I was reading the definition of inner product  and they state the following:  Let $V$ be a vector space over a field $\mathbb F$ ( $\mathbb R$ or $\mathbb C$). A map $\left&lt; \cdot , \cdot \right&gt;: V \times V \to \mathbb F$ is called an inner product if, for all $x,y,z \in V$ and $\alpha \in \mathbb F$:  $\left&lt; \alpha x , y \right&gt; =  \alpha \left&lt; x , y \right&gt;$ and $\left&lt; x + z , y \right&gt; = \left&lt; x , y \right&gt; + \left&lt; z , y \right&gt;$ $\left&lt; x , y \right&gt; = \overline{\left&lt; y , x \right&gt;}$ $\left&lt; x , x \right&gt; &gt; 0$, if $x \neq 0$.   My question is about property 3. Let $V$ be a vector space over $\mathbb C$, then $\left&lt; \cdot , \cdot \right&gt;: V \times V \to \mathbb C$, this means that for all $x \in V$, $\left&lt; x , x \right&gt;$ is a complex number. So what do they mean by $\left&lt; x , x \right&gt; &gt; 0$? Do they mean that the inner product of $x$ with itself is allawys a real number, so if $\left&lt; \cdot , \cdot \right&gt;$ is an inner product then $\Im(\left&lt; x , x \right&gt;) = 0$? If not, what does it mean for a complex number to be greater than $0$ in this context?
121109
Let $V$ be a vector space over a field $F=\mathbb{R}/\mathbb{C}.$ Then an inner product on V is defined to be a function $(,):V\times V \rightarrow \mathbb{F}$ satisfying the following property:  Linearity in the first coordinate.  Conjugate-Symmetry.  Positivity: For all $x \in V$, if $x$ is nonzero then $(x,x)&gt;0$.   My Questions:  If I  take $\mathbb{F}$ to be $\mathbb{C}$. Since it is not an ordered field, the idea of positivity does not make sense. But I find this definition everywhere. Am I missing something out?  Every Inner Product defined on $V$ (an $\mathbb{R}$-vector space) is bilinear. Does every bilinear map $f: V\times V\rightarrow \mathbb{R}$  give rise to an inner product on $V$?
72156
Question While looking over the exercise $3.F-34$ in Linear Algebra Done Right, I encountered the following paragraph  Suppose $V$ is finite dimensional. Then $V$ and $V'$ are isomorphic, but finding an isomorphism from $V$ onto $V'$, generally, requires choosing a basis of $V$. In contrast, the isomorphism from $V$ to $V''$ does not require a choice of basis and thus is considered more natural.  and these questions showed up in my mind: $1$. Does the word natural just means that we don't need to choose a basis? I have seen the word canonical is used in the similar manner too. Is there a more precise definition for natural or canonical? $2$. Assuming the answer to question $1$ is Yes, then why there is no natural isomorphism from $V$ onto $V'$? $3$. I think that there is a relation between the answer to question $2$ and the proof of Riesz representation theorem. So, if we cannot find a natural isomorphism between $V$ and $V'$ then it means that we cannot prove Riesz representation theorem without choosing a basis of $V$. Is this true?  Complementary Informations The Isomorphism from $V$ onto $V^{''}$. Suppose $V$ is a finite dimensional vector space. Consider the following map $$ \Lambda(v)(\phi)=\phi(v),  \qquad \forall v \in V, \,\, \forall \phi \in V^{'} $$ then $\Lambda$ is an isomorphism from $V$ onto $V''$. Riesz Representation Theorem. Suppose $V$ is a finite dimensional linear space equipped with an inner product and $\phi$ is a linear functional on $V$. Then there is a unique vector $v_0 \in V$ such that $$\phi(v) = {\langle v,v_0 \rangle}_{V}, \qquad \forall v \in V$$  Other Related Posts I found the following posts related to this question on MSE and MO. , , , .
71019
Consider the vectorspace of all real $m \times n$ vectors and define an    inner product $\langle A,B\rangle = \operatorname{tr}(B^T   A)$.    "tr" stands for "trace"        which is the sum of the diagonal entries of a matrix.   How do you prove that $\operatorname{tr}(B^T   A)$ is indeed a inner product?  Kind regards
112862
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form $$ \mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t}  $$ where $\mathbb{Z}^n$ now means $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$.  From an intuition perspective, and in the sense of sets, is this more or less the same as $\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}$? (Bear in mind I am normally using these ideas in relation to homology groups.)
37976
(1) If $X$ and $Y$ are two sets, we define the Cartesian product $X \times Y$ as the set of ordered pairs $(x,y)$, such that $x \in X$ and $y \in Y$.  (2) On the other hand [Folland, Real Analysis, page 4], if $\{X_\alpha\}_{\alpha \in A}$ is infinite indexed family of sets, their Cartesian product $$ \prod_{\alpha \in A}X_\alpha $$ is defined as the set of maps $f: A \to \bigcup\limits_{\alpha \in A} X_\alpha$ such that $f(\alpha) \in X_\alpha$ for every $\alpha \in A$.  After saying this, Folland remarks:     it should be noted, and promptly forgotten, that when $A = \{1,2\}$, the previous definition  of $X_1 \times X_2$ [that's (1) above] is set-theoretically different from the present definition of $\prod_1^2 X_j$ [that's (2) above]. Indeed, the latter concept depends on the mappings, which are defined in terms of the former one.   I am not grasping this remark. Specifically, here are my questions.  Question 1: How is (2) set-theoretically different from (1)? A simple illustrative example?  Question 2: If (1) is extended to infinite families, which definition would be stronger? A simple illustrative example?  Question 3: Why should this be "promptly forgotten"?  I'll probably have more questions depending on the type of answers I'll get to these.  Thanks!
