In inner product spaces what is the motivation behind conjugate symmetry axiom? Well if it is a real inner product I could make sense of symmetry, but here I couldn't make sense of conjugate symmetry Any insight toward it or any geometric ideas would be helpful
121130
There is an axiom of inner product spaces that states:   $\overline{\langle x,y\rangle } = \langle y,x\rangle$   Basically (without any conceptual understanding) it seems like all you have to do when you swap the order of the arguments in an inner product space is take their conjugate.   How does this make any sense? I know if we are dealing with an inner product space over $\mathbb{R}$ then the conjugate of a real number is just the real number itself so there is no change. But how does this make sense over the field $\mathbb{C}$?
54213
I'm stuck on this problem for a long time. I can't find a proper solution for this without using complex calculations. I hope some one can help me with this problem:          Let $ (\;|\;)$ be the standard inner product on $\mathbb C^{2}$. Prove that there is no non-zero linear operator $T$ on $\mathbb C^{2}$ such that $(\alpha|T\alpha) = 0\;$ for every $\alpha$ in $\mathbb C^{2}$. Generalize.
93393
@BrianM.Scott 's answer to this question  suggests that there is no standard concept of symmetry for 3-, 4-, N-dimensional arrays, in constrast to the case for 2-D arrays, as in linear algebra for matrices.  Are there alternative definitions of symmetry for higher-dimensional arrays?  Are there specific definitions that are widely used in certain contexts, e.g. in tensor calculus?  (I don't have a specific need; I'm trying to help implement a symmetry test for a matrix library [ for the Clojure language].  Since the library allows higher-dimensional arrays, there's a question about whether there is a natural choice for what the symmetry test should return for higher-dimensional arrays.)
36814
I need to prove that if $V$ is a finite dimensional vector space over a field K with a non-degenerate inner-product and $W\subset V$ is a subspace of V, then: $$ (W^\bot)^\bot=W $$ Here is my approach:  If $\langle\cdot,\cdot\rangle$ is the non-degenerate inner product of $V$ and $B={w_1, ... , w_n}$ is a base of $V$ where ${w_1, ... , w_r}$ is a base of $W$ then I showed that  $$ \langle u,v\rangle=[u]^T_BA[v]_B $$ for a symmetric, invertible matrix $A\in\mathbb{R}^{n\times n}$. Then $W^\bot$ is the solution space of $A_rx=0$ where $A_r\in\mathbb{R}^{r\times n}$ is the matrix of the first $r$ lines of $A$. Is all this true?  I tried to exploit this but wasn't able to do so. How to proceed further?
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I wanted to find a counter example to show that the completeness of the inner product space is necessary in . Please give an example of a bounded linear functional $T$ on an incomplete inner product space $X$ which do not have any inner product representation i.e. there does not exist any $z$ in $X$ s.t. $T(x)= \langle ,z\rangle$ for all $x$ in $X$.
3709
So, we know that there is no "natural" isomorphism between a finite dimensional vector space $X$ and its dual $X^{\vee}$. However, given a basis for $X$ $(e_1, \dots, e_n)$ we can always construct a (dual) basis $(e^1, \dots, e^n)$ for $X^{\vee}$ that satisfies $e^i(e_j) = \delta^i_j$. It follows that $V \approx V^{\vee}$ simply because they have the same finite dimension $n$. Now, it seems to me that we can construct an isomorphism by assigning $$ e_i \mapsto e^i $$ and extending by linearity. That is, for an arbitrary vector $v \in X$ given by $v = \sum v^ie_i$ we can specify a linear map   $$\phi:V \rightarrow V^{\vee}$$  by  $$ \phi(\sum v^ie_i) := \sum v^ie^i $$ It is a quick exercise to check that $\phi$ is an isomorphism. Now, although this construction seems correct, there is something that doesn't quite sit right with me about it. For one thing, there is no way to really make sense of the summation convention as the right side of the function will always contain two superscripts thus requiring an explicit summation sign. The other thing that feels rather off is actually taking a lower index and moving it to an upper index, i.e., $e_i \mapsto e^i$ (which, of course, is the reason that the summation convention doesn't work).  So, with this background my questions are   Is there any sense in which the above isomorphism is "favored" or "canonical"? Is there another way to construct an isomorphism between a vector space and its dual, that would "conserve indexes", for lack of a better way to state it. My guess here is that the answer is no but becomes possible if one assumes $V$ has an inner product.
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$\langle A,B \rangle := Trace (AB^t)$ is an inner product over the vector space $M_n(\mathbb R)$ of all real matrices of size $n$ , I would like to know whether this inner-product has any Geometrical or Physical significance (interpretation) or not ? Please shed some light . Thanks in advance
31665
I am trying to answer the following question:     Which Lie groups have a Lie algebra admitting an $\text{Ad}$-invariant inner product?   First of all, all compact Lie groups satisfy this condition because one can average an arbitrary inner product on the algebra over the group.  This means that all Lie groups that have the same Lie algebra as some compact group also satisfy the condition. Indeed, the group is then locally isomorphic to a compact group, and taking "the same" inner product on the Lie algebra, we see that this inner product is invariant under all $\text{Ad}_g$ for $g$ in some neighbourhood of the identity, but this neighbourhood generates the whole group, from which the claim follows. Another approach would be to show that an inner product is $\text{Ad}$-invariant precisely if it is $\text{ad}$-invariant (like , although there may be an issue with connectedness of the group) and conclude that the Lie algebra completely characterizes whether or not an invariant inner product exists.  I suspect these are all the Lie groups that answer the question, because  (the second property, at "this property characterizes compact Lie algebras"). However, the reference given is just "SpringerLink", which is of no use.     Could somebody confirm that the answer to the above question is "all Lie groups with a compact Lie algebra"?   I have consulted multiple sources, but none of them completely establish the result I'm looking for. For example, Lie groups beyond an introduction by Knapp establishes what I already know (that it works for groups with a compact Lie algebra), and establishes the converse only in case the inner product is actually the Killing form.
