Dimension (vector space) Puns

Given the vectors v1 = (3,2,7,1,1) and v2 = (3,1,6,1,1) my intuition says they are linearly independent since one is not a scaled up version of the other. However if we solve the system

3a + 2b + 7c + d + e = 0

3a + b + 6c + d + e = 0

We subtract the 2nd from the 1st equation

b + c = 0 <=> b = -c

We rewrite the system

3a -2c + 7c + d + e = 0

3a -c + 6c + d + e = 0

Both equations are equal to each 3a+5c+d+e. That means 0=0 so any solutions are true. So are they dependent in the end? What to do from here?

Hello I can't understand how do we find the dimension for a vector space when we are dealing with matrices.

e.g dimension of diagonal M[2x2], its basis set of vectors, has 2 vectors just,but the dimension for the vector space is 4,am I wrong? and what is the concept that I'm misunderstanding about this?

See title. I understand this is how the dimension of a given polynomial field is defined but can someone explain the definition/provide a digestible proof for it? Thanks.

i'm coming from the fact that sin(x) and sin(x+1) are linerly independent, aren't they?

edit: they aren't lineraly independent with respect to the inner product integral from -1 to 1. but you still can't write one of them as a linear combination of the other, right?

Sorry about come here and hope you can help me, but I don't know how to proof that!!! :( Thanks a lot.

I intuitively expected the space of formal power series to have countably infinite dimension over the reals.

This does not seem to be the case. My logic is: Since the dimension of a space is greater than the number of eigenvalues of any linear operator on the space, and we can differentiate (the taylor series of)[; e^{ax} ;] for any real [;a;] to get an uncountable number of linearly independent eigenvectors in the space, the space must have uncountably infinite dimension.

On the other hand, there are only countably many terms in any power series, and each one of those can be assigned a single real, we arrive at countably many dimensions.

One of the above is clearly wrong, but I'm not sure which one. I think an answer may have to do with expressing x^r, where r is real, as a series, but this would not resolve the contradiction. If the dimension is uncountable, is there an intuitive reason why this is the case?

I'd like an answer different than "because we can form a basis with 2 solutions of the DE". I want to use the answer to the title to conclude that there are 2 independent solutions.

I'm not really sure where to start with this problem. I get that if B is a basis for V, then |B| = n, but that's just by definition and may not really be that helpful...

Thanks for any help!

Oh and also, C = set of complex numbers and R = set of real numbers.

Just a random question that I just thought of: can the dimension of a vector space be a real number ?I've always studied mathematics where the dimension is a natural number 3,7,32, ...

Would it be possible to imagine a space of dimension 4.32 for example ?

If you could give me any pointers why not, it would be much appreciated.

If so, please can you prove it. If not please can you try explain why? Is it true for only certain spaces and metrics?

Hi,

For some context, this isn't a school assignment, just something I was thinking about.

I'm trying to figure out how I can position K number of vectors (all starting from the origin) such that the angle between all of them at a maximum and equal.

For example, in 2 dimensions, you can equally space K vectors by an angle easily computed as 2*pi / K.

In 3 (or higher) dimensions it gets a little harder for me to understand because we are dealing with solid angles. Also, I feel like there has to be some minimum K based on the dimension.

I'm wondering if anyone knows of any theorems around this or has any suggestions what I should research to help me with this?

Thanks

Help me with my HW plz, also, could any1 tell me about a good lineal algebra book where i can read of span, dimension, bases and linear independence?

Can anyone explain how this is done? Specifically using the True False problem below. The answer is supposedly false, and the dim is 6. Not trying to cheat on hw just trying to prep for a test.

https://preview.redd.it/qxgdehb9sgt11.jpg?width=756&format=pjpg&auto=webp&s=92a4f4289acd1535a3794f15b70528c7c1357393

Suppose that V1 is m-dimensional and V2 is n-dimensional. Prove that dim(V1×V2) = m + n.

As background, we're given: V1 × V2 as a set is the set of all ordered pairs (x1, x2) where x1 ∈ V1 and x2 ∈ V2.

The set follows standard vector addition (like (x1,x2)+(y1,y2)=(x1+y1, x2+y2) and scalar multiplication and has a zero vector.

How do I even begin this?

Find the dimension of the vector space V and give a basis for V. V = {p(x) in P2 : p(0) = 0}

How do I go about this? I see that p(0)=0 would mean no constants. I am stumped, not sure where to start at all. Thanks for the help in advance!

