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## New answers tagged inner-products

### Why is a norm the square root of an inner product?

Jean’s answer covers everything but the one question of “why the square root?” This has a simple answer: otherwise, the norm would not be homogenous. In other (less formal) words, the units would be ...
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### Why is a norm the square root of an inner product?

"The norm is the square root of the inner product" is an oversimplification. An inner product takes two vectors, a norm takes just one vector. If $\langle \bullet, \bullet \rangle$ is an ...

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For c) as you mentioned, you want to write $A= \text{Re}(A)+i\text{Im}(A)$, where $\text{Re}(A)$ and $\text{Im}(A)$ are self-adjoint. Similarly as for complex numbers, you take $\text{Re}(A)=\tfrac{1}{... 1 vote ###$\lambda \in \mathbb{C}$is an eigenvalue of the operator$A$, then$\text{Re}(\lambda) = 0$AND$H$is a complex vector space, then$A = iB$. (b) Follow your steps in case it is a complex inner product space: $$\langle Ax, x \rangle = \langle x, A^*x \rangle = \langle x, -Ax \rangle = -\overline{\langle Ax, x \rangle} \\ \langle Ax, x \... • 113k 0 votes Accepted ### If the operator A is self-adjoint and \lambda and \mu are distinct eigenvalues of A, then their corresponding eigenspaces are orthogonal. (a) You have that (A^*x,x)=(x,Ax)=\overline{(Ax,x)}=(Ax,x). Here the hypothesis that (Ax,x) is real was used in the last equation. Therefore, ((A-A^*)x,x)=0, for all x\in H. Let me call (x,y)... • 99 2 votes Accepted ### Prove that (H,{.,.}) is a Hilbert space. I think 1. and 2. are easy to see. For 3. note that$$\{x,y\}=\langle Ux+A^2Ux,Uy\rangle=\langle Ux, Uy\rangle+\langle A^2U x,Uy\rangle=\langle x,y \rangle+\langle AUx,AUy\rangle=\langle y,x\rangle+\... 1 vote Accepted ### a Cauchy Schwarz application Define$x_i^2 = a_i^2+b_i^2, x_i >0$, then Using Multinomial theorem, $$\left(x_1+x_2+\cdots+x_n\right)^2=\sum_{ \substack{ 0 \le j_1, j_2, \ldots, j_n \le 2\\ j_1+j_2+\cdots +j_n = 2}} \frac{2}{... • 3,260 0 votes ### How to understand dot product is the angle's cosine? Instead of asking "How can one see that a dot product gives the angle's cosine between two vectors?" what about asking "Why is the relationship between a dot product and the angle ... • 473 5 votes Accepted ### True or False: Inner product on \mathbb{R}^2 satisfying a specific norm. The actual parallelogram law is an identity, not an inequality:$$\|v+y\|^2+\|v-y\|^2=2\|v\|^2+ 2\|y\|^2$$and that is not satisfied in your example, hence the norm is not associated with an inner ... • 1,883 1 vote Accepted ### Orthogonal projection is bounded First you need the fact that P^*=P. Indeed,$$ \langle P^*(u+w),u+w\rangle=\langle u+w,P(u+w)\rangle=\langle u+w,u\rangle=\langle u,u\rangle=\langle u,u+w\rangle=\langle P(u+w),u+w\rangle.$\$ It ...
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