# Dimension and basis for a subspace

1. Oct 10, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
$\mathbb{H} = \{(a,b,c) : a - 3b + c = 0,~b - 2c = 0,~2b - c = 0 \}$

2. Relevant equations

3. The attempt at a solution
This definition of a subspace gives us the vector $(3b - c,~2c,~2b) = b(3,0,2) + c(-1,2,0)$. This seems to suggest that a basis is {(3, 0, 2), (-1, 2 0)}, and that the subspace is 2-dimensional. However, if I take a different approach and solve the homogeneous system given by the subspace, we have that the only a, b and c that satisfy the system is (0, 0 , 0), which is to say that the subspace has no basis and is zero-dimensional. What am I doing wrong to get these two different answers?

2. Oct 10, 2016

### Staff: Mentor

The entire expression would have been $(a,b,c) = (3b - c,~2c,~2b) = b(3,0,2) + c(-1,2,0)$ since you substituted for $a,b,c$.
And this can only hold for $b=2c=4b$, i.e. $b=0$ and so on.

3. Oct 10, 2016

### Mr Davis 97

But isn't it true by the calculations shown that the subspace is all linear combinations of (3,0,2) and (-1,2,0)? And since these two vectors are linear independent, the form a basis?

4. Oct 10, 2016

### Staff: Mentor

No. Only one linear combination $0 \,\cdot\, (3,0,2) + 0\,\cdot\, (-1,2,0)$ satisfies $(a,b,c) = b \,\cdot\,(3,0,2) + c\,\cdot\, (-1,2,0)$

5. Oct 10, 2016

### Staff: Mentor

To add to what @fresh_42 said, the three equations that define H are independent; the only solution is (a, b, c) = (0, 0, 0).

6. Oct 10, 2016

### Mr Davis 97

So in general, given a problem such as this, should I first always try to calculate the general solution to the several equations given, and then make claims about basis and dimension?

7. Oct 10, 2016

### Staff: Mentor

Apart from $(1,0,0)$, which you have forgotten, you correctly found $(3,0,2)$ and $(−1,2,0)$.
These are three vectors, which are linearly independent (see post #5), i.e. they span the entire vector space. This means in return, that there are no dimensions left for a non-trivial solution.

In terms of formulas, it reads as follows:

Let $\varphi\, : \, \mathbb{V} \rightarrow \mathbb{V}$ be a linear map, then $\dim \mathbb{V} = \dim kernel \;\varphi + \dim image \; \varphi$. The three vectors above arranged as a matrix represent $\varphi$. The image is $\mathbb{V}$, i.e. the rank of the matrix or dimension of the image is the dimension of $\mathbb{V}$, three in this case. Therefore the kernel $\mathbb{H}$ has to be $\{0\}$ and its dimension - also sometimes called the defect of $\varphi$ - is zero as well.

So to answer your question: It is always a good idea to find out as much as possible about the three values in the formula above.

8. Oct 10, 2016

### Mr Davis 97

Thanks. I have a similar but slightly different question.
If I were to try to show that H is a subspace, that is to try to show that it has the zero vector, that is it closed under addition and scalar multiplication, what is the "general" or "algorithmic" way to do this? Specifically, how would I represent an arbitrary element of H to show that is it both closed under addition and scalar multiplication?

9. Oct 10, 2016

### andrewkirk

H is the nullspace of the matrix $\begin{pmatrix}1&-3&1\\0&1&-2\\0&2&-1\end{pmatrix}$ formed from the equations given in the problem. It is a theorem of linear algebra, quite easy to prove, that the nullspace of a $n\times n$ real matrix is a subspace of $\mathbb R^n$.

If you want to know whether H is trivial, all that's needed is to do row reduction of the matrix and see if you end up with one or more zero rows. Alternatively, you can calculate the determinant to see if it's zero. If either of those happens, H will be nontrivial.

10. Oct 10, 2016

### Mr Davis 97

Thanks , so that is one easy way. What about if I didn't want to use any notion of matrix or nullspace, and just wanted to prove it's a subspace by the definition of subspace? I am specifically wondering how you would show that it's closed under addition and scalar multiplication.

11. Oct 11, 2016

### andrewkirk

Well the three equations all have a RHS of 0. So if you have $(a,b,c)$ that satisfies them, then so will $k(a,b,c)$, since that is $(ka,kb,kc)$ and substituting the components of that in the equations will give a RHS of $k\times 0=0$ for each equation.

