[LinAlg] Find the dimension of the subspace

[bornofflame]

Homework Statement

Find the dimension of the subspace of all vectors in $\mathbb{R}^3$ whose first and third entries are equal.

Homework Equations

The Attempt at a Solution

So I arrived at two solutions and I'm not entirely sure which is the valid one.

#1

Let $H \text{ be a subspace of } \mathbb{R}^3$
$H = \left\{ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mid a, b, c \in \mathbb{R}^3, a = c \right\}$

$a \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \rightarrow \mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$
Dim = # elements in $\mathcal{B}$ = 3
#2

Let $H \text{ be a subspace of } \mathbb{R}^3$
$H = \left\{ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mid a, b \in \mathbb{R}^3 \right\}$

$a \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \rightarrow \mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\}$
Dim = # elements in $\mathcal{B}$ = 2

I believe that the first one is correct, or at least more correct if it's wrong, since to span $\mathbb{R}^3$ we would need at 3 vectors.

 

[LCKurtz]

You don't know ahead of time you need span $R^3$. In your first example, is your first vector in the subspace?

 

[Mark44]

Homework Statement

Find the dimension of the subspace of all vectors in $\mathbb{R}^3$ whose first and third entries are equal.

Homework Equations

The Attempt at a Solution

So I arrived at two solutions and I'm not entirely sure which is the valid one.

#1

Let $H \text{ be a subspace of } \mathbb{R}^3$
$H = \left\{ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mid a, b, c \in \mathbb{R}^3, a = c \right\}$

$a \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \rightarrow \mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$
Dim = # elements in $\mathcal{B}$ = 3
#2

Let $H \text{ be a subspace of } \mathbb{R}^3$
$H = \left\{ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mid a, b \in \mathbb{R}^3 \right\}$

$a \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \rightarrow \mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\}$
Dim = # elements in $\mathcal{B}$ = 2

I believe that the first one is correct, or at least more correct if it's wrong, since to span $\mathbb{R}^3$ we would need at 3 vectors.

Your first solution is wrong, because it doesn't take into account that the first and third entries of any vector in the subspace have to be equal.

Also, the basis for a proper subspace of a vector space won't have as many vectors as the dimension of the vector space.

 

[LCKurtz]

Not to mention that the "arguments" leave a lot to be desired.

 

[member 587159]

There should be a bell ringing when you read "first coordinate and last coordinate are equal"-> this means that you can forget one coordinate, and should make you think about an isomorphism with a 2 dimensional space

 

[bornofflame]

Your first solution is wrong, because it doesn't take into account that the first and third entries of any vector in the subspace have to be equal.

Ahh! Not sure why I didn't realize that when I was writing it, but as soon as I saw your words I was like, yup.

Also, the basis for a proper subspace of a vector space won't have as many vectors as the dimension of the vector space.

I didn't realize this. Is this a fact that I should just gravitate towards based on the theorems, definitions and such that I've learned or this a theorem or something that I have yet to encounter.

There should be a bell ringing when you read "first coordinate and last coordinate are equal"-> this means that you can forget one coordinate, and should make you think about an isomorphism with a 2 dimensional space

Oh, it rang. Not loud enough I guess or I wouldn't have made the mistake with the first one that Mark44 pointed out. I'm not sure what you mean by forget about one coordinate: Because the two coordinates are the same, the vector can be looked at as being a member of $\mathbb{R}^2$, since $a = c, x_1 = x_3$, etc?
So then a linear mapping from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ in a sense (b/c we're dealing with a subspace of $\mathbb{R}^3$) where each b in $\mathbb{R}^m$ is mapped to exactly one x in $\mathbb{R}^n$

Here for my own benefit:
Isomorphism: A one-to-one linear mapping from one vector space onto another.

one-to-one: A mapping $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that each b in $\mathbb{R}^m$ is the image of at most one } x in $\mathbb{R}^n$

onto: A mapping $T:R^n \rightarrow R^m$ such that each b in $R^m$ is the image of at least one x in $R^n$

 

[Mark44]

Also, the basis for a proper subspace of a vector space won't have as many vectors as the dimension of the vector space.

I didn't realize this. Is this a fact that I should just gravitate towards based on the theorems, definitions and such that I've learned or this a theorem or something that I have yet to encounter.

None of the above. It's fairly easy to understand based just on a bit of geometry.

