## Math Genius: An array of polynomials can be considered as a tensor?

if $${1, x, x^2, x^3, dots, x^{n-1}}$$ is the base of a vector space, can I say that is $$bar{P} = [ P_1, P_2, dots, P_m]$$ a tensor? Where each $$P_i$$ is a combination of the monomial base. If it’s, What is its dimension?

I’d like to thank in advance.

Ok, if you bound the degree of polynomials in an array by, say $$d$$, and since
$$P_i=P_i(x)=a_{i0}+a_{i1}x+…+a_{id}x^d,$$
where the $$a_{ij}$$ are taken from a field, say $$mathbb R$$, then the vector space of your $$tilde P$$ objects, collected into a set, let’s called $$V$$,
will have $$dim V=(d+1)^m$$.

For, if we denote by $$mathbb R_d[x]$$ the abelian group of polynomials of degree $$d$$ at most, then
$$V=mathbb R_d[x]timescdotstimes mathbb R_d[x],$$
is the cartesian product of $$m$$ factors, with each factor having
$${1,x,x^2,…,x^d}$$ as basis.

Elements of a vector space are also dubbed rank one contravariant tensors.

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## Math Genius: How to get the left and right rotation quaternions between two \$mathbb{R}^4\$ unit vectors?

This question is somewhat similar to this, but in higher dimension.

I have two non-collinear unit vectors in $$mathbb{R}^4$$, $$bf a$$ and $$bf b$$. I know from Wikipedia that there is at least one pair of quaternions $$Q_L$$ and $$Q_R$$ that is able to rotate $$bf a$$ to $$bf b$$, i.e.:
$${bf b} = Q_L {bf a} Q_R$$
(where multiplication is the Hamilton product)

Questions:

1. What is an easy to understand and fast to compute way to find $$Q_L$$ and $$Q_R$$?
2. What are the rotation quaternions that moves $$bf a$$ towards $$bf b$$, but by a given angle $$theta$$ ?

Edit: the comments made me realize I need further constraints to get the rotation I need. So, $$bf a$$ and $$bf b$$ spans a subspace. The rotation I need is closed in this subspace. I think this restriction reduces the number of possible solutions from infinite to two, (as there are two pairs of quaternions that describes the same rotation).

Let $$a,b$$ be two unit quaternions, for simplicity $$a,bnepm1$$ and $$bnepm a$$.

There are many pairs $$(p,q)$$ of quaternions for which $$b=paq$$, e.g. $$(ba^{-1},1)$$ or $$(1,a^{-1}b)$$.

You can even parametrize them all by solving for $$p$$ in terms of $$q$$ or vice-versa.

There is a unique pair (up to $$pm$$) for which $$F(x)=pxq$$ acts as a 2D rotation in the plane spanned by $$a,b$$ but fixes all points in the plane orthogonal to it, and that is $$(sqrt{ba^{-1}},sqrt{a^{-1}b})$$.

The square roots are defined by halving the convex angle of polar form, i.e. $$sqrt{e^{thetamathbf{u}}}=e^{(theta/2)mathbf{u}}$$ if $$thetain[0,pi)$$; in practice you can calculate this using half-angle formulas on real/imag parts:

$$sqrt{r+mathbf{v}}=sqrt{frac{1+r}{2}}+sqrt{frac{1-r}{2}}frac{mathbf{v}}{|mathbf{v}|}.$$

(I define quaternions as sums of scalars $$r$$ and 3D vectors $$mathbf{v}$$; these are real/imaginary parts.)

Here’s how to get the formula. First define the left/right multiplication maps $$L_p(x)=px$$ and $$R_p(x)=xp$$. If we can create a rotation $$G(x)$$ which sends $$1$$ to $$a^{-1}b$$ and is a nonrotation in the complent, then we can define $$F=L_acirc Gcirc L_a^{-1}$$. The effect of the $$L_a^{-1}$$ is to send the $$ab$$-plane to the $$1(a^{-1}b)$$-plane, and the former’s complement to the latter’s complement. After we rotate purely in the $$1(a^{-1}b)$$-plane with $$G$$ we go back to the $$ab$$-plane with $$L_a$$.

If $$p=e^{thetamathbf{u}}$$ then $$L_p$$ and $$R_p$$ both act as a rotation by $$theta$$ in the $$1mathbf{u}$$-plane and its complement, the only difference being opposite directions in the complement (this follows algebraically from the fact orthogonal vectors anticommute). Therefore if we were to compose $$L_{sqrt{p}}$$ and $$R_{sqrt{p}}$$ we get a rotation from $$1$$ to $$p$$ in the $$1mathbf{u}$$-plane and a nonrotation in the complement.

