2
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Try to get to your point quickly, and then go into detail. Your title is very unspecific, and is often the subject of irritating questions, so a lot of people who might be able to help you will possibly not wade into your long long question to find out what you are asking. $\endgroup$ – Thomas Andrews Jun 4 '15 at 3:07
  • $\begingroup$ thanks, I'll get on that. Not sure about what I can replace the title with. $\endgroup$ – Rory Grice Jun 4 '15 at 3:11
  • 1
    $\begingroup$ I would suggest bolding important sections; the general consensus is that using all caps is equivalent to shouting. You can also use a "#" to make text large; adding more "#"s makes text progressively smaller from there. $\endgroup$ – HDE 226868 Jun 8 '15 at 0:39
  • $\begingroup$ thanks, I considered highlighting, but I couldn't figure out how. $\endgroup$ – Rory Grice Jun 8 '15 at 0:40
3
$\begingroup$

In general, determinant is not defined for "matrices" with infinitely many rows and columns. It's not surprising that you're running into paradoxes if you try to use identities that work for finite matrices on these "matrices". There are some rather special cases where you can define a determinant for an infinite matrix. For example, see this Math Overflow discussion. I see no evidence that you are in those cases.

$\endgroup$
8
  • $\begingroup$ -1 this says nothing except that what Op is doing is hard and it's an open problem. I don't think those were the questions Op was asking. $\endgroup$ – Zach466920 Jun 20 '15 at 14:19
  • $\begingroup$ @Zach466920 Read the answer again. "What you're doing is not well defined" is a perfectly good response to "Why am I getting an answer that doesn't make sense?" $\endgroup$ – Potato Jun 21 '15 at 10:32
  • $\begingroup$ @Potato then it's certainly a comment $\endgroup$ – Zach466920 Jun 21 '15 at 14:22
  • $\begingroup$ @Zach466920 It's definitely an answer to the question being asked. It correctly identifies the reason why the asker is getting an answer that doesn't make sense. I'm having trouble understanding why you think otherwise. (Note that it's been accepted.) $\endgroup$ – Potato Jun 21 '15 at 15:36
  • $\begingroup$ @Potato this is the only answer, sometimes people just give in and up vote to avoid feeling guilty. Op asks to find the determinant of a matrix, this post argues that you can't, which does not answer the question. In addition, it provides no proof. That's why it doesn't answer the question and should be a comment. This post probably deterred the Op from finding an answer based on opinion, something that I find to be inexcusable. $\endgroup$ – Zach466920 Jun 21 '15 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.