For your problem, the vector space was $\mathbb R^3$, which is of dimension 3, and requiring a basis with three vectors. There can be subspaces whose dimension is zero, one, two, or three. The subspace in your problem is a plane in space (necessarily passing through the origin), so its dimension is two, and a basis would consist of two linearly independent vectors. The ones you listed in your 2nd attempt would serve this purpose.

Subspaces of dimension 1 would be lines in space passing through the origin. Any nonzero vector along this line could be a basis for this subspace. You could also have a subspace of dimension 0 (a single point -- the origin). This one isn't very interesting.

In higher dimensions, say $\mathbb R^n$, if you have one less equation than the number of variables and end up with one free variable, the subspace is a "hyperplane" embedded in the higher dimension space. With one free variable, a basis will require n - 1 linearly independent vectors.

 

[LCKurtz]

@bornofflame: At the risk of repeating myself, even if you now somewhat understand this problem, if it is a homework problem what you have shown is not a satisfactory proof to hand in. It is a small step from what you have written to make a correct argument that your subspace is of dimension 2 and you have a basis for it, but you haven't shown it in such a way that I am convinced you really understand it. If your argument was clear, you would have known the first try was incorrect.

 

[bornofflame]

None of the above. It's fairly easy to understand based just on a bit of geometry.

For your problem, the vector space was $\mathbb R^3$, which is of dimension 3, and requiring a basis with three vectors. There can be subspaces whose dimension is zero, one, two, or three. The subspace in your problem is a plane in space (necessarily passing through the origin), so its dimension is two, and a basis would consist of two linearly independent vectors. The ones you listed in your 2nd attempt would serve this purpose.

Subspaces of dimension 1 would be lines in space passing through the origin. Any nonzero vector along this line could be a basis for this subspace. You could also have a subspace of dimension 0 (a single point -- the origin). This one isn't very interesting.

In higher dimensions, say $\mathbb R^n$, if you have one less equation than the number of variables and end up with one free variable, the subspace is a "hyperplane" embedded in the higher dimension space. With one free variable, a basis will require n - 1 linearly independent vectors.

As soon as you mentioned that subspaces could could have dimensions less than the vector space that they occupy I realized that this was something that I should have know, like you said, based on just a bit of geometry. I'm annoyed b/c I did not realize that and I should have.

@bornofflame: At the risk of repeating myself, even if you now somewhat understand this problem, if it is a homework problem what you have shown is not a satisfactory proof to hand in. It is a small step from what you have written to make a correct argument that your subspace is of dimension 2 and you have a basis for it, but you haven't shown it in such a way that I am convinced you really understand it. If your argument was clear, you would have known the first try was incorrect.

Unfortunately, you're correct; I don't believe that I understand it well enough otherwise I would have realized that, as you say. This is a homework problem but for my own benefit since the professor doesn't collect or go over any of the homework. So knowing that this argument wouldn't cut it now is better to hear now then to see later on a test b/c I wasn't aware. I've spread myself a bit thin this semester and so I feel like my learning has been on the superficial side as I go from subject to subject. There are still some fundamentals that I am shakey on, such as isomorphism, one-to-one, etc, but I'm working on it.

That aside, using what's been pointed out/what I know:
I understand that in order to find the dimensions of a vector/sub space, you need to find the basis because however many vectors are in a basis is equal to the dimension for that space. In order to create any vector whose first and third entries are the same, we would only need two vectors; one for the first and second entries, and one for the second entries. These two vectors are linearly independent and would form the basis for a subspace whose vectors contain entries where the first and third are equal, and the only way to get to the zero vector using these vectors is if the weights are zero. With that in mind, the dimension would be two because the number of vectors in the basis is two.

Is that adequate?

 

[Mark44]

In order to create any vector whose first and third entries are the same, we would only need two vectors; one for the first and second third entries, and one for the second entries.

Fixed the above.
A system of equations that represents the vectors in your subspace would be:
x = x
y = y
z = x
Here x and y are arbitrary (free variables), but z is dependent on x. From this system of equations you can go directly to a vector equation, as you did in your work in post #1.
$\begin{bmatrix}x \\ y \\z \end{bmatrix} = \begin{bmatrix}x \\ y \\x \end{bmatrix} = x\begin{bmatrix}1 \\ 0 \\1 \end{bmatrix} + y\begin{bmatrix}0 \\ 1 \\0 \end{bmatrix}$, and there's your basis for the subspace. The two vectors are obviously linearly independent, since neither is a multiple of the other. The subspace, a plane in $\mathbb R^3$ is spanned by these two vectors, and is generated by the set of all linear combinations of these two vectors (which is just restating "spanned by" in different words) .