Thus $$F(x)=abig(sqrt{a^{-1}b}(a^{-1}x)sqrt{a^{-1}b}big)$$. Let $$q=asqrt{a^{-1}b}a^{-1}$$. Then $$q^2=ba^{-1}$$ so $$q=pmsqrt{ba^{-1}}$$ (square roots of nonreal quaternions are unique up to $$pm$$; you can prove this with polar form). Since $$sqrt{a^{-1}b}$$ has positive real part, so does its conjugate $$q$$, so $$q$$ must be $$+sqrt{ba^{-1}}$$.

Conclude $$F(x)=sqrt{ba^{-1}}xsqrt{a^{-1}b}$$.

A more symmetric version of the above is how I intuited the formula: $$L_{sqrt{ba^{-1}}}$$ and $$R_{sqrt{ab^{-1}}}$$ both rotate $$a$$ halfway to $$b$$, but rotate in opposite directions in the complement.

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## Math Genius: Eyeballing direction of acceleration in curve of fuction from \$Bbb R^1\$ to \$Bbb R^n\$

This question emanates directly from This previous question I made, but from a more general perspective.

Given some function $$f(t)colon mathbb{R^1}tomathbb{R^n}$$ parametrized in some arbitrary way that draws some curve, can I eyeball the acceleration from the curve? can I at least eyeball where the acceleration would NOT point to?

For example, the following curve in RED is the image of some function $$f(t)colon mathbb{R^1}tomathbb{R^2}$$. The velocity is marked in ORANGE at some point and is tangent to the curve (that we already know). The purple lines indicate possible acceleration vectors for that curve in that point. The question:

Given only the image of the function (the curve). Are all the purple acceleration vectors equally possible?
Can I discard some general direction? Why?

Here I’m interested only in direction, since I understand that the magnitude is impossible to infer.

The relevant formula you want is this: If $$upsilon$$ is the speed of the particle, then

$$vec a = upsilon’ vec T + kappaupsilon^2 vec N,$$

where $$vec T$$ and $$vec N$$ are the unit tangent and principal normal, respectively. It follows that the acceleration vector must be in the half-plane determined by the tangent line and the principal normal. (To derive this, differentiate $$upsilonvec T$$ with respect to time.)

## Math Genius: Eyeballing direction of acceleration in curve of fuction from \$Bbb R^1\$ to \$Bbb R^n\$

This question emanates directly from This previous question I made, but from a more general perspective.

Given some function $$f(t)colon mathbb{R^1}tomathbb{R^n}$$ parametrized in some arbitrary way that draws some curve, can I eyeball the acceleration from the curve? can I at least eyeball where the acceleration would NOT point to?

For example, the following curve in RED is the image of some function $$f(t)colon mathbb{R^1}tomathbb{R^2}$$. The velocity is marked in ORANGE at some point and is tangent to the curve (that we already know). The purple lines indicate possible acceleration vectors for that curve in that point. The question:

Given only the image of the function (the curve). Are all the purple acceleration vectors equally possible?
Can I discard some general direction? Why?

Here I’m interested only in direction, since I understand that the magnitude is impossible to infer.

The relevant formula you want is this: If $$upsilon$$ is the speed of the particle, then

$$vec a = upsilon’ vec T + kappaupsilon^2 vec N,$$

where $$vec T$$ and $$vec N$$ are the unit tangent and principal normal, respectively. It follows that the acceleration vector must be in the half-plane determined by the tangent line and the principal normal. (To derive this, differentiate $$upsilonvec T$$ with respect to time.)

## Math Genius: Prove that a parametric equation’s range is subset of a cartesian equation

$$r(t)=t(t-2)^3i +t(t-2)^2j$$ where $$r:Bbb R to Bbb R^2$$

$$C = {(x,y)∈Bbb R^2 mid y^4=x^3+2x^2y}$$

I have difficulty with this question. This is where I am up to.

1. Find $$dy/dx$$ of the cartesian curve $$dy/dx = (3x^2+4xy)/(4y^3-2x^2)$$;
2. $$(4y^3-2x^2) = 0, y= (2x^{2/3})/2$$;
3. Substitute $$y$$ into cartesian equation and find $$x=-27/16, 0$$.

I’m not sure where to go from here, can anyone clarify if I am on the right track, thanks!

I think you’re making it too complicated.

If $$x(t)=t(t-2)^3$$ and $$y(t)=t(t-2)^2$$,

then it’s easy to show that $$y^4=x^3+2x^2y$$:

$$t^4(t-2)^8=t^3(t-2)^9+2t^2(t-2)^6t(t-2)^2$$

because the right side is $$t(t-2)^8[t^2(t-2)+2t^2]=t(t-2)^8[t^3].$$

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## Math Genius: How to show: Matrix \$H_v = I – 2vv^{top}\$ with \$||v|| =1\$ to the reflection on hyperplane \$v^{bot}\$ is symetrical and orthogonal and \$det H_v = -1\$?