71904
Do these three kinds of vector spaces, those with an inner-product, those with a norm and those with a metric, are the same sets of vector spaces? At least for finite dimensional vector spaces all of these coincide?   It would be great to know finite dimensional counter-examples if they exist any and if anyone can link to some lecture notes explaining this point.     Like are there examples of inner product spaces which cannot have a metric or metric spaces which cannot have a norm or various other such possible conflicts between these 3 properties.
66521
I realize that every inner product defined on a vector space can give rise to a norm on that space.  The converse, apparently is not true.  I'd like an example of a norm which no inner product can generate.
22490
I was just wondering whether anyone knows an answer to the following:  Suppose that ${\mathbb F}$ is a subfield of a field ${\mathbb G}$ and that $v_1,\ldots ,v_k$ are linearly independent vectors in ${\mathbb F}^n$ (over $\mathbb F$). Is it necessarily true that $v_1,\ldots ,v_k$ are also linearly independent when considered as vectors in ${\mathbb G}^n$ (over $\mathbb G$})?  Any help would be much appreciated. Thanks!
35285
Is there a canonical way to define on any vector space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ a norm ? (Or, if there isn't, can someone give me an example of a vector space over $\mathbb{K}$ that is not normable ?)  I have now looked through several books on the subject but nowhere is something like this mentioned and I also wasn't able to find a way to construct such norm (or to find a counterexample).
24
Let $H$ be a vector space equipped with an inner product $(\cdot, \cdot)$ and $f:H\to H',\ f(x)=(\cdot,x)$ surjective. Now, why $H$  is a Hilbert space? The other direction is clear by Riesz' representation theorem but what about this?
73086
Suppose $V$ is an inner product space and $U_1, U_2$ are subsets, then I'd like to show that $U_1 \subset U_2$ if and only if $U^{\perp}_2 \subset U^{\perp}_1$. I can show the first forward direction, by first taking $w \in U^{\perp}_2$. Then, $w$ must be a vector where $&lt;w, u_2&gt; = 0$ for all $u_2 \in U_2$. But, since $U_1 \subset U_2$, then $&lt;w, u_1&gt; = 0$ for all $u_1 \in U_1$. Hence, $w \in U^{\perp}_1$.   I am not sure how to show the opposite direction. My proof is as follows:  Take $u_1 \in U_1$. Now, let $w \in U^{\perp}_2$. Then, $&lt;w,u_2&gt; = 0$ for all $u_2 \in U_2$. But, $U^{\perp}_2 \subset U^{\perp}_1$, and so $w \in U^{\perp}_1$ as well. Hence, $&lt;w,u_1&gt; = 0$ for all $u_1 \in U_1$. After this, I am not sure how to complete the proof. Am I in the right direction? Thanks!
103768
To be clear on this, I know what is the definition of an inner product space and some properties and theorems about them. What I am asking for is an intuition for this definition in the complex case. In the real case, the intuition (or at least one of them) is  geometric: The inner product of two vectors is the length of the projection of the first to the second scaled by the norms of both vectors so that it is symmetric (modulo some details). In particular I (and everybody else) think of "inner product zero" as geometric orthogonality and of orthonormal bases as, well, orthonormal bases and so on. The question is, what should I think about when working with complex (or should I say hermitian?) inner product spaces? what is the "meaning" of the complex number associated to two vectors called their inner product?  I will be happy to hear all kinds of answers. For example, what physical phenomena does it model or in what mathematical situations does in "naturally" appear. Answers that stress the "nice structure" resulting are also welcome, yet I feel that by itself  it is a bit unsatisfying.
111637
Suppose $(e_1,e_2,...,e_n)$ is an orthonormal basis of the inner product space $V$ and $v_1,v_2,...,v_n$ are vectors of $V$ such that $$||e_j-v_j||&lt; \frac{1}{\sqrt{n}}$$ for each $j \in \left\{1,2,...,n \right \}$. Prove that $(v_1,v_2,...,v_n)$ is a basis of $V$.  I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?
3994
Let $X$ be an inner product space, and let $M$ be a non-empty subset of $X$. Then we have the following:   (a) If the space of $M$ is dense in $X$, then $M^\perp = \{0 \}$, that is, $x \in X$, $x \perp M$ imply $x =0$.   (b) If $X$ is complete and $M^\perp = \{0 \}$, then the span of $M$ is dense in $X$.   In (b), what if the space $X$ is not complete? In such a case, can we have a set $M$ such that $M^\perp = \{0\}$, but the span of $M$ is not dense in $X$?
106390
I noticed something when I was doing a proof of the BAC-CAB rule, and wanted to check if my intuition was correct.   First, when I actually multiplied out $\vec{B} (\vec{A} \cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B}) $ I noticed that I got the same thing. That is, the two quantities are equal. This makes sense since they are dot products. But that further implies that the expression goes to zero.  Well, that means that $\vec{A} \times (\vec{B} \times \vec{C})$ would have to be zero too. And when I thought about it it occurred to me that $\vec{B} \times \vec{C}$ describes a vector perpendicular to both B and C; it's orthogonal. But $\vec{A} \times$ (any vector) will describe a vector perpendicular to (any vector) and in this case, that's $\vec{B} \times \vec{C}$. That is going to be parallel with the vector described by $\vec{B} \times \vec{C}$ itself and that further implies that the cross product of the two is equal to zero. Which is equal to the BAC-CAB quantity above.   So, what is the big honking flaw in my reasoning here? :-)
15850
Quick question: Can I define some inner product on any arbitrary vector space such that it becomes an inner product space? If yes, how can I prove this? If no, what would be a counter example? Thanks a lot in advance.
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