Read the too long didnt read version in bold before going into the finer detail.
The overall point is that when I decompose this matrix to try and find its determinant I get an answer that doesn't make any sense. I have a thought on why it doesn't make any sense, but that thought also doesn't make any sense to my lecturer.
I was playing with the determinant of a matrix with a countable infinity of columns and rows:
Trying to find the determinant of this matrix with countably infinite rows and columns.
\begin{bmatrix} v_{1} & v_{2} & v_{3} & v_{4}& v_{5} & \cdots \\ e^{a\times ln(1)} & e^{a\times ln(2)} & e^{a\times ln(3)} & e^{a\times ln(4)} & e^{aln(2)} & \cdots\\ e^{b\times ln(1)} & e^{b\times ln(2)} & e^{b\times ln(3)} & e^{b\times ln(4)} & e^{bln(2)} & \cdots\\ e^{c\times ln(1)} & e^{c\times ln(2)} & e^{c\times ln(3)} & e^{c\times ln(4)} & e^{cln(2)} & \cdots\\ e^{d\times ln(1)} & e^{d\times ln(2)} & e^{d\times ln(3)} & e^{d\times ln(4)} & e^{d\times ln(2)} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}
$v_{n}$ is a unit vector in infinite dimensional space, expanding along the first row we see that each unit vector is multiplies by the determinant of some other matrix filled with values of some number $e^{a\times ln(b)}$ and is clearly non zero as each row and column of this matrix is linearly independent. therefore the determinant of this system is non zero.
Decompose it into two matrices also with infinite columns and rows that when multiplied together. This uses a taylor series to expand the exponentials along rows and columns. $e^x = \sum_{n=1}^\infty {x^n\over n!}$
Matrix one on the left(also countable infinity of rows and columns):
Decompose it into the two matrices. The first one on the left. The only problem is that the first of the decomposed matrices has a zero determinant.
\begin{bmatrix} v_{1} & 0 & v_{2} & 0 & v_{3} & 0 & \cdots \\ 0 & a^0 & 0 & a^1 & 0 & a^2 & \cdots\\ 0 & b^0 & 0 & b^1 & 0 & b^2 & \cdots\\ 0 & c^0 & 0 & c^1 & 0 & c^2 & \cdots\\ 0 & d^0 & 0 & d^1 & 0 & d^2 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}
Multiplied by matrix two on the right(also countable infinity of rows and columns):
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ (ln(1))^0\over 0! & (ln(2))^0\over 0! & (ln(3))^0\over 0! & (ln(4))^0\over 0! & (ln(5))^0\over 0! & (ln(6))^0\over 0! & \cdots\\ 0 & 1 & 0 & 0 & 0 & 0 & \cdots\\ (ln(1))^1\over 1! & (ln(2))^1\over 1! & (ln(3))^1\over 1! & (ln(4))^1\over 1! & (ln(5))^1\over 1! & (ln(6))^1\over 1! & \cdots\\ 0 & 0 & 1 & 0 & 0 & 0 & \cdots\\ (ln(1))^2\over 2! & (ln(2))^2\over 2! & (ln(3))^2\over 2! & (ln(4))^2\over 2! & (ln(5))^2\over 2! & (ln(6))^2\over 2! & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}
I like to think of this as an 'interrupted identity matrix' as it has rows of zeros with a one, like an identity matrix to produce an 'unchanged' row, rather than producing an unchanged matrix
Using $det(AB) = det(A)\times det(b)$
We can calculate the original determinant by finding the determinants of the other two 'factors' of that matrix obtained by the taylor expansion. However looking at the first matrix 'factor' can be calculated as zero by expanding down the first column giving:
\begin{bmatrix} a^0 & 0 & a^1 & 0 & a^2 & \cdots\\ b^0 & 0 & b^1 & 0 & b^2 & \cdots\\ c^0 & 0 & c^1 & 0 & c^2 & \cdots\\ d^0 & 0 & d^1 & 0 & d^2 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}
Along the unit vector $v_{1}$. Here we see that there are multiple zero columns to expand down the determinant to give zero. by $det(AB) = det(A)\times det(b)$ this makes the determinant of the original matrix zero, something that we previously ruled out as the rows and columns were all linearly independent.
The only counterpoint I can think of to this is that the matrices that multiply to give the original matrix, although both still have countably infinite rows and columns, are larger than the original matrix. This would mean that $det(AB) = det(A)\times det(b)$ does not apply as the matrices of different sizes.