I have several questions

1. How can I show that a Matrix $$H_v = I – 2vv^{top}$$ with $$||v|| =1$$ to the reflection on the hyperplane $$v^{bot}$$ is symetrical and orthogonal and the determinant is $$det H_v = -1$$?
2. And is every symetrical orthogonal matrix $$A$$ with $$A = -1$$ formed $$A=H_v$$ for a $$v$$?

3. How to show that every orthogonal matrix $$A$$ is a product of reflection matrices $$H_v$$ on the hyperplane $$v^{bot}$$

4. How can one describe every quadratic upper triangular matrices $$A$$ which are orthogonal?

I had to work a bit on it, and with the help of other answers on this board I came up with some answers:

1. $$I$$ is symetric. The dot product §vv^{top}§ gives a symmetric matrix as result. The $$2$$ in front is irrelevant. It’s a mutpile so that still counts.
It is $$(AB)^{top} = B^{top} A^{top}$$, so $$I^{top} = I$$ and
$$(vv^{top})^{top} = vv^{top}$$. And with that $$H^{top} = (I- vv^{top})^{top} = I^{top} – (vv^{top})^{top} = I – vv^{top} = H$$.
$$H$$ is therefor symmetrical.

2. counter example matrix [-1 0 0; 0 -1 0; 0 0 -1]

3. Let $$u,v in mathbb{R^n}$$ be colmn vectors, their dot product is then $$langle u,v rangle = u^{top} cdot v$$. Let $$A$$ be a orthogonal Matrix, then:

$$langle A cdot u,A cdot v rangle = (A cdot u)^{top} cdot (A cdot v) = u^{top} cdot A^{top} cdot A cdot v = u^{top} cdot A^{-1} cdot A cdot v = u^{top} cdot v$$
This shows that the transformation keeps the dot product. no change of angles is happening. This is in general the product of reflections. Bonus: Every reflection can be made with a series of householder reflections. So it is about householder reflections and matrices, too.

1. There are no upper triangular matrices which are orthogonal. Under the main diagonal there are only zeros, and multiplied with another upper tirangular matrix, the result will again be an upper triangular matrix. Thus the Inverse of the upper triangular matrix $$R$$ is an upper triangular matrix. The transpose on the other hand is an lower triangular matrix $$R^{top} = L$$
But from that follows $$R^{top} neq R^{-1}$$

## Math Genius: missing basis and finding coordinates of vector with respect to basis

Vectors $$w_1, w_2, w_3$$ form an orthogonal basis for $$R^3$$. Given that $$w_1 = begin{pmatrix} 2\3\5 end{pmatrix}$$, what are the coordinates of the vector $$v=begin{pmatrix}0\1\2end{pmatrix}$$ with respect to the basis?

I am not sure how to begin this problem. I believe I should find $$w_2$$ and $$w_3$$ first, and I know that $$w_1 cdot w_2 = 0$$, $$w_2 cdot w_3 = 0$$ and $$w_3 cdot w_1 = 0$$, but I’m not quite sure how that helps me exactly, given that there are 6 unknown variables (entries) from the basis. Any help would be really appreciated.

Hint: Can you think of any nonzero vector that is orthogonal to $$(2,3,5)$$? Let this be $$w_2$$.

Then take $$w_3=w_1times w_2$$.

As stated, the problem cannot be solved. Suppose that $$v=alpha_1w_1+alpha_2w_2+alpha_3w_3$$. Then the coefficients $$alpha_2$$ and $$alpha_3$$ cannot be both different from $$0$$ (since $$v$$ is not a multiple of $$w_1$$). If $$alpha_2ne0$$, then$$v=alpha_1w_1+alpha_2w_2+alpha_3w_3=alpha_1w_1-alpha_2(-w_2)+alpha_3w_3$$and $${w_1,w_2,w_3}$$ is still an orthogonal basis, but now you have different coefficients. And the same thing happens if $$alpha_3ne0$$.

For $$w_2$$ you have that
eqalign{ & w_{,2} = left( {matrix{ a cr b cr c cr } } right);; :quad 2a + 3b + 5c = 0quad Rightarrow cr & Rightarrow quad w_{,2} = left( {matrix{ a cr b cr { – left( {2a + 3b} right)/5} cr } } right) cr}
meaning that you are left two degree of freedom as it should be.

