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Jan
1
revised Forming contrapositive
added 319 characters in body
Jan
1
answered Forming contrapositive
Dec
31
reviewed Approve Is the complement of an open ball in a Banach space connected?
Dec
31
answered Non-zero eigenvalues of $AA^T$ and $A^TA$
Dec
31
answered help on a specific power series expansion, i cannot see what the author did here
Dec
31
comment Integral of an exponential
Where do you get stuck when using the hint?
Dec
31
accepted $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$?
Dec
31
answered $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$?
Dec
31
accepted How should I think about reflexive spaces? What “property” do I get from a space being reflexive?
Dec
31
comment inequality for point on a sphere
@Dr.SonnhardGraubner: Why would that cause trouble? This would only imply that $\sqrt{1-a} < \sqrt{a}$, which is perfectly fine.
Dec
31
comment inequality for point on a sphere
I have looked through the solution given on pages 250 and 251 but I don't see any part where $1-2a < 0$ would cause trouble. Can you explain why exactly you think this would cause trouble?
Dec
31
revised inequality for point on a sphere
formatting
Dec
31
reviewed Looks OK Question about loss of generality in proofs
Dec
31
awarded  Enlightened
Dec
31
awarded  Nice Answer
Dec
31
comment Finding an example
You can set $f_2 = \lambda f_1$ for some constant $\lambda$ and they will be distinct functions, but the ratio still constant.
Dec
31
comment Finding an example
Take $f_1 = f_2 \neq 0$.
Dec
31
comment Find an orthogonal matrix $V$ such that $V^{T}B(\gamma)V=diag(1+\gamma n,1,1,\cdots,1)$
No, the spectral theorem only gives existence of such an eigenbasis. But you already know how to diagonalise matrices, and due to the spectral theorem, you will be able to choose the eigenvectors such that they are orthogonal.
Dec
31
comment Find an orthogonal matrix $V$ such that $V^{T}B(\gamma)V=diag(1+\gamma n,1,1,\cdots,1)$
You can see from your definition of $B(\gamma)$ that $B(\gamma)$ is symmetric, and then, from the spectral theorem, it follows that there is such an orthogonal basis of eigenvectors. Also, from the diagonal matrix in your post, you can read off the eigenvalues of $B(\gamma)$ and conclude that the matrix is also positive definite.
Dec
31
comment Find an orthogonal matrix $V$ such that $V^{T}B(\gamma)V=diag(1+\gamma n,1,1,\cdots,1)$
Do you know how to diagonalise matrices?