## anonymous 5 years ago example finding the basis and dimension of the subspace of X of a 2 x 2 matrix?

1. anonymous

What exactly do you need? Do you have a specific question?

2. anonymous

If you post your question and I'm not around, I'll get an e-mail about it and I'll help you out as soon as I can.

3. anonymous

The question is --- finding the basis and dimension of the subspace of X of a 2 x 2 matrix?

4. anonymous

The basis will be dependent upon the form of the subspace. I mean, if we're taking a normal 2 x 2 matrix, M = |a b| |c d| the standard basis is the set: B = { |1 0|, |0 1|, |0 0|, |0 0|} |0 0| |0 0| |1 0| |0 1| and the dimension is, by definition, the cardinality of the basis (i.e. the number of elements in the basis). So $\dim(M_{2 \times 2})=4$

5. anonymous

You can get basis sets that have a different number of elements (and therefore, dimension) depending on special conditions. For example, if the 2 x 2 matrix is to be symmetric, then you'd have, S = |a b| = a|1 0| + b|0 1| + c|0 0| |b c| |0 0| |1 0| |0 1|. Therefore the set { |1 0| , |0 1| , |0 0| } |0 0| |1 0| |0 1| spans the subset of all symmetric 2 x 2 matrices, and can be shown to be linearly independent, and so forms a basis for this subset. The dimension in this case would be 3.

6. anonymous

Incidentally, it's easy to show that these basis elements are linearly independent, since by the definition of linear independence, a set of vectors in linearly independent if the equation $c_1v_1+c_2v_2+...+c_nv_n=0$has only the trivial solution,$c_1=c_2=...=c_n=0$

7. anonymous

For the first basis, letting each of the v_i be one of the matrices, $c_1v_1+c_2v_2+c_3v_3+c_4v_4$ = |c_1 c_2| |c_3 c_4| = |0 0| |0 0| if and only if$c_1=c_2=c_3=c_4=0$the trivial solution.

8. anonymous

You can do similar with the second basis.