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I'm attempting to compute the value of an arbitrarily large matrix (order $n$). In simplifying the determinant, I was able to express its value in terms of the determinants of smaller matrices of the same form. Specifically,

$$ D_n = 2 \cos \theta D_{n-1} - D_{n-2}, $$

where $D_n$ is the determinant of the original order $n$ matrix. The form of the original matrix isn't critical to this question (just the recursive relationship), but if you're curious

$$ D_n = \begin{vmatrix} \cos \theta & 1 & 0 & 0 & & & 0 \\ 1 & 2 \cos \theta & 1 & 0 & \dots & \dots & 0 \\ 0 & 1 & 2 \cos \theta & 1 & & & 0 \\ 0 & 0 & 1 & 2 \cos \theta & & & 0 \\ & & \vdots & & \ddots & & \vdots \\ & & \vdots & & & 2 \cos \theta & 1 \\ 0 & 0 & 0 & 0 & \dots & 1 & 2 \cos \theta \\ \end{vmatrix}. $$

I've defined $D_{n-1}$ in terms of $D_n$ by removing the last row and column from $D_n$.

My question is: where does this recursive relationship end? In Mathematical Methods in the Physical Sciences by boas, section 3.3, Boas states:

The value of det A if A is a 1 by 1 matrix is just the value of the single element

So, $D_1=\cos\theta$.

But, what about the 0x0 matrix $D_0$? Is this just 0? Is it defined at all?

MattHusz
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    The determinant of a $0\times 0$ matrix is usually considered to be equal to $1$. – JMoravitz Oct 08 '20 at 16:40
  • If you take determinant to be defined as the “product” of eigenvalues, then $0\times0$ matrix determinant is going to be an empty product; conventionally, that’s taken as equal to $1$. – Jack LeGrüß Oct 08 '20 at 16:43
  • @JMoravitz yes that does. Thanks for pointing that out. – MattHusz Oct 08 '20 at 16:43

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