Basis of set of skew symmetric nxn matrices

[indigogirl]

Hi,

I am having trouble with the question above. In general, I have trouble with questions like:

What is the basis for all nxn matrices with trace 0? What is the dimension?

What is the basis of all upper triangular nxn matrices? What is the dimension?

Please help!

 

[HallsofIvy]

Start counting. I hope you know that the dimension of the space of all nxn matrices is n² because you can take anyone of the n² entries 1, the other 0, to get a basis matrix.

There is no such thing as "the" basis for a vector space- any vector space has an infinite number of bases, each having the same number (the dimension) of vectors in it.

For the dimension of "all nxn matrices with trace 0", start by looking at small n. For n= 2, a 2x2 matrix is of the form
[tex]\left[\begin{array}{cc}a & b \\ c & d \end{array}\right][/tex]
Since we could choose anyone of the four entries, a, b, c, d equal to 1 its dimension is, as I said above, 2²= 4. What about "traceless" matrices, with trace 0? Now we require that a+ d= 0. I could still choose b, c to be anything I want but now I must have d= -a. I can choose a to be whatever I want but then d is fixed- I have 3 arbitrary choices so the dimension is 3. A possible basis is
[tex]\left{\left[\begin{array}{cc}1 & 0 \\0 & -1\end{array}\right], \left[\begin{array}{cc}0 & 1 \\0 & 0\end{array}\right], \left[\begin{array}{cc}0 & 0 \\1 & 0\end{array}\right]\right}[/tex]
where I have chosen a, b, c, in turn to be 1, others 0, with d= -a.

Okay, in a general axa matrix I could choose any of the n² entries arbitrarily, but in a trace 0 matrix, I am not free to choose all the diagonal entries arbitrarily. Since I have one equation that must be satisfied, I could choose all but one of the diagonal entries arbitrarily, then solve for the last- I have n²- 1 arbitrary choices.

For upper triangular matrices, much the same thing. A general nxn matrix has n² entries, n of them on the diagonal leaving n²- n "off diagonal" entries. Exactly half of those, (n²- n)/2, are above the diagonal and half below. With an upper triangular matrix, the entries below the diagonal must be 0 so I can't choose them arbitrarily. I can choose all the n entries on the diagonal and the (n²- n)/2 entries above the diagonal arbitrarily, a total of n+ (n²- n)/2= (n²+ n)/2 choices. Again, you can construct a basis (not "the" basis) by choosing each entry, in turn, to be 1, all others 0.

 

[tacman]

Hi there,

I understand that you are having trouble with questions related to the basis of certain sets of nxn matrices. Let me try to break it down for you and provide some guidance.

Firstly, a basis is a set of linearly independent vectors that span a vector space. In the case of matrices, the basis would be a set of linearly independent matrices that can be used to represent any other matrix in that particular set.

Now, let's tackle the first question regarding the basis for all nxn matrices with trace 0. The trace of a matrix is the sum of its diagonal elements, and for an nxn matrix, it will always be a single value. So, in order for the trace to be 0, all the diagonal elements must be 0. This means that the basis for this set of matrices would be all nxn matrices with 0 on the diagonal and any non-zero entries in the other positions. As for the dimension, it would be n^2 - n, since we have n^2 total entries in an nxn matrix, but n of them are fixed at 0.

Moving on to the second question, the basis for all upper triangular nxn matrices would be all matrices with 0 in the lower triangular positions (below the main diagonal) and any non-zero entries in the upper triangular positions (above the main diagonal). The dimension for this set would be n(n+1)/2, since there are n(n+1)/2 total entries in an nxn upper triangular matrix and n of them are fixed at 0.

I hope this helps clarify things for you. Remember, the key is to understand the properties and characteristics of the set in question and think about what types of matrices would be needed to represent any other matrix in that set. Best of luck!