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

$rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{bmatrix}\Rightarrow$ 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 ,$n=3\Rightarrow [V]= \begin{bmatrix}<br /> a & b & b \\<br /> b & a & b \\<br /> b & b & a \\<br /> \end{bmatrix}$

$rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{bmatrix}\Rightarrow$ 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 ,$n=3\Rightarrow [V]= \begin{bmatrix}<br /> a & b & b \\<br /> b & a & b \\<br /> b & b & a \\<br /> \end{bmatrix}$

$rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{bmatrix}\Rightarrow$ 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 $n=3\Rightarrow [V]= \begin{bmatrix}<br /> a & b & b \\<br /> b & a & b \\<br /> b & b & a \\<br /> \end{bmatrix}$

$rref[V]=\begin{bmatrix}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{bmatrix}\Rightarrow$ 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.