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This is definition      "A real linear $V$ is said to have an inner product if for each pair of elements $x$ and $y$ in $V$, there corresponds a unique real number $(x.y)$ satisfying what we know as axioms for inner product."    What does it mean? $(x.y)=xy$ and $xy$ is a real number?   When $(x.y)$ is complex, it is called complex space, does that mean that $(x.y)=xy=$ complex?   I don't understand what inner product is.
120471
Ok so I read the following article the other day:  and I wanted to learn more about lie groups. Using my exceptional deduction skills I thought "oh it must have something to do with groups" So I picked up a copy of Dummit and Foote's book on abstract algebra and skimmed through it. It didn't say anything about Lie groups however. $E_8$ is coming to be rather famous so maybe other people are interested in this question too. Lets suppose I wanted to learn about lie groups. What books should I read to be ready to learn about Lie groups and what is a good book that talks about Lie groups. I'm guessing its a combination of group theory (representation theory in specific) and also differential geometry. Is this correct? Thank you very much for your time.
93091
First, let me admit that I suffer from a fundamental confusion here, and so I will likely say something wrong. No pretenses here, just looking for a good explanation.   There is a theorem from linear algebra that two vector spaces are isomorphic if and only if they have the same dimension. It is also well-known that two sets have the same cardinality if and only if there exists a bijection between them. Herein lies the issue...  Obviously $|\mathbb{R}| = |\mathbb{R^2}| = \mathfrak c.$ This is often stated as "there are as many points in the plane as there are on a line." Why, then, are $\mathbb{R}$ and $\mathbb{R^2}$ not isomorphic?  It makes intuitive sense that they shouldn't be. After all, I can only "match up" each point in $\mathbb{R}$ with the first coordinate of $\mathbb{R^2}.$ I cannot trust this intuition, however, because it fails when considering the possibility of a bijection $f : \mathbb{N} \rightarrow \mathbb{Q}!$  Even more confusing: As real vector spaces, $\mathbb{C}$ is isomorphic to $\mathbb{R^2}.$ However there is a bijection between $\mathbb{C}$ and $\mathbb{R}$ (just consider the line $\rightarrow$ plane example as above).  If you can explain the error in my thinking, please help!
1217
Seems like I , I think I am missing something important.     Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:         $T^2 = T$   $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;         Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.   Now, I know these things:   The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in). To do this I can prove that:   For every $v$ in $ImT^{\perp}$, $T(v) = 0$ Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$. $T$ is  (according to Wikipedia) The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point). What else?    I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things.  Thanks!
123760
I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:  $u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$  Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$   $\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?   Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$  And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$  We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$  This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?   //not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!   EDIT: corrected notation a little.
52327
I was just wondering why we don't ever define multiplication of vectors as individual component multiplication.  That is, why doesn't anybody ever define $\langle a_1,b_1 \rangle \cdot \langle a_2,b_2 \rangle$ to be $\langle a_1a_2, b_1b_2 \rangle$?  Is the resulting vector just not geometrically interesting?
108527
Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
101631
In the beginning of linear algebra courses, there are vectors in $\mathbb R^n$ and the dot product is introduced. We learn that if the dot product of two vectors is zero, then these vectors are called orthogonal and there is a right angle between them. Therefore, we understand perpendicularity from the definition of orthogonality.   In the last chapters, all previous notions are generalized as vector spaces and inner products are introduced. Then we learn vectors $\vec u$ and $\vec v$ are orthogonal if $\langle \vec u, \vec v \rangle=0$ in inner product spaces. Our previous knowledge guides us to seek a right angle between them.   Let $P$ be the vector space of first degree polynomials. Let $p(x)=ax+b$ and $q(x)=cx+d$. Define $\langle p(x), q(x) \rangle$ as  $$\langle p(x), q(x) \rangle=ac+bd$$  Then if we consider $p(x)=x-2$ and $q(x)=4x+2$, we find that $\langle p(x), q(x) \rangle=0$. But if we draw them, we see that there is $30.96$ degrees between them. So orthogonality in inner product spaces does not necessarily mean perpendicularity. If not perpendicularity, what should we understand from orthogonality in inner product spaces?
80058
I am starting reading about Hodge theory and while reading the definition of abstract Hodge structure a very basic question came to my mind... What is the definition of the conjugate of a subspace of a complex vector space?  I found online the definition of conjugate of an entire vector space, but when we do the Hodge decompostition we want $V_{\mathbb{C}}=\oplus_{p,q \in \mathbb{Z}}V^{p,q}$ with the requirement $\overline{V^{q,p}}=V^{p,q}$. So I want to take a conjugate inside the space $V_{\mathbb{C}}$. I can see it if my space is of differential forms, but what is the definition for an abstract one?  Also, if I take the tensor product of two vector spaces, is this conjugation induced componentwise?  I know the question is pretty basic, but I couldn't find any reference for such definitions.