# Determine if the subset is a subspace of the indicated vector space.

• May 23rd 2013, 11:15 AM
Lambin
Determine if the subset is a subspace of the indicated vector space.
PROBLEM:

Determine if the subset is a subspace of the indicated vector space. Give reasons for your answers.

1. The zero vector together with all vectors $x=(x_1,x_2)$ in $\mathbb{R}^2$ for which $\frac{x_2}{x_1}$ has a constant value.

ATTEMPT:

The book states that the subset satisfies the subspace criterion, therefore it is a subspace of the indicated vector space. However, I have reason to believe it to be otherwise. First,

$W=(x| 0+x);\frac{x_2}{x_1}$ and $x=(x_1, x_2)$.

Here, in set notation, I have defined that the subset in question, W, has a zero vector "together" with all vectors, x. By "together", I presume they mean the linear combination of the zero vector with all vectors.

Next, I observe that $x_2/x_1$ has a constant value. What does it mean to be a constant value? I asked myself. To be a constant value, $x_2$ and $x_1$ must also be constant values; they do not contain variable terms. In addition, if we suppose that $x_1$ be equal to 0, then the quotient is undefined—which also is not a constant value. The latter is my argument for why this subset is not a subspace of the indiciated vector space​.

Example:

The vector x = 0 + (0,2) would not belong in W because $\frac{1}{0}$ is undefined. Which also qualifies that it is not a constant value.

If we imagined two vectors which do belong in W, namely x = 0 + (1,1) and y = 0 + (-1,1). These two vectors indeed belong in W because (1,1) and (-1,1) are vectors in $\mathbb{R}^2$ that have constant values $x_2/x_1$ together with the zero vector.

However, the linear combination of the two produces a vector that does not belong in W, thereby violating the subspace critera,

Subspace criterion. If every vector of the form

${\alpha}_1x_1+{\alpha}_2x_2$

belongs to W whenever $x_1$ and $x_2$ belong to W, and ${\alpha}_1$ and ${\alpha}_2$ are arbitrary sclars, then W is a subspace of V.

I take this rule strictly because the subspace criterion states that is is a subspace if every vector is of the form.

Thoughts?
• May 23rd 2013, 12:45 PM
HallsofIvy
Re: Determine if the subset is a subspace of the indicated vector space.
Quote:

Originally Posted by Lambin
PROBLEM:

Determine if the subset is a subspace of the indicated vector space. Give reasons for your answers.

1. The zero vector together with all vectors $x=(x_1,x_2)$ in $\mathbb{R}^2$ for which $\frac{x_2}{x_1}$ has a constant value.

ATTEMPT:

The book states that the subset satisfies the subspace criterion, therefore it is a subspace of the indicated vector space. However, I have reason to believe it to be otherwise. First,

$W=(x| 0+x)$

I have no idea what this means. You appear to be defining W to be the set of all x "such that 0+ x" is what? There is no condition here. Am I at least correct that x is a vector in $R^2$, not a vector?

Quote:

$;\frac{x_2}{x_1}$

Quote:

and $x=(x_1, x_2)$.
Okay, so x is in $R^2$

Quote:

Here, in set notation, I have defined that the subset in question, W, has a zero vector "together" with all vectors, x.
No, that was not what you wrote. What you should have written is $W= \{0\}\cup \{(x_1 x_2)| \frac{x_2}{x_1}= c\}$ for some fixed value of c.

Quote:

By "together", I presume they mean the linear combination of the zero vector with all vectors.
You are aware that a "linear combination of the zero vector" with any vector is just that vector times a scalar aren't you? And, in particular, that " $0+ (x_1, x_2)$" is just $(x_1, x_2)$?

Quote:

Next, I observe that $x_2/x_1$ has a constant value. What does it mean to be a constant value? I asked myself. To be a constant value, $x_2$ and $x_1$ must also be constant values; they do not contain variable terms.
That makes no sense. $x_1$ and $x_2$ are numbers, there is no question of "constant" or "variable" for numbers. What this means is that if $(x_1, x_2)$ and $(x_3, x_4)$ are two vectors in this set then $\frac{x_1}{x_2}= \frac{x_3}{x_4}$

Quote:

In addition, if we suppose that $x_1$ be equal to 0, then the quotient is undefined—which also is not a constant value. The latter is my argument for why this subset is not a subspace of the indiciated vector space​.

Example:

The vector x = 0 + (0,2) would not belong in W because $\frac{1}{0}$ is undefined. Which also qualifies that it is not a constant value.
All you are saying is that if one of the components is 0, the other must also be 0.

Quote:

If we imagined two vectors which do belong in W, namely x = 0 + (1,1) and y = 0 + (-1,1). These two vectors indeed belong in W because (1,1) and (-1,1) are vectors in $\mathbb{R}^2$ that have constant values $x_2/x_1$ together with the zero vector.

However, the linear combination of the two produces a vector that does not belong in W, thereby violating the subspace critera,

Subspace criterion. If every vector of the form

${\alpha}_1x_1+{\alpha}_2x_2$

belongs to W whenever $x_1$ and $x_2$ belong to W, and ${\alpha}_1$ and ${\alpha}_2$ are arbitrary sclars, then W is a subspace of V.

I take this rule strictly because the subspace criterion states that is is a subspace if every vector is of the form.

Thoughts?
The problem is that you are completely mistaken as to what " $\frac{x_2}{x_1}$ has a constant value" means. You seem to be under the impression that it simply means "is a finite number" and that is not true. "Constant" means "always has the same value". (1, 1) and (-1, 1) can NOT be in such a subspace because the ratio of 1 to 1 is 1 and the ratio of -1 to 1 is -1. Those are not the same number so the $\frac{x_2}{x1}$ is NOT constant. I strongly recommend you review basic definitions since you are using a number of words, including "constant" in incorrect ways, as well as using meaningless notation such as "[tex]0+ (x_1, x_2)[tex]".
• May 23rd 2013, 12:51 PM
Lambin
Re: Determine if the subset is a subspace of the indicated vector space.
AFTERTHOUGHT:

I may have misinterpreted what was meant by "the zero vector together with all vectors", it could also mean that the subset must contain the zero vector along with all vectors. It could also mean that "together" is either referring to the linear combination of the zero vector with all vectors, or (to be less insular) the product of the zero vector with all vectors.

That's irrelevant, however. Even if we considered all cases to be true, the issue still stands. $0/0$ is of an indeterminate form, and not a constant value. And so, the zero vector can't possibly be contained in the subset with the condition that $\frac{x_2}{x_1}$ be a constant value.
• May 23rd 2013, 01:06 PM
Lambin
Re: Determine if the subset is a subspace of the indicated vector space.
Thanks for replying. Mathematics is not my native language, so I apologize for my mistakes. :p

Yes, I was fully aware that $0+(x_1, x_2)=(x_1,x_2)$, but I try to make a habit not to trivialize any problems no matter how intuitive it appears. Also, the book often attempts to be clever and would indeed ask if 0 + x is still a vector, so that it reminds the reader to regard the axioms and not mechanically run through the problem.

Thanks again.