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 Jan1 revised Forming contrapositive added 319 characters in body Jan1 answered Forming contrapositive Dec31 reviewed Approve Is the complement of an open ball in a Banach space connected? Dec31 answered Non-zero eigenvalues of $AA^T$ and $A^TA$ Dec31 answered help on a specific power series expansion, i cannot see what the author did here Dec31 comment Integral of an exponential Where do you get stuck when using the hint? Dec31 accepted $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$? Dec31 answered $U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$? Dec31 accepted How should I think about reflexive spaces? What “property” do I get from a space being reflexive? Dec31 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. Dec31 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? Dec31 revised inequality for point on a sphere formatting Dec31 reviewed Looks OK Question about loss of generality in proofs Dec31 awarded Enlightened Dec31 awarded Nice Answer Dec31 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. Dec31 comment Finding an example Take $f_1 = f_2 \neq 0$. Dec31 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. Dec31 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. Dec31 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?