This was a long question and is, knowing me, quite likely riddled with mistakes. If you see one please suggest an edit and I will get onto it.
ADDITION ONE: I asked my lecturer if one countable infinity can be greater than another and he told me that they can't as by definition if an infinity can be mapped to the integers it is countable. Therefore if they can both be mapped to the integers then they can be mapped to each other. So my question is now, If the two infinities are the same then why do they get a different result?
ADDITION TWO: I am now considering expanding along the first row and then decomposing the matrices formed by the expansion to remove the $v_{n}$ term rows to remove the zero determinant error. However the zero determinant in the first decomposition makes me think that this might also have problems even if it produces a non-zero determinant. Unless this issue is addressed I have little faith in the answer that I get from this method even if it is rigours.
Going by terms we get:
$v_{1} \times det \begin{bmatrix} e^{a\times ln(2)} & e^{a\times ln(3)} & e^{a\times ln(4)} & e^{aln(2)} & \cdots\\ e^{b\times ln(2)} & e^{b\times ln(3)} & e^{b\times ln(4)} & e^{bln(2)} & \cdots\\ e^{c\times ln(2)} & e^{c\times ln(3)} & e^{c\times ln(4)} & e^{cln(2)} & \cdots\\ e^{d\times ln(2)} & e^{d\times ln(3)} & e^{d\times ln(4)} & e^{d\times ln(2)} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$
$-v_{2} \times det \begin{bmatrix} e^{a\times ln(1)} & e^{a\times ln(3)} & e^{a\times ln(4)} & e^{aln(2)} & \cdots\\ e^{b\times ln(1)} & e^{b\times ln(3)} & e^{b\times ln(4)} & e^{bln(2)} & \cdots\\ e^{c\times ln(1)} & e^{c\times ln(3)} & e^{c\times ln(4)} & e^{cln(2)} & \cdots\\ e^{d\times ln(1)} & e^{d\times ln(3)} & e^{d\times ln(4)} & e^{d\times ln(2)} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$
$+v_{3} \times det \begin{bmatrix} e^{a\times ln(1)} & e^{a\times ln(2)} & e^{a\times ln(4)} & e^{aln(2)} & \cdots\\ e^{b\times ln(1)} & e^{b\times ln(2)} & e^{b\times ln(4)} & e^{bln(2)} & \cdots\\ e^{c\times ln(1)} & e^{c\times ln(2)} & e^{c\times ln(4)} & e^{cln(2)} & \cdots\\ e^{d\times ln(1)} & e^{d\times ln(2)} & e^{d\times ln(4)} & e^{d\times ln(2)} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$
$\vdots$
Then breaking those down into taylor expansions to get rid of the 'interrupted identity matrix' that was produced by the vectors that produced the zero determinant.
$v_{1} \times det \begin{bmatrix} e^{a\times ln(2)} & e^{a\times ln(3)} & e^{a\times ln(4)} & e^{aln(2)} & \cdots\\ e^{b\times ln(2)} & e^{b\times ln(3)} & e^{b\times ln(4)} & e^{bln(2)} & \cdots\\ e^{c\times ln(2)} & e^{c\times ln(3)} & e^{c\times ln(4)} & e^{cln(2)} & \cdots\\ e^{d\times ln(2)} & e^{d\times ln(3)} & e^{d\times ln(4)} & e^{d\times ln(2)} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$
$v_{1} \times det \begin{bmatrix} a^0 & a^1 & a^2 & a^3 & \cdots\\ b^0 & b^1 & b^2 & b^3 & \cdots\\ c^0 & c^1 & c^2 & c^3 & \cdots\\ d^0 & d^1 & d^2 & d^3 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} \times det \begin{bmatrix} ln(1)^0\over 0! & ln(2)^0\over 0! & ln(3)^0\over 0! & ln(4)^0\over 0! & \cdots\\ ln(1)^1\over 1! & ln(2)^1\over 1! & ln(3)^1\over 1! & ln(4)^1\over 1! & \cdots\\ ln(1)^2\over 2! & ln(2)^2\over 2! & ln(3)^2\over 2! & ln(4)^2\over 2! & \cdots\\ ln(1)^3\over 3! & ln(2)^3\over 3! & ln(3)^3\over 3! & ln(4)^3\over 3! & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$
and thats as far as I can get it.