Then for $$w_3$$ you can take
$$w_{,3} = lambda ,w_{,1} times w_{,2}$$
where you can limit $$c$$ to be positive if you need to keep the chirality according to the “right-hand” rule.
You have in total two degree of freedom and a ascale parameter.

After that you can express $$v$$ in such a basis, leaving the parametrs free.

## Math Genius: If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.

If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.

My Attempt:
Let \$vec {a}\$ and \$vec {b}\$ be two vectors such that \$|vec {a}|=|vec {b}|\$

Magnitude of Resultant:
\$\$=sqrt {a^2+b^2+2abcos theta}\$\$
\$\$=sqrt {2a^2+2abcos theta}\$\$

How do I proceed further?

From \$|a+b|^2=|a|^2\$ and \$|a|=|b|\$ we have
\$\$|a|^2=|a+b|^2=|a|^2+|b|^2+2langle a,brangle=2|a|^2+2langle a,brangle,\$\$
so \$langle a,brangle=-|a|^2/2\$. Hence the cosine of the angle between \$a\$ and \$b\$ using \$|a|=|b|\$ again is
\$\$frac{langle a,brangle}{|a||b|}=frac{-1}{2}.\$\$

The resultant vector \$mathbf{OC}\$ is the diagonal of the rhombus whose adjacent sides are \$mathbf{OA}, mathbf{OB}\$. Since its length is equal to either, we have an equilateral triangle \$mathbf{OBC}\$. Thus the angle between the vectors is \$120^circ\$

Let,\$R\$ be the resultant of the sum of the two vectors \$vec A\$ and \$vec B\$ such that \$|vec A|=|vec B|\$.So,\$\$R=sqrt{|vec A|^2+|vec B|^2+2|vec A||vec B|costheta}\$\$

where \$theta\$ is the angle between \$vec A\$ and \$vec B\$.So,

\$\$R=sqrt{2|vec A|^2+2|vec A|^2costheta}\$\$.

Now magnitude of resultant \$R\$ is equal to either of the vectors \$vec A\$ and \$vec B\$.So,

\$\$|A|=sqrt{2|vec A|^2+2|vec A|^2costheta}\$\$
\$\$implies|vec A|^2=2|vec A|^2+2|vec A|^2costheta\$\$
\$\$impliesfrac{-|vec A|^2}{2|vec A|^2}=cos theta\$\$
\$\$impliestheta=cos^{-1}-frac{1}{2}\$\$.

Hope this helps!!

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## Math Genius: Convert vector into matrix

I have a vector $$vinmathbb{R}^{mntimes 1}$$ and I would like to make a matrix out of it in such form:
$$V=begin{bmatrix}v_{1} & v_{2} & dots & v_{m} \ vdots & vdots & dots & vdots\ v_{(n-1)m+1} & v_{(n-1)m+2} & dots & v_{mn} end{bmatrix}$$

Is there a mathematical way to do so?

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## Math Genius: why is the third side of this vector triangle || v – u ||

I’m new to linear algebra, and am self-learning from a book by Kuldeep Singh. While talking about vector dot products he shows this diagram below.

What I do not understand is why in figure 2.14 the third side (the “$$a$$” side) would equal $$|| v – u ||$$. He uses this fact to derive a formula for angles between vectors, but gives no explanation for that initial premise that the third side would equal $$|| v – u ||$$.

Can someone explain it to me? Consider this.

You start from point A. Then go to point B. Then point C from there and then back to point A.

A $$rightarrow$$ B then B $$rightarrow$$ C then C $$rightarrow$$ A

A $$rightarrow$$ B = ||u||

B $$rightarrow$$ C = ||x|| (we have to find x)

C $$rightarrow$$ A = -||v|| (minus because direction is opposite. We are going from C $$rightarrow$$ A and not A $$rightarrow$$ C)

Since it start and comes back to same point, their sum should be 0.

||u|| + ||x|| – ||v|| = 0

||x|| = ||v – u||

Hope it helps

$$mathbf{u}$$ and $$mathbf{v}$$ are the vectors representing sides $$overline{AB}$$ and $$overline{AC}$$ respectively. The difference, $$mathbf{v}-mathbf{u}$$, is just the vector you’d have to add to $$mathbf{u}$$ to get $$mathbf{v}$$. That is, $$mathbf{u}+(mathbf{v}-mathbf{u})=mathbf{v}$$.

If you’re aware that vector addition is like attaching each vector together head-to-tail, then it should be clear that $$mathbf{v}-mathbf{u}$$ is the vector for $$overline{BC}$$.

$$|mathbf{v}-mathbf{u}|$$ is just the magnitude of the difference vector, which makes it the length of $$overline{BC}$$.

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