Linear algebra: Find the span of a set

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SUMMARY

The span of the set U = {2, cos x, sin x: x ∈ ℝ} consists of linearly independent functions that span a three-dimensional space of real functions, not ℝ³. For the set V = {(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b ∈ ℝ}, the vectors span ℝ³ if a and b are non-zero. However, since V is a subset of ℝⁿ, the vectors actually span ℝⁿ-1 when considering the dimensionality of the space.

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gruba
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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


- Span
-Subset

The Attempt at a Solution


[/B]
Objects in U :2,\cos x,\sin x are linearly independent, so they span \mathbb{R^3}.

Let ,n=3\Rightarrow [V]= \begin{bmatrix}<br /> a &amp; b &amp; b \\<br /> b &amp; a &amp; b \\<br /> b &amp; b &amp; a \\<br /> \end{bmatrix}

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

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

Is this correct?
 
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gruba said:

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


- Span
-Subset

The Attempt at a Solution


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

rref[V]=\begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> \end{bmatrix}\Rightarrow vectors in V span \mathbb{R^3}, if a,b\neq 0.
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 V\subset\mathbb{R^n}\Rightarrow vectors span \mathbb{R^{n-1}}.

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

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


- Span
-Subset

The Attempt at a Solution


[/B]
Objects in U :2,\cos x,\sin x are linearly independent, so they span \mathbb{R^3}.
They aren't even in ##\mathbb{R}^3##; they belong to a space of real functions.
 
To clarify/correct my "Yes" in post #2, the functions span a three-dimensional space of functions, not ##\mathbb{R}^3##.
 
Objects in U :2,\cos x,\sin x are linearly independent, so they span \mathbb{R^3}.
Yes, these vectors are linearly independent.
No, these vectors span a space isomorphic to \mathbb{R}^3

Regarding V. If a=b, then all of those vectors are linearly dependent. If a=0, then the vectors will span a null space. If a\neq 0, then the system of vectors will be reduced to only one vector and so it would span \mathbb{R}^1\subset \mathbb{R}^n
If a\neq b, 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.
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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