Linear algebra: Find the span of a set

Homework Statement Find the span of U=\{2,\cos x,\sin x:x\in\mathbb{R}\} (U is the subset of a space of real functions) and V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}Homework Equations-Vector space span-Linear independence-RankThe Attempt at a SolutionObjects in U :2,\cos x,\sin x are linearly independent, so they span \mathbb{R^3}.What's the
  • #1
gruba
206
1

Homework Statement


Find the span of [itex]U=\{2,\cos x,\sin x:x\in\mathbb{R}\}[/itex] ([itex]U[/itex] is the subset of a space of real functions) and [itex]V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}[/itex]

Homework Equations


-Vector space span
-Linear independence
-Rank

The Attempt at a Solution


Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
Let ,[itex]n=3\Rightarrow [V]= \begin{bmatrix}
a & b & b \\
b & a & b \\
b & b & a \\
\end{bmatrix}[/itex]

[itex]rref[V]=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\Rightarrow [/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].

But because [itex]V\subset\mathbb{R^n}\Rightarrow [/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

Is this correct?
 
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  • #2
gruba said:

Homework Statement


Find the span of [itex]U=\{2,\cos x,\sin x:x\in\mathbb{R}\}[/itex] ([itex]U[/itex] is the subset of a space of real functions) and [itex]V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}[/itex]

Homework Equations


-Vector space span
-Linear independence
-Rank

The Attempt at a Solution


Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
Let ,[itex]n=3\Rightarrow [V]= \begin{bmatrix}
a & b & b \\
b & a & b \\
b & b & a \\
\end{bmatrix}[/itex]

[itex]rref[V]=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\Rightarrow [/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].

But because [itex]V\subset\mathbb{R^n}\Rightarrow [/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

Is this correct?

No. Having ##a,b \neq 0## is not enough. Look at the determinant of your ##3 \times 3## matrix.
 
  • #3
Ray Vickson said:
No. Having ##a,b \neq 0## is not enough. Look at the determinant of your ##3 \times 3## matrix.

Thanks, conditions [itex]a\neq b,a\neq -2b,a\neq 0,b\neq 0[/itex] must be satisfied.
But are the span of [itex]U,V[/itex] correct?
Also, what happens if the previous conditions aren't satisfied?
 
  • #4
gruba said:
Thanks, conditions [itex]a\neq b,a\neq -2b,a\neq 0,b\neq 0[/itex] must be satisfied.
But are the span of [itex]U,V[/itex] correct?
Also, what happens if the previous conditions aren't satisfied?

That is a good question for you to think about. It might make a difference whether you have ##a = b## or ##a = -2b##.
 
  • #5
gruba said:

Homework Statement


Find the span of [itex]U=\{2,\cos x,\sin x:x\in\mathbb{R}\}[/itex] ([itex]U[/itex] is the subset of a space of real functions) and [itex]V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}[/itex]

Homework Equations


-Vector space span
-Linear independence
-Rank

The Attempt at a Solution


Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
Let ,[itex]n=3\Rightarrow [V]= \begin{bmatrix}
a & b & b \\
b & a & b \\
b & b & a \\
\end{bmatrix}[/itex]

[itex]rref[V]=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\Rightarrow [/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].

But because [itex]V\subset\mathbb{R^n}\Rightarrow [/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

Is this correct?
But I think the question is on the span of ##U## , and , in your post you have that U is a subset of a function space. Function spaces are usually infinite-dimensional; I cannot think of any function space that is not.
 
  • #6
gruba said:

Homework Statement


Find the span of [itex]U=\{2,\cos x,\sin x:x\in\mathbb{R}\}[/itex] ([itex]U[/itex] is the subset of a space of real functions) and [itex]V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}[/itex]

Homework Equations


-Vector space span
-Linear independence
-Rank

The Attempt at a Solution


Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
What's the definition of span? The functions 2, cosine, and sine aren't elements of ##\mathbb{R}^3##, so how can they span ##\mathbb{R}^3##?

Let [itex]n=3\Rightarrow [V]= \begin{bmatrix}
a & b & b \\
b & a & b \\
b & b & a \\
\end{bmatrix}[/itex]

[itex]rref[V]=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\Rightarrow [/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].

But because [itex]V\subset\mathbb{R^n}\Rightarrow [/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

Is this correct?
No, it's not correct. In the ##n=3## case, are you claiming the three vectors span both ##\mathbb{R}^2## and ##\mathbb{R}^3##? How can that possibly work? The elements of ##\mathbb{R}^2## are ordered pairs while the elements of ##\mathbb{R}^3## are 3-tuples.
 

FAQ: Linear algebra: Find the span of a set

What is the definition of "span" in linear algebra?

The span of a set of vectors is the set of all possible linear combinations of those vectors. In other words, it is the set of all possible vectors that can be created by multiplying each vector in the set by a scalar and adding them together.

How do you find the span of a set of vectors?

To find the span of a set of vectors, you can use Gaussian elimination or matrix operations to reduce the set of vectors to its simplest form, known as the echelon form. The span of the set will then be equal to the number of non-zero rows in the echelon form.

Is the span of a set of vectors unique?

No, the span of a set of vectors is not unique. It can vary depending on the choice of vectors in the set. However, the dimension of the span, which is the number of vectors in the set, will remain the same.

Can the span of a set of vectors be larger than the number of vectors in the set?

Yes, it is possible for the span of a set of vectors to have a dimension larger than the number of vectors in the set. This can happen if the vectors in the set are linearly dependent, meaning that one or more vectors can be written as a linear combination of the others.

How is the span of a set related to the concept of linear independence?

The span of a set of vectors is closely related to the concept of linear independence. If a set of vectors is linearly independent, then the dimension of its span will be equal to the number of vectors in the set. On the other hand, if the set of vectors is linearly dependent, then the dimension of its span will be less than the number of vectors in the set.

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