Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be vectors in the space $\mathbb{E}^3$. Then it’s easy to check by calculation that the following equation holds: $$\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=(\mathbf{a}\cdot \mathbf{c}) \mathbf{b}-(\mathbf{a}\cdot \mathbf{b}) \mathbf{c}.$$  However, the textbook says the equation can be explained geometrically, and I wonder how. All that has come to me so far is that the vector $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$ lies on the plane spanned by $\mathbf{b},\mathbf{c}$ (given that they are not parallel), but what exactly can represent the double cross product geometrically? Please help.
105408
As in the title, I was wondering if the formula: $$a\times (b\times c)=b(a\cdot c)-c(a \cdot b)$$ for $\mathbb R ^3$ cross product has some geometrical interpretation. I've recently seen a proof (from Vector Analysis - J.W. Gibbs) that's not at all difficult to understand, however I would hardly remember the steps of the proof (and I keep forgetting the correct order of A,B,C's), since it appears to me just as algebraic manipulation. So, why is this true? Or is it just an accident?
16780
If $\mathbf{a} \wedge \mathbf{x}$= $\mathbf{a} \wedge \mathbf{b}$ and $\mathbf{a}\cdot \mathbf{(x-b)}=2a^2$, show that $\mathbf{x=2a+b}$.  I take cross product throughout  $$\mathbf{a} \wedge \mathbf{a} \wedge \mathbf{x} = \mathbf{a} \wedge \mathbf{a} \wedge \mathbf{b}$$  $$\mathbf{(a.x).a-(a.a).x}=\mathbf{(a.b).a-(a.a).b}$$  $$\mathbf{(a.x).a-|a|^{2}.x}=\mathbf{(a.b).a-|a|^2.b}$$  $$\mathbf{|a|^2(x-b)-(a.x).a-(a.b).a}=0$$  Then,I do not know how to proceed.  A second part in the question is  Given $\mathbf{x}$^$\mathbf{a}$=$\mathbf{b-x}$  Prove  $(a^2+1)\mathbf{x}=\mathbf{a}$^$\mathbf{b+b(a.b)a}$  I have been using this same dot product and crossproduct method, but don't get it.
28439
I'm  about tensor products of modules and I don't understand the following paragraph on page 2:     Even though elements of $M \times N$ and $M \oplus N$ are written in the same way, as pairs $(m, n)$, bilinear functions $M \times N \rightarrow P$ should not be confused with linear functions $M \oplus N \rightarrow P $. For example, addition as a function $ R \oplus R \rightarrow R $ is linear, but as a function $R \times R \rightarrow R$ it is not bilinear. Multiplication as a function $R \times R \rightarrow R$ is bilinear, but as a function $R \oplus R \rightarrow R$ it is not linear. Linear functions are generalized additions and bilinear functions are generalized multiplications.   My questions are:   Why the distinction between $\times$ and $\oplus$? Aren't they the same for finite sums (products)? I don't understand how addition can be linear but not bilinear. Surely, if $(a,b) \mapsto a + b$ then it doesn't matter in which argument it is linear. (Thinking of e.g. $\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$) And if $(a,b) \mapsto ab$ then how can it be linear in both arguments (bilinear) but not be in one (linear). It's difficult to write this sentence as it makes no sense to me whatsoever.   Many thanks for your help!
14074
Under what conditions on the sets $A$ and $B$ will it be true that $A\times B= B\times A$?   I think it should be set $A$ $=$ set $B$ but don't know how to prove it or is there any other way for those two cross product to be true?
85485
I need to determine if the following set of vectors are are linearly independent or not.  1) $S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3$, it is known that $(1,2,3) \notin \text{span}\{\mathbf u , \mathbf v , \mathbf w\}$  2) $S=\{\mathbf u + \mathbf v,\mathbf u - \mathbf v , \mathbf u - 2\mathbf v +\mathbf w\}\subseteq \mathbb R^n$ where $\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^n$ is a linearly independent set.  3) $S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3$, where it is known that $(1,2,-3),(1,-3,2),(2,-1,5),(0,0,1)$ all belong to $\text{span}\{\mathbf u , \mathbf v , \mathbf w\}$.  I honestly have no idea how to start this, all I know is that the if the vectors $\mathbf u , \mathbf v , \mathbf w$ are linearly independent, then $a\mathbf u + b\mathbf v + c\mathbf w=0$ has only the trivial solution.  Any help here will be appreciated, many thanks!
