# Dimension of these vector spaces?

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• Feb 2nd 2014, 03:44 AM
Phyba
Dimension of these vector spaces?
Hello!

I don't know how to deduce the dimension of these vector spaces (Crying). Can anybody help me ? I want to know how to think and understand each dimension.

Determine the dimension of each of the http://upload.wikimedia.org/math/0/c...9c36e07886.png -vector spaces of matrices with real coefficients and order n x n:

- the space of all these matrices --->I know that the dimension of this one is n2 , the number of elements
- the space of diagonal matrices ---> would this be n?
- the space of the upper triangular matrices ----->I would say this is n2/2 + n/2 (half of the matrix plus the other half diagonal), but I'm not sure
- the space of the symmetric matrices
- the space of the antisymmetric matrices

Don't know about the last two

Thanks!
• Feb 2nd 2014, 06:45 AM
Plato
Re: Dimension of these vector spaces?
Quote:

Originally Posted by Phyba
Determine the dimension of each of the http://upload.wikimedia.org/math/0/c...9c36e07886.png -vector spaces of matrices with real coefficients and order n x n:
- the space of all these matrices --->I know that the dimension of this one is n2 , the number of elements CORRECT
- the space of diagonal matrices ---> would this be n? YES
- the space of the upper triangular matrices ----->I would say this is n2/2 + n/2 (half of the matrix plus the other half diagonal), but I'm not sure
- the space of the symmetric matrices
- the space of the antisymmetric matrices.

The triangular numbers are $\frac{n(n+1)}{2}$. That also counts the upper triangular matrices.

If you think about symmetric matrices, every upper triangular matrix can be made into a symmetric matrix. So how many are there?

For the anti-symmetric matrix, $A=-A^T$, it is necessary that the diagonal elements must be zero. There are $\frac{n(n-1)}{2}$ strictly upper triangular elements. Again we can use each of those into an antisymmetric matrix.
• Feb 2nd 2014, 07:02 AM
HallsofIvy
Re: Dimension of these vector spaces?
Quote:

Originally Posted by Phyba
Hello!

I don't know how to deduce the dimension of these vector spaces (Crying). Can anybody help me ? I want to know how to think and understand each dimension.

Determine the dimension of each of the http://upload.wikimedia.org/math/0/c...9c36e07886.png -vector spaces of matrices with real coefficients and order n x n:

- the space of all these matrices --->I know that the dimension of this one is n2 , the number of elements

Yes. For example, if n= 4, any such matrix is of the form [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}+ d\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}[tex]. The "dimension" of a vector space is the size of a basis for that space. Do you see how this shows a basis for the space of 2 by 2 matrices?

Quote:

- the space of diagonal matrices ---> would this be n?
Yes. You can "independently" choose the n numbers on the diagonal and all other numbers are 0. In the case of 2 by 2 matrices,
$\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$

Quote:

- the space of the upper triangular matrices ----->I would say this is n2/2 + n/2 (half of the matrix plus the other half diagonal), but I'm not sure
Yes. You can choose any of the n numbers on the diagonal ("half diagonal"?). Since there are $n^2$ numbers in the entire matrix so $n^2- n$ not on the diagonal so $\frac{n^2- n}{2}$ in the upper half. That gives a total of $n+ \frac{n^2- n}{2}= \frac{2n+ n^2- n}{2}= \frac{n^2+ n}{2}= \frac{n^2}{2}+ \frac{n}{2}$ as you say.

Quote:

- the space of the symmetric matrices
Exactly the same as the upper triangular matrices. Choose an upper triangular matrix, then replace the "0"s in the lower half with the corresponding number from the upper half. You still have $\frac{n^2}{2}+ \frac{n}{2}$ choices.

Quote:

- the space of the antisymmetric matrices
Again, exactly the same. Start with a symmetric matrix and multiply each of the numbers in the lower half by -1. You still have exactly $\frac{n^2}{2}+ \frac{n}{2}$ independent choices.

Quote:

Don't know about the last two

Thanks!
• Feb 2nd 2014, 07:03 AM
Phyba
Re: Dimension of these vector spaces?
So.. triangular, symmetric, and antisymmetric have all a dimension n(n+1)/2 ! Thank you very much, I can see now that symmetric and antisymmetric are almost the same except for the signs below the diagonal.

You helped me a lot!
• Feb 2nd 2014, 07:40 AM
SlipEternal
Re: Dimension of these vector spaces?
Quote:

Originally Posted by Phyba
So.. triangular, symmetric, and antisymmetric have all a dimension n(n+1)/2 ! Thank you very much, I can see now that symmetric and antisymmetric are almost the same except for the signs below the diagonal.

You helped me a lot!

Actually, Plato said that antisymmetric is $\dfrac{n(n-1)}{2}$, not $\dfrac{n(n+1)}{2}$. This is because for triangular and symmetric, the diagonal may be nonzero. For antisymmetric, the diagonal must be zero.
• Feb 3rd 2014, 10:54 AM
Deveno
Re: Dimension of these vector spaces?
For any matrix A, we have the following (easy to prove) facts:

1. A + AT is symmetric.
2. A - AT is anti-symmetric.
3. A = (1/2)(A + AT) + (1/2)(A - AT)
4. If A is symmetric AND anti-symmetric, A = 0.

Hopefully the above should convince you that the vector space of all nxn matrices (over a field where 1 + 1 is not 0) is the direct sum of the subspaces of symmetric and anti-symmetric matrices.

That, then, implies that the subspace of anti-symmetric matrices has dimension:

n2 - n(n+1)/2 = (2n2 - n2 - n)/2 = (n2 - n)/2 = n(n-1)/2

Here are bases for the symmetric and anti-symmetric subspaces for n = 3.

A basis for symmetric matrices:

$\left\{ \begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}, \begin{bmatrix}0&0&0\\0&1&0\\0&0&0 \end{bmatrix}, \begin{bmatrix}0&0&0\\0&0&0\\0&0&1 \end{bmatrix}, \begin{bmatrix}0&1&0\\1&0&0\\0&0&0 \end{bmatrix}, \begin{bmatrix}0&0&1\\0&0&0\\1&0&0 \end{bmatrix}, \begin{bmatrix}0&0&0\\0&0&1\\0&1&0 \end{bmatrix} \right\}$

A basis for anti-symmetric matrices:

$\left\{\begin{bmatrix} 0&1&0\\-1&0&0\\0&0&0 \end{bmatrix}, \begin{bmatrix}0&0&1\\0&0&0\\-1&0&0 \end{bmatrix}, \begin{bmatrix}0&0&0\\0&0&1\\0&-1&0 \end{bmatrix}\right\}$

Not surprisingly, 6 + 3 = 9.