Hey guys! I’ve been trying to work on this question for quite a while, but haven’t been able to solve it. Any help is aporeciated. The question asks to prove that if V is a vector space over the field of the complex numbers with dimV = n, then the same vector space V over the field of the real numbers has dimV = 2n. It seems quite logical since complex numbers can be thought of as ordered pairs of real numbers (a,b) such that the complex number is a+bi. But I don’t know how I would formally prove it or what reasoning/theorems would be useful. Thanks!

I'm a bit confused here.

I was reading this Reddit post and posted this:

> Follow-up question. I had the suspicion that for a set of k equidistant points in a metric space M of dimension d, k<=d+1. I briefly toyed around with proving that if k equidistant points exist in M, than k+1 equidistant points exist in a space M' when M' is an extension of M in a natural way, got stuck on defining 'natural', then felt embarrassed when I realized the discrete metric is an incredibly trivial counter-example. Then I wondered what happened when M was an inner product space, and realized all my intuition came from Euclidean spaces where the inner product induces a norm. I have no idea what's going on here. I figured out a proof for Euclidean spaces, but that proof is less than enlightening (induction meets messy algebra). Can anyone shed some light?

So I understand that functions scale and add linearly so its a vector space.

But if you would actually see the vector, which and how does it point? For example f(x) = sin (x). Or f(x) = x^2.

I dont know how v = ( x,y,z, ... )^T looks like. And what does the x in f(x) correspond to. I first thought that the vector was just in R^2, like (x , f(x) ) but that can't be right.

Thanks in advance

Geometry is an art derived from the planet itself. It is what was studied by all civilizations and existed since the beginning of society. Primary and secondary education only focus on Euclidian geometry and often limits analysis to 2 dimension. For the masses, that is all that’s required.

For people seeking a superior understanding of this intricate system;

For people who want to journey beyond the plane;

For people who want to see the possibilities beyond their imagination;

For people who want a vector to guide them into a new dimension;

------------------—-

They need to shift their geometric cognition.

They need to supplement their synthetic geometry with a system of coordinates that allows for this understanding {3D Cartesian, Cylindrical, and Spherical Coordinates}

They need to embrace a perspective where position and orientation is defined by vectors {position, normal, and directional vectors}

They need to wire their minds to accept the temporal in addition to the spatial {parametric}.

They need to project the 3D world they feel onto the 2D world they see.

—————————

At the same time, there are those who only desire to live in the finite world.

All of calculus relies on the limit, a human construct, a weak attempt at trying to touch infinity.

Calculus attempts to create logical lines from free-flowing curves {derivative}, yet those creations cannot be constructed by the hands of humans. >!You can use compass to make a tangent to circle but not other curves.!<

—————————

Most universities entangle the infinitesimal nature of calculus with the horizon of the vector space into a course called Multivariable Calculus.

Though a powerful weapon, there are those who do not wish to wield it.

Those who want travel across a space curve without bothering with its density {line integral}.

Those who want to sit on a hyperbolic parabolic without bothering with the volume below {iterated integral}.

Those who want to climb the mountain of multi-variable functions without worrying about finding its peak {second derivative test}.

Those who want to skate inside a parabolic cylinder without worrying about how much area they covered {surface integral}.

Those who want an elegant shape without flooding it with turbulent waters {flux, Stokes’ Theorem, Divergence Theorem}.

—————————

To those who are like me. Those who want to ride across the vector space without bothers. Those who want to experience the pure beauty of surfaces and curves. **We n

Hello, I'm currently in my first year of comp sci at Uni and we're studying linear algebra. So far so good, I don't have much of a problem understanding the basics apart from one thing that keeps bugging me. This week we had a lecture on a basis vectors i (1, 0) and j (0, 1). I understand that those vectors can be used to represent a certain space in 2D when multiplied by any 2D vector v.

My question is, why do we need to represent the 2D space by two vectors (i and j) rather than a single vector with the same coordinates e.g. i = (1,1)? What is the exact purpose of defying 2D space by two 2D vectors rather than one?

I'm well aware that this question might sound stupid for anyone familiar with linear algebra but I'm having bit of a trouble grasping it such that I feel comfortable with this notion of basis vectors and I can't find a clear answer to my question.