Similarly, for each of the three equations, if you have a solution $(a,b,c)$ and another solution $(d,e,f)$ then $(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$ will be a solution since substituing the components of that latter vector in one of the defining equations of H is the same as adding together two instances of the equation, one in which $(a,b,c)$ has been substituted and the other in which $(d,e,f)$ has been. So the RHS is the sum of the two RHS, which is 0+0=0.

Finally, (0,0,0) is a solution of the equations since they are all homogeneous.

The concept of matrices and nullspaces just allows us to say all that more concisely.

12. Oct 11, 2016

### Math_QED

Let V be a subspace. It is easy to show from the definition of a subspace that all you have to do is:

You take vectors $a,b \in V$ and scalars $\alpha , \beta \in \mathbb{K}$ and then you show that $\alpha a + \beta b \in V$

13. Oct 11, 2016

### Mr Davis 97

But how would I represent an element in V in this specific situation?

14. Oct 11, 2016

### Staff: Mentor

In $\mathbb{H} = \{(a,b,c) : a - 3b + c = 0,~b - 2c = 0,~2b - c = 0 \}$ you have defined an element by
$$\begin{bmatrix}a \\ b \\c \end{bmatrix} \in \mathbb{H} \Longleftrightarrow \begin{bmatrix}a \\ b \\c \end{bmatrix} = \begin{bmatrix} 3b-c \\ 2c \\2b \end{bmatrix}\Longleftrightarrow \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -a+3b-c \\ -b+2c \\2b-c \end{bmatrix}$$

With that you have to show for $\begin{bmatrix}a \\ b \\c \end{bmatrix} , \begin{bmatrix}a' \\ b' \\c' \end{bmatrix} \in \mathbb{H}$ and $\alpha , \beta \in \mathbb{K}$ that $\alpha \,\cdot\, \begin{bmatrix}a \\ b \\c \end{bmatrix} \, + \beta \,\cdot\, \begin{bmatrix}a' \\ b' \\c' \end{bmatrix}$ satisfies the same three component equations, i.e.
$$\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -a''+3b''-c'' \\ -b''+2c'' \\2b''-c'' \end{bmatrix}$$
with $a''= \alpha a + \beta a'\, , \, b''= \alpha b + \beta b'\, , \, c'= \alpha c + \beta c'$

15. Oct 11, 2016

### Staff: Mentor

I assume that V is the set mentioned in post #7. If there is no description of V, there's no way to represent elements of V.

16. Oct 11, 2016

### Mr Davis 97

I meant H, the potential subspace in the original post. I'm an still not completely certain how I would show that it is a subspace. For example, in other similar problems, where we might have a set defined as $\mathbb{W} = \{(a,b,c) : a - 3b + c = 0\}$, it is simply a matter of showing that $(3b - c,b,c) + (3b'-c',b',c') = (3(b+b') - (c + c'),b+b',c+c')$, and hence the sum of two elements in W is still in W. For H, it seems that it is not this simple. And of course, I could just say that the equations form a matrix whose nullspace is a subspace, but I want to show that it is a subspace by satisfying the individual axioms, particularly the closure axioms.

17. Oct 11, 2016

### andrewkirk

That's what post #11 does.

18. Oct 11, 2016

### Mr Davis 97

Ah okay, I now actually understand how your argument goes. Just one more question; why didn't you make explicit substitutions in the addition of the two example vectors, as I did in post #16? I'm just trying to see why your way of doing it slightly differs from the way I did the example in post #16.

19. Oct 12, 2016

### Staff: Mentor

This thread is getting a bit on the long side, and it involves at least two different problems. andrewkirk's comment about his work in post #11 refers back to the original question, with the subspace H. As it turned out, that subspace consisted solely of the zero vector, and is an example of what is called a trivial subspace (of R3). Making that determination requires solving the system of equations, by the use of algebra, or by the use of matrix row reduction, or by calculating the determinant of the matrix of coefficients of the three equation (and noting that the determinant is nonzero).

If the subspace consists of only the zero vector, it's very simple to show that this subspace:
1) includes the zero vector (duh!)
2) is closed under vector addition -- $\vec{0} + \vec{0} = \vec{0}$, which is obviously in this set.
3) is closed under scalar multiplication -- $k\vec{0} = \vec{0}$, obviously in this set.

The work in post #16 involved a different subspace; namely $\{(a, b, c) \in R^3 : a - 3b + c = 0\}$.
This set is a two-dimensional subspace of R3, which some quick algebra shows is spanned by $\{\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix} \}$. These vectors form a basis for this subspace.