MHB Determining if a set is a subspace.

[AngrySnorlax]

Hey there guys, its AngrySnorlax here again with another problem. I posted here before when I was having an issue and the responses I got were extremely helpful because there was a specific step that I just could not grasp that was explained to me and I am hoping that is the same situation here with these problems.

I am trying to determine how to tell if a set is a subspace. The problem reads like this:
Determine if the described set is a subspace. If so, give a proof. If not, explain why not. Unless stated otherwise, a, b, and c are real numbers.

The subset of {R}^{3} consisting of vectors of the form
$$\left[\begin{array}{c}a \\ 0 \\ b \end{array}\right]$$

The answer at the back of the book reads this:
This is a subspace, equal to span
$$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$$$$\left[\begin{array}{c}0 \\ 0 \\ 1 \end{array}\right]$$

Now I am aware that there are three steps, but I am not entirely sure on how to use these steps.

1) S must contain a 0 vector.
2) If u and v are in S, then u+v is also in S.
3) If r is a real number and u is in S, the ru is also in S.

I am totally lost in applying these steps! I have an idea on how to apply the first one, but even still I don't feel confident in how I am applying it. So once again, I would be eternally grateful for any help :)

 

[topsquark]

Hey there guys, its AngrySnorlax here again with another problem. I posted here before when I was having an issue and the responses I got were extremely helpful because there was a specific step that I just could not grasp that was explained to me and I am hoping that is the same situation here with these problems.

I am trying to determine how to tell if a set is a subspace. The problem reads like this:
Determine if the described set is a subspace. If so, give a proof. If not, explain why not. Unless stated otherwise, a, b, and c are real numbers.

The subset of {R}^{3} consisting of vectors of the form
$$\left[\begin{array}{c}a \\ 0 \\ b \end{array}\right]$$

The answer at the back of the book reads this:
This is a subspace, equal to span
$$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$$$$\left[\begin{array}{c}0 \\ 0 \\ 1 \end{array}\right]$$

Now I am aware that there are three steps, but I am not entirely sure on how to use these steps.

1) S must contain a 0 vector.
2) If u and v are in S, then u+v is also in S.
3) If r is a real number and u is in S, the ru is also in S.

I am totally lost in applying these steps! I have an idea on how to apply the first one, but even still I don't feel confident in how I am applying it. So once again, I would be eternally grateful for any help :)

1) Is there an a, b such that $$\left[\begin{array}{c}a \\ 0 \\ b \end{array}\right] = \left[\begin{array}{c}0 \\ 0 \\ 0 \end{array}\right]$$

2) Let [math]u = \left[\begin{array}{c}c \\ 0 \\ d \end{array}\right] [/math] and [math]v = \left[\begin{array}{c}e \\ 0 \\ f \end{array}\right] [/math]. Is u + v in the form [math]\left[\begin{array}{c}a \\ 0 \\ b \end{array}\right] [/math]?

3) I'll let you figure this out from here. Let us know if you still have troubles.

-Dan

 

[AngrySnorlax]

I still don't get it. I think I may be over thinking it or I just haven't found the right way to think about it yet.

 

[topsquark]

I still don't get it. I think I may be over thinking it or I just haven't found the right way to think about it yet.

What are you not getting? Without knowing we are simply guessing as to how to help.

Do you understand that the zero vector is [math]\left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ] [/math]? Do you understand that this is expressible as [math]\left [ \begin{matrix} a \\ 0 \\ b \end{matrix} \right ] [/math] for a = b = 0, and is thus a member of the proposed subspace?

-Dan

 

[Deveno]

Hey there guys, its AngrySnorlax here again with another problem. I posted here before when I was having an issue and the responses I got were extremely helpful because there was a specific step that I just could not grasp that was explained to me and I am hoping that is the same situation here with these problems.

I am trying to determine how to tell if a set is a subspace. The problem reads like this:
Determine if the described set is a subspace. If so, give a proof. If not, explain why not. Unless stated otherwise, a, b, and c are real numbers.

The subset of {R}^{3} consisting of vectors of the form
$$\left[\begin{array}{c}a \\ 0 \\ b \end{array}\right]$$

The answer at the back of the book reads this:
This is a subspace, equal to span
$$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$$$$\left[\begin{array}{c}0 \\ 0 \\ 1 \end{array}\right]$$

Now I am aware that there are three steps, but I am not entirely sure on how to use these steps.

1) S must contain a 0 vector.
2) If u and v are in S, then u+v is also in S.
3) If r is a real number and u is in S, the ru is also in S.

I am totally lost in applying these steps! I have an idea on how to apply the first one, but even still I don't feel confident in how I am applying it. So once again, I would be eternally grateful for any help :)

The condition (1) may be replaced by $S \neq \emptyset$, because then, given:

$u \in S$, condition (3) implies $-u = (-1)u \in S$, and condition (2) implies that since $u,-u \in S$, so is $u + -u = 0$.

This is worth remembering, and it is clear your set is non-empty, so we really needn't worry overmuch about the 0-vector. Of course, it is fairly clear taking $a = b = 0$, that:

$\begin{bmatrix}0\\0\\0\end{bmatrix}$ is of the form $\begin{bmatrix}a\\0\\b\end{bmatrix}$.

The two closure conditions (2) and (3) are far more important. Here is how you would verify (2):

Let $u = \begin{bmatrix}a\\0\\b\end{bmatrix}$ and $v = \begin{bmatrix}a'\\0\\b'\end{bmatrix}$.

Then $u + v = \begin{bmatrix}a\\0\\b\end{bmatrix} + \begin{bmatrix}a'\\0\\b'\end{bmatrix} = \begin{bmatrix}a+a'\\0\\b+b'\end{bmatrix}$

and clearly $a+a',b+b'$ are real numbers if $a,a',b,b'$ are.

Intuitively, what we mean by $S$, here, is "all 3-vectors with 2nd coordinate zero". Said this way, it is obvious the 0-vector has second coordinate zero (ALL its coordinates are zero, including the second one), and that adding two vectors with second coordinate zero, still leaves the second coordinate zero (because we add "coordinate-to-coordinate"), and that multiplying such a vector by any scalar STILL leaves the second coordinate zero (if our scalar is $r$, the second coordinate will be $r\cdot 0 = 0$).