How to write basis for symmetric nxn matrices

In summary, to write down a basis for the space of nxn symmetric matrices, you can start by defining the canonical basis for the space of nx1 vectors and then use it to write the matrix with a 1 in the (i,j) position and 0 everywhere else. Then, for a general n, you can write out the n matrices with a 1 along the main diagonal and all other entries 0, and then add in n(n-1)/2 additional matrices by placing a 1 in one of the n(n-1)/2 entries above the main diagonal and a corresponding 1 below the main diagonal. This results in a total of n(n+1)/2 basis matrices. For example, for n=
  • #1
Opus_723
178
3

Homework Statement



Write down a basis for the space of nxn symmetric matrices.



The Attempt at a Solution



I just need to know what the notation for this sort of thing is. I understand what the basis looks like, and I was even able to calculate that it would have dimension (n/2)(n+1), but I can't think of how you're supposed to compactly write down an answer for this that isn't simply a rambling paragraph. For a given n, I could simply write down a whole bunch of matrices with ones and zeros in appropriate places. But for a general n, I don't know what the format would be.
 
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  • #2
You could start by defining the canonical basis for the space of nx1 vectors, say [itex]e_i[/itex] = the column vector with a 1 in the i'th position and 0 everywhere else. You can use this to succinctly write the matrix that has a 1 in the (i,j) position and 0 everywhere else, and from there it's easy enough to write a basis for the space of nxn symmetric matrices.
 
  • #3
First the are the n matrices with 1 at a single entry along the main diagonal and all other entries 0. Now there are [itex]n^2- n[/itex] entries NOT on the main diagonal and so n(n-1)/2 entries above the diagonal. Put a 1 into anyone of those and a 1 in the corresponding position below the main diagonal. That will give you the [itex]n+ \frac{n(n-1)}{2}= \frac{2n+ n^2- n}{2}= \frac{n^2+ n}{2}= \frac{n(n+1)}{n}[/itex] basis matrices.

If n= 2, 2(2+1)/2= 3 and the three basis matrices are
[tex]\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}[/tex]

Now, you try to write out the 3(3+1)/2= 6 basis matrices for the 3 by 3 symmetric matrices.
 

FAQ: How to write basis for symmetric nxn matrices

What is a basis for symmetric nxn matrices?

A basis for symmetric nxn matrices is a set of linearly independent matrices that can be used to represent any symmetric nxn matrix through linear combinations.

How do I determine the size of the basis for symmetric nxn matrices?

The size of the basis for symmetric nxn matrices is equal to the number of distinct eigenvalues of the matrix. This can be determined by finding the characteristic polynomial of the matrix and counting the number of distinct roots.

How do I find the basis for symmetric nxn matrices?

The basis for symmetric nxn matrices can be found by first diagonalizing the matrix to obtain a matrix with only eigenvalues on the diagonal. Then, for each distinct eigenvalue, construct a matrix with 1 in the corresponding diagonal entry and 0 elsewhere. These matrices will form a basis for symmetric nxn matrices.

Can any set of matrices be a basis for symmetric nxn matrices?

No, not all sets of matrices can be a basis for symmetric nxn matrices. The matrices must be linearly independent and must have the same dimension as the original matrix (nxn). Additionally, the matrices must be symmetric, meaning they are equal to their own transpose.

Can the basis for symmetric nxn matrices change?

Yes, the basis for symmetric nxn matrices can change if the eigenvalues of the matrix change. Additionally, if the matrix is not diagonalizable, the basis may also change. However, the size of the basis will always remain the same for a given matrix.

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