Linear algebra: Find the span of a set

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gruba
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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


- Span
-Subset

The Attempt at a Solution


[/B]
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}<br /> a & b & b \\<br /> b & a & b \\<br /> b & b & a \\<br /> \end{bmatrix}[/itex]

[itex]rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \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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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


- Span
-Subset

The Attempt at a Solution


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

[itex]rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{bmatrix}\Rightarrow[/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].
I don't know how valid your argument is, here. It's given that ##V \subset \mathbb{R}^n##. Can you extend this to a statement about an n-dimensional space instead of a 3-dimensional space?
gruba said:
But because [itex]V\subset\mathbb{R^n}\Rightarrow[/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

Is this correct?
Why do you conclude that the vectors span ##\mathbb{R}^{n - 1}##?
 
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


- Span
-Subset

The Attempt at a Solution


[/B]
Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
They aren't even in ##\mathbb{R}^3##; they belong to a space of real functions.
 
Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].
Yes, these vectors are linearly independent.
No, these vectors span a space isomorphic to [itex]\mathbb{R}^3[/itex]

Regarding [itex]V[/itex]. If a=b, then all of those vectors are linearly dependent. If a=0, then the vectors will span a null space. If [itex]a\neq 0[/itex], then the system of vectors will be reduced to only one vector and so it would span [itex]\mathbb{R}^1\subset \mathbb{R}^n[/itex]
If [itex]a\neq b[/itex], then all of the vectors are in fact linearly independent: one can construct an n x n matrix of said vectors and reduce it to a diagonal matrix.
 
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