!

Ahh no I think I got that wrong in my mind. We have $f \circ \psi$ (if $f$ is the function i want to integrate and $\psi$ the parametrization) in the integral so the inequivality just carries over, right?

Oh, okay... thank you. Is it hard to proove? I do'nt see how I would

For lebesgue integrals we know that $f \leq g$ a.e. implies $\int f \leq \int g$. Is my intuition correct, that this does not in general hold for integrals over submanifolds (because the inequivality can be destroyed by the parametrizations) ?

Bye. Thank you again

What are you doing if I may ask? (working, studying..?)

Well but I just cannot realize how some genius like you guys can look at such things and after 5 seconds say "ah ok thats easy" :D But learning consistently is the key I guess

Thank you very much. I will try to bring this clean on paper and hopefully remember this kind of "trick". I mean I even looked at "could i get the eigenvalues of this thing?" And gave up after 5 minutes because I did not see a way of doing this. Sometimes I feel like I should quit mathematics at all

How would that look like? I feel like I forgot everything about linear algebra in the period of one semester... we also hat jordan forms and such funny things but I really dont remember anything anymore

Sorry for your time guys

yes, thank you

And then the product is the solution..... what the heck

$\lambda -1$ has to match the eigenvalues from above so $\lambda \in \{ 1, ||a||^2 + 1 \}$

Yes

wait

$(I + aa^T)x = \lambda x \Leftrightarrow x + (aa^T)x = \lambda x \Leftrightarrow (aa^T)x = (\lambda - 1) x$

I mean:

this is latex math xD

wait what am i even doing

Well $(I+aa^T)x = x + (aa^T)x = x$ equiv. to $(aa^T)x = x$ so they are the same as of $aa^T$

I mean the determinant of $aa^T$ is zero. What can one say about the determinant of sums of matrices?

It should be zero then

Well symmetric matrices are diagonalisable

Wasnt that the case for symmetric matrices? I dont know, thaught I heared that in la

The determinant of $aa^T$ should be the product of the eigenvalues? (So 0)

So we have the eigenvalues $||a||^2$ with multiplicity $1$ and $0$ with mult. $n-1$

of $aa^T$, but then my original $A$ is just $I$ so this case is not interesting

Well if $a$ is zero the matrix is $0$ and the determinant is trivially also $0$ :D So lets assume $a$ is not zero so the kernel is $n-1$ dimensional, so looking at $a^Tx$ there should be many zeros as eigenvalues in the situation above

otherwise the kernel would be $n$ dimensional

if $a$ is not zero

It is, because the image is not trivial

So the kernel is up to $n-1$ dimensional

Okay so the "matrix" is $a^T$ which has one row, so the image can be highest one-dimensional

Ah yes we have a number as result..

its not?

So this linear map is represented by the matrix where we have the $a^T_i$ on the diagonal

Wrote it before you edited the question ^^

if $x$ is not zero it is one dimensional (maximal)

wait no a is not a matrix (ok technically yes but no)

That x is in the kernel of $a^T$

When $a^Tx = 0$ I guess?

Yes I forgot the square there

@Astyx So in that case i remain with only one eigenvalue $||a||$ with multiplicity $n$ ?