57486
Let $A,B,C \in \ M(3,\mathbb{R})$ be such that $A$ commutes with $B $and $B$ commutes with $C$. $B$ is not scalar matrix. Then, $A$ commutes with $C$.   Is this true or false? I think it is not true, but cannot think of a counterexample. Any suggestions?
122731
The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:  $$\left| \begin{matrix} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{matrix}\right|$$  Is there any physical significance to this matrix, or is it just some mathematical trick?
22539
$\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector?  Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?   So why is cross product defined in the way that it is? I am mainly interested in the right hand rule defintion too as it is out of reach?
28943
I read that the cross product can't be generalized to $\mathbb R^n$. Then I found that in $n=7$ there is a Cross product:   Why is it not possible to define a cross product for other dimensions $ \ge 4$?
7294
I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this?  $$(a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?$$  Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks.
116228
Let's say I have two simple vectors: $[0, 1]$ and $[1, 0]$.  Their Kronecker product would be $[0, 0, 1, 0]$.  Let's say I have only the Kronecker product. How can I find the two initial vectors back?  If my two vectors are written as : $[a, b]$ and $[c, d]$, the (given) Kronecker product is:  $$[ac, ad, bc, bd] = [k_0, k_1, k_2, k_3]$$ So I have a system of four non linear equations that I wish to solve: $$\begin{align*} ac &amp;= k_0\\ ad&amp;= k_1\\ bc&amp;= k_2\\ bd &amp;=k_3. \end{align*}$$  I am looking for a general way to solve this problem for any number of initial vectors in $\mathbb{C}^2$ (leading my number of variables to $2n$ and my equations to $2^n$ if I have $n$ vectors).  So here are a few specific questions:  What is the common name of this problem?   If a general solution is known, what is its complexity class?  Does the fact that I have more and more equations when $n$ goes up compared to the number of variables help?  (Note: I really didn't know what to put as a tag.)
85313
Let us look at two linearly independent vectors $v_1$ and $v_2$, which span a plane in $\mathbb{R}^n$ through the origin. Let us form a matrix $V$ out of these vectors so that the first column of $V$ is the vector $v_1$ and the second column is the vector $v_2$. Show that the the matrix $P=V(V^TV)^{-1}V^T$ projects all vectors $x\in\mathbb{R}^n$ orthogonally onto the plane spanned by $v_1$ and $v_2$.
58472
$$ [(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})]\cdot(\vec{c}+\vec{a})=2\vec{c}\cdot(\vec{b}\times\vec{a})$$  I'm supposed to prove that this is true for all vectors $a,b,c$, but I keep getting $4(\vec{a}\times \vec{b})\cdot \vec{c}=0$, which is obviously not true.   Seems to me like there is a mistake in the problem, that it should be $\vec{a}\times\vec{b}$ instead of $\vec{b}\times\vec{a}$. Or I'm doing it wrong?  I'm using facts that $\vec{a}\times\vec{a}=0$ and $[\vec{a},\vec{b},\vec{c}]=[\vec{c},\vec{b},\vec{a}]=-[\vec{b},\vec{a},\vec{c}]$.
