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Mar
29
awarded  Popular Question
Mar
25
comment Limit of the absolute value of a function
@C-S yes i reread it. Thanks !!
Mar
23
comment Needing a double inclusion to determine the kernel of a matrix
So you agree that we need the other inclusion and what we have by that computation is just the inclusion of the kernel in the span and it is not an equality ?
Mar
23
comment Needing a double inclusion to determine the kernel of a matrix
@AlexHalm If we know that dimension of $\ker f$ is one then it's ok we don't need the other inclusion but here we don't seem to know the dimension of $\ker f$, aren't we ?
Mar
23
asked Needing a double inclusion to determine the kernel of a matrix
Mar
23
comment Limit of the absolute value of a function
I think this is not true take $f(n)=(-1)^n$. While $\lim_{n\to \infty}|f(n)|=1$ but $f(n)$ does not have a limit at infinity.
Mar
22
awarded  Popular Question
Mar
11
awarded  Popular Question
Mar
11
awarded  Popular Question
Mar
5
revised Decomposition of the image of a projection.
deleted 6 characters in body
Mar
5
accepted Decomposition of the image of a projection.
Mar
5
comment Decomposition of the image of a projection.
Thanks alot it is clear now and i will edit $q^2=q$.
Mar
5
asked Decomposition of the image of a projection.
Mar
4
comment Universal spaces are homotopy equivalent
So when we write $B\mathbb Z=S^1$ it is not an actual equality and we would rather write $B\mathbb Z\simeq S^1$ meaning that $B\mathbb Z$ is a space that is homotopy equivalent to $S^1$ but not necessarily equal to $S^1$.
Mar
4
accepted Universal spaces are homotopy equivalent
Mar
4
accepted Corresponding a vector subspace to a point of the space.
Mar
4
comment Corresponding a vector subspace to a point of the space.
So in this case you define the grassamannian as being the set $[0,\pi)$ which parametrizes the set $\mathbb R P^1$ of $1-$dimensional subspaces of $\mathbb R^2$. And in this point of view, $[0,\pi)$ is the grassamannian and not $\mathbb R P^1$. Is my understanding correct ?
Mar
4
comment Corresponding a vector subspace to a point of the space.
Yes but here you have two well defined sets: $[0,2\pi)$ and the circle $S^1$ and a well defined map. I see that the analogous of $S^1$ is the set $B$ of all $n-$dimensional subspaces of $W$ but what is exactly $G$ that you take as the analogous of $[0,2\pi)$ and the map that takes an element of $G$ to a unique well defined $n-$dimensional subspace of $W$ ?
Mar
4
comment Corresponding a vector subspace to a point of the space.
Could you please explain the meaning of the notion of a Grassmannian parametrizes the $n-$dimensional subspces of $W$. What do you mean by parametrises ?
Mar
4
asked Corresponding a vector subspace to a point of the space.