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Basis of set of skew symmetric nxn matrices

by indigogirl
Tags: basis, matrices, skew, symmetric
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indigogirl
#1
Sep12-06, 07:30 AM
P: 9
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!
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HallsofIvy
#2
Sep12-06, 09:18 AM
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Start counting. I hope you know that the dimension of the space of all nxn matrices is n2 because you can take any one 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 any one 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.


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