Question

example finding the basis and dimension of the subspace of X of a 2 x 2 matrix?

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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 \[\dim(M_{2 \times 2})=4\]

Answer 5

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.

Answer 6

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,\[c_1=c_2=...=c_n=0\]

Answer 7

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.

Answer 8

You can do similar with the second basis.