Can a Scalar Multiple of a Basis Still Form a Basis?

  • Thread starter Thread starter Rijad Hadzic
  • Start date Start date
  • Tags Tags
    Basis
AI Thread Summary
A basis for a vector space can have multiple representations, including scalar multiples of its vectors. If {v_1, v_2, ..., v_n} is a basis, then the set {cv_1, cv_2, ..., cv_n} (where c is a nonzero scalar) also forms a basis due to its linear independence and ability to span the same space. The confusion arises from the misconception that a vector space can only have one basis; in reality, different bases can represent the same vector space. The key is that any linear combination of a basis that maintains independence will still span the space. Understanding these properties clarifies the nature of bases in vector spaces.
Rijad Hadzic
Messages
321
Reaction score
20

Homework Statement


I'm using this problem, and although I got the correct answer my question is really about the properties of a basis, not the problem that I'm going to state..

anyways,

Let = {v_1, v_2, ..., v_n } be a basis for a vector space V. Let c be a nonzero scalar. Show that the set {cv_1, cv_2, ..., cv_n } is also a basis for V

Homework Equations

The Attempt at a Solution


Since the set {cv_1, cv_2, ..., cv_n } has the same dimensions as set {v_1, v_2, ..., v_n }, I now only need to check for linear independence (or span, but it makes no sense to check for the more difficult condition.)

So [ k1cv1 + k2cv2 + ... kncvn = 0

the set consists of vectors {v_1, v_2, ..., v_n }, which was stated to be a basis, thus implying that any set with these vectors must = 0 for each constant,since c is a nonzero scalar, and this must mean that k1 through kn are all zero.

Does this seem like a logical/correct conclusion?

Anyways, my real question is:

It says that c is any nonzero scalar, but this doesn't make sense to me.

I was under the impression that a vector space can only have one basis. This problem contradicts that belief. Maybe my understanding of the terminology is wrong, but if a basis is a scalar multiple of another basis, they are essentially the same basis, right?

Sorry I know my wording for this question doesn't really make sense, I hope you guys can understand what I'm saying.

Through finishing this question, my train of thought is as follows:

Since a basis is
1) a linear combo which will = 0 if and only if the constant multiple is = 0, (its linearly independent)
2) spans every vector in the vector space (meaning every single vector in the vector space can be formed by the basis)

that means that you can have "multiple" basis, all being scalar multiplies of each other, because by taking a scalar multiple of those multiple basis, you can construct any vector in the vector space V.

Again I apologize because reading this I truly understand how stupid my questions sound but I'm having a hard time grasping these concepts, so please bear with me.

Also I posted this in precalculus since the problem doesn't involve any understanding of calculus. I hope this is in the correct section and I ask of the mods to move it if its not.
 
Physics news on Phys.org
You are over-thinking this. Consider any vector in the space, write it in terms of the given basis vectors {v_1, v_2, ..., v_n }, convert that expression to a linear combination of {cv_1, cv_2, ..., cv_n }. That does it.

You are wrong about there only being one set of basis vectors. Not only any nonzero multiplier, but lots of other linear combinations of a basis will also be a basis. Like {v_1, v_2+v_1, ..., v_n+v_1} will be a basis.
 
Rijad Hadzic said:
I now only need to check for linear independence (or span, but it makes no sense to check for the more difficult condition.)

So [ k1cv1 + k2cv2 + ... kncvn = 0

the set consists of vectors {v_1, v_2, ..., v_n }, which was stated to be a basis, thus implying that any set with these vectors must = 0 for each constant,since c is a nonzero scalar, and this must mean that k1 through kn are all zero.

Does this seem like a logical/correct conclusion?
I don't think so.
Let's consider a simpler space, ##\mathbb R^2## and a set of two vectors: ##\vec u = <1, 0>## and ##\vec v = <2, 0>##.
I notice that the equation ##c_1\vec u + c_2 \vec v = \vec 0## is true when ##c_1 = c_2 = 0##. Does this imply that 1) ##\vec u## and ##\vec v## are linearly independent, and 2) that these two vectors form a basis for ##\mathbb R^2##? After all, there are two vectors.
 
Mark44 said:
I don't think so.
Let's consider a simpler space, ##\mathbb R^2## and a set of two vectors: ##\vec u = <1, 0>## and ##\vec v = <2, 0>##.
I notice that the equation ##c_1\vec u + c_2 \vec v = \vec 0## is true when ##c_1 = c_2 = 0##. Does this imply that 1) ##\vec u## and ##\vec v## are linearly independent, and 2) that these two vectors form a basis for ##\mathbb R^2##? After all, there are two vectors.

Isn't that different from this cause though, because it's not explicitly stated that u and v form a basis?
 
Rijad Hadzic said:
Isn't that different from this cause though, because it's not explicitly stated that u and v form a basis?
My point is that for any vectors u and v, we can always set up the equation ##c_1\vec u + c_2\vec v = \vec 0##, and find that ##c_1=0## and ##c_2 = 0##, whether or not u and v are linearly independent.
 
My earlier reply was base on what you said here.
Rijad Hadzic said:
So [ k1cv1 + k2cv2 + ... kncvn = 0

the set consists of vectors {v_1, v_2, ..., v_n }, which was stated to be a basis, thus implying that any set with these vectors must = 0 for each constant,since c is a nonzero scalar, and this must mean that k1 through kn are all zero.

I didn't notice that you had said this, later on.
Rijad Hadzic said:
Since a basis is
1) a linear combo which will = 0 if and only if the constant multiple is = 0, (its linearly independent)
That "if and only if" part is a subtlety that a lot of students miss. Clearly you get it, so my earlier post wasn't needed.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top