90174
First of all, I am very comfortable with the tensor product of vector spaces. I am also very familiar with the well-known generalizations, in particular the theory of monoidal categories. I have gained quite some intuition for tensor products and can work with them. Therefore, my question is not about the definition of tensor products, nor is it about its properties. It is rather about the mental images. My intuition for tensor products was never really geometric. Well, except for the tensor product of commutative algebras, which corresponds to the fiber product of the corresponding affine schemes. But let's just stick to real vector spaces here, for which I have some geometric intuition, for example from classical analytic geometry.   The direct product of two (or more) vector spaces is quite easy to imagine: There are two (or more) "directions" or "dimensions" in which we "insert" the vectors of the individual vector spaces. For example, the direct product of a line with a plane is a three-dimensional space.  The exterior algebra of a vector space consists of "blades", as is nicely explained in the .  Now what about the tensor product of two finite-dimensional real vector spaces $V,W$? Of course $V \otimes W$ is a direct product of $\dim(V)$ copies of $W$, but this description is not intrinsic, and also it doesn't really incorporate the symmetry $V \otimes W \cong W \otimes V$. How can we describe $V \otimes W$ geometrically in terms of $V$ and $W$? This description should be intrinsic and symmetric.  Note that  basically asked the same, but received no actual answer. The answer given at  discusses where tensor products are used in differential geometry for more abstract notions such as tensor fields and tensor bundles, but this doesn't answer the question either. (Even if my question gets closed as a duplicate, then I hope that the other questions receive more attention and answers.)  More generally, I would like to ask for a geometric picture of the tensor product of two vector bundles on nice topological spaces. For example, tensoring with a line bundle is some kind of twisting. But this is still some kind of vague. For example, consider the Möbius strip on the circle $S^1$, and pull it back to the torus $S^1 \times S^1$ along the first projection. Do the same with the second projection, and then tensor both. We get a line bundle on the torus, okay, but how does it look like geometrically?  Perhaps the following related question is easier to answer: Assume we have a geometric understanding of two linear maps $f : \mathbb{R}^n \to \mathbb{R}^m$, $g : \mathbb{R}^{n'} \to \mathbb{R}^{m'}$. Then, how can we imagine their tensor product $f \otimes g : \mathbb{R}^n \otimes \mathbb{R}^{n'} \to \mathbb{R}^m \otimes \mathbb{R}^{m'}$ or the corresponding linear map $\mathbb{R}^{n n'} \to \mathbb{R}^{m m'}$ geometrically? This is connected to the question about vector bundles via their cocycle description.
105887
While reading Advanced Quantum Mechanics by J.J. Sakurai, chapter: Relativistic Quantum Mechanics of Spin-1/2 Particles, section 3.2 the Dirac Equation, the author states the following identity: $$\textbf{p}\times\textbf{A}=-i\hbar\left(\nabla\times\textbf{A}\right)-\textbf{A}\times\textbf{p}$$ where $\textbf{p}$ is the momentum operator and $\textbf{A}$ is the vector potential. Problem: We know that $\textbf{p}\equiv-i\hbar\nabla$. So $\textbf{p}\times\textbf{A}=-i\hbar\left(\nabla\times\textbf{A}\right)$. How then the second term appeared in the first equation? Could anyone please explain how this relation is derived?
92931
In my class, there is no distinction made between,   $$ C_{ab}{}^{b} $$ and  $$ C^{b}{}_{ab}. $$ All I know, and read about so far, is the distinction of covariant and contravariant, form/vector, etc. etc. But what is this slanted business all about?
19860
I have been trying to solve for $\vec{r}$:  $\vec{a}\times \vec{r} + \lambda \vec{r} = \vec{c}$   And I thought of using a triple scalar product, I dot both sides with $\vec{a}$:  $\vec{a} \cdot (\vec{a}\times \vec{r}) + \lambda \vec{a} \cdot \vec{r} = \vec{a} \cdot \vec{c}$  And now since the triple scalar product is cyclic $\vec{a} \cdot (\vec{a}\times \vec{r}) = \vec{r} \cdot (\vec{a}\times \vec{a}) = 0$, so the main equation becomes:  $\lambda \vec{a} \cdot \vec{r} = \vec{a} \cdot \vec{c}$  And I do not know how to proceed from here. I cannot use the dot product again, $\vec{a}$ cannot be canceled out. I am not sure whether this approach is right in isolating the $\vec{r}$ but it seems closer to the answer.