These two vectors are linearly independent and would form the basis for a subspace whose vectors contain entries where the first and third are equal

 

[LCKurtz]

Unfortunately, you're correct; I don't believe that I understand it well enough otherwise I would have realized that, as you say. This is a homework problem but for my own benefit since the professor doesn't collect or go over any of the homework. So knowing that this argument wouldn't cut it now is better to hear now then to see later on a test b/c I wasn't aware. I've spread myself a bit thin this semester and so I feel like my learning has been on the superficial side as I go from subject to subject. There are still some fundamentals that I am shakey on, such as isomorphism, one-to-one, etc, but I'm working on it.

That aside, using what's been pointed out/what I know:
I understand that in order to find the dimensions of a vector/sub space, you need to find the basis because however many vectors are in a basis is equal to the dimension for that space. In order to create any vector whose first and third entries are the same, we would only need two vectors; one for the first and second entries, and one for the second entries. These two vectors are linearly independent and would form the basis for a subspace whose vectors contain entries where the first and third are equal, and the only way to get to the zero vector using these vectors is if the weights are zero. With that in mind, the dimension would be two because the number of vectors in the basis is two.

Is that adequate?

That is adequate enough that I now believe you understand the issues. But I assume your class is a math class and your teacher would expect equations instead of a paragraph of prose. So you can start by calling your subspace M. Then let $\begin{bmatrix} a \\ b \\ a \end{bmatrix}$ be an arbitrary element of M. Then observe, as you have, $\begin{bmatrix} a \\ b \\ a \end{bmatrix} = a\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}+b\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. That shows that your two vectors span M. Then, similarly, give an equation showing that the only way the zero vector can be a similar linear combination of your two vectors is if $a=0,~b=0$. That will show they are independent and those are the two things showing you have a basis and the dimension is 2. It is just as you have described above, but written mathematically.

 

[bornofflame]

Fixed the above.
A system of equations that represents the vectors in your subspace would be:
x = x
y = y
z = x
Here x and y are arbitrary (free variables), but z is dependent on x. From this system of equations you can go directly to a vector equation, as you did in your work in post #1.
$\begin{bmatrix}x \\ y \\z \end{bmatrix} = \begin{bmatrix}x \\ y \\x \end{bmatrix} = x\begin{bmatrix}1 \\ 0 \\1 \end{bmatrix} + y\begin{bmatrix}0 \\ 1 \\0 \end{bmatrix}$, and there's your basis for the subspace. The two vectors are obviously linearly independent, since neither is a multiple of the other. The subspace, a plane in $\mathbb R^3$ is spanned by these two vectors, and is generated by the set of all linear combinations of these two vectors (which is just restating "spanned by" in different words) .

Just that "small" step of writing it down as a system of equations using x, y, and z and turning that into a vector was quite helpful. Thank you.

That is adequate enough that I now believe you understand the issues. But I assume your class is a math class and your teacher would expect equations instead of a paragraph of prose.

Absolutely. I'm still quite the beginner with LaTex so, in the time I had, it was quicker to write it out like that.

So you can start by calling your subspace M. Then let $\begin{bmatrix} a \\ b \\ a \end{bmatrix}$ be an arbitrary element of M. Then observe, as you have, $\begin{bmatrix} a \\ b \\ a \end{bmatrix} = a\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}+b\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. That shows that your two vectors span M. Then, similarly, give an equation showing that the only way the zero vector can be a similar linear combination of your two vectors is if $a=0,~b=0$. That will show they are independent and those are the two things showing you have a basis and the dimension is 2. It is just as you have described above, but written mathematically.

Thank you for that.

 

[Ray Vickson]

Just that "small" step of writing it down as a system of equations using x, y, and z and turning that into a vector was quite helpful. Thank you.Absolutely. I'm still quite the beginner with LaTex so, in the time I had, it was quicker to write it out like that.Thank you for that.

You don't need LaTeX to write the equations correctly. For example, if you write vectors as rows instead of columns, it is pretty easy to write
(a,b,a) = a(1,0,1) + b(0,1,0), and that says it all.

Of course, LaTeX looks better, and I would join others in urging you to start using it, but the more important issue is understanding, not formatting and typesetting.