Originally Posted by

**Hartlw** Your terminology is confusing

You have expressed a complex matrix (an mxn matrix with complex components) as a linear combination of mxnx2 linearly independent complex matrices, so the basis dimension is 12. Period.

There is another way of looking at it which is a matter of convention.

If you assume that the matrix is an ordered mxn n-tuple (a vector), you can write it out as an mxn n-tuple (vector).

if aij is a typical component of A, then:

A $\displaystyle \equiv$ (a11,a12...,a1n, a21,a22,....,a2n,........,am1, am2,......,amn)

The ntuple has mxn components. If they are real, they represent the real matrix as an mxn dimensional vector space with mxn base vectors (1,0,0,0,0,0...), (0,1,000000), (0,0,1,0000) .......

If the matrix is complex, then the vector form of A has complex components and it takes two real numbers to specify each component, so now the dimension is 2xmxn.

A set of 2xmxn base vectors is now: [(0,1),(0,0),(0,0),...........], [(1,0),(0,0),(,0),.......], [(0,0),(1,0),(0,0,0.....], [(0,0),(0,1)....], .......

The above converts the question to:

What is the dimension of a vector (n-tuple) with complex components? Another way to look at (which I,ve never heard of) is write double the number of components with the understanding that the first component is the real part of a complex number a1 and the second component is the imaginary part of a complex number a1, the THIRD component is the real part of a2, the fourth componet is the imaginary component of a2, etc.

It would be interesting to examine vectors with complex components.

Personally, I think my previous post was very concise and clear.

(You are trying to say something like a complex number is an ordered pair over a real Field, but that's reduntant when you specify the aij as complex.)