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

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In summary: Anyway, it's right... for the right reasons!In summary, the conversation discusses the properties of a basis for a vector space. It is shown that for any nonzero scalar c, the set of vectors {cv_1, cv_2, ..., cv_n} is also a basis for the same vector space. The concept of linear independence and span is discussed, and it is concluded that there can be multiple sets of basis vectors in a vector space, as long as they have the same dimensions. The conversation also mentions a simpler example in which the equation c_1 * u + c_2 * v = 0 is always true, regardless of whether u and v are linearly independent or form a basis.
  • #1
Rijad Hadzic
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 

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

What is a basis?

A basis is a set of vectors that can be used to represent any vector in a vector space. It is the foundation of linear algebra and is used to create a coordinate system in a vector space.

How is a basis determined?

A basis is determined by finding a set of linearly independent vectors in a vector space. This means that none of the vectors in the basis can be written as a linear combination of the other vectors in the basis.

What is the importance of a basis in linear algebra?

A basis is important because it allows us to easily represent and manipulate vectors in a vector space. It also allows us to solve systems of linear equations and understand the geometric properties of vector spaces.

Can a vector space have more than one basis?

Yes, a vector space can have infinitely many bases. This is because there are multiple sets of linearly independent vectors that can be used to represent any vector in a vector space.

How is a basis related to dimension?

The number of vectors in a basis is equal to the dimension of the vector space. This means that the more vectors in a basis, the higher the dimension of the vector space. For example, a basis with 3 vectors would represent a 3-dimensional vector space.

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