Linear algebra: Find the span of a set

[gruba]

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
-Rank

The Attempt at a Solution

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?

 

[Ray Vickson]

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
-Rank

The Attempt at a Solution

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?

No. Having $a,b \neq 0$ is not enough. Look at the determinant of your $3 \times 3$ matrix.

 

[gruba]

No. Having $a,b \neq 0$ is not enough. Look at the determinant of your $3 \times 3$ matrix.

Thanks, conditions a\neq b,a\neq -2b,a\neq 0,b\neq 0 must be satisfied.
But are the span of U,V correct?
Also, what happens if the previous conditions aren't satisfied?

 

[Ray Vickson]

Thanks, conditions a\neq b,a\neq -2b,a\neq 0,b\neq 0 must be satisfied.
But are the span of U,V 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 $a = -2b$.

 

[WWGD]

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
-Rank

The Attempt at a Solution

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?

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.

 

[vela]

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
-Rank

The Attempt at a Solution

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

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 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?

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.