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I've come across with the following proof of the Laplace expansion:

Let

$\Delta=\sum_{j=1}^n (-1)^{1+j} a_{1j}\bar M_j^1$

and

$\tilde{\Delta}= \sum_{j=1}^n (-1)^{i+j} a_{ij}\bar M_j^i$

We'll prove by induction that $\tilde{\Delta}=\Delta$.

For a $2 \times 2$ matrix, clearly, it holds:

$\tilde{\Delta_2}=\sum_{j=1}^n (-1)^{i+j} a_{ij}\bar M_j^i=-a_{21}a_{12}+a_{22}a_{11}=\Delta_2$

Suppose that for $n -1$ matrix $\tilde\Delta_{n-1}=\Delta_{n-1}$ is true. Then:

$$\tilde{\Delta_n}=\sum_{j=1}^n (-1)^{i+j} a_{ij}\bar M_j^i=\sum_{j=1}^n (-1)^{i+j} a_{ij}\left(\sum_{k<j}(-1)^{1+k}a_{1k}\bar M_{jk}^{i1}+\sum_{k>j}(-1)^k a_{1k}\bar M_{jk}^{i1}\right) $$

Gathering all the coefficients of $\bar M_{j_0k_0}^{i\,\,1}$

$$j_0>k_0\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}+(-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0}=(-1)^{i+j_0+k_0+1}(a_{ij_0}a_{1k_0}-a_{ik_0}a_{1j_0})=(-1)^{i+j_0+k_0+1}M_{j_0k_0}^{1\,\,i}$$

$$j_0<k_0\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{k_0}a_{1k_0}+(-1)^{i+k_0}a_{ik_0}(-1)^{j_0+1}a_{1j_0}=(-1)^{i+j_0+k_0+1}(a_{ik_0}a_{1j_0}-a_{ij_0}a_{1k_0})=(-1)^{i+j_0+k_0+1}M_{j_0k_0}^{1\,\,i}$$

$$\tilde{\Delta_n}=\sum_{j\ne k}(-1)^{i+j+k+1}M_{jk}^{1i}\bar M_{jk}^{1i}$$

$$\Delta=\sum_{j=1}^n (-1)^{1+j} a_{1j}\bar M_j^1=\sum_{j=1}^n (-1)^{1+j} a_{1j}\left(\sum_{k<j}(-1)^{i+k-1}a_{ik}\bar M_{jk}^{1i}+\sum_{k>j}(-1)^{i+k-2}a_{ik}\bar M_{jk}^{1i}\right) $$

Gathering all the coefficients of $\bar M_{j_0k_0}^{i\,\,1}$

$$j_0>k_0\colon\; (-1)^{1+j_0}a_{1j_0}(-1)^{i+k_0-1}a_{ik_0}+(-1)^{1+k_0}a_{1k_0}(-1)^{i+j_0-2}a_{ij_0}=(-1)^{j_0+i+k_0}(a_{1j_0}a_{ik_0}-a_{1k_0}a_{ij_0})=(-1)^{i+j_0+k_0+1}M_{j_0k_0}^{1\,\,i}$$

$$j_0<k_0\colon\; (-1)^{1+j_0}a_{1j_0}(-1)^{i+k_0-2}a_{ik_0}+(-1)^{1+k_0}a_{1k_0}(-1)^{i+j_0-1}a_{ij_0}=(-1)^{k_0+i+j_0}(a_{1k_0}a_{ij_0}-a_{1j_0}a_{ik_0})=(-1)^{i+j_0+k_0+1}M_{j_0k_0}^{1\,\,i}$$

$${\Delta_n}=\sum_{j\ne k}(-1)^{i+j+k+1}M_{jk}^{1i}\bar M_{jk}^{1i}=\tilde{\Delta_n} ~~ \blacksquare $$

I just can't understand the part where they "gather coefficients":

$$j_0>k_0\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}+(-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0}=\\ =(-1)^{i+j_0+k_0+1}(a_{ij_0}a_{1k_0}-a_{ik_0}a_{1j_0})=(-1)^{i+j_0+k_0+1}M_{j_0k_0}^{1\,\,i}$$

I'm okay with the $ (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}$ part, which is simply what we get after plugging $j_0$ and $k_0$ into the previous formula of $\tilde{\Delta_n}$ (the sum of two sums) for a single and specific iteration. But where does the $$(-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0}$$ come from? I just can't see anything like this before $\bar M_{j_0k_0}^{i\,\,1}$.

I'm trying to figure this out for many hours and I'm completely lost. I would appreciate any explanation!

user175788
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  • Just curious, which book is this proof from? – Phonon Sep 15 '14 at 20:36
  • @Phonon - it is taken from wikipedia. I tried to find the same proof in books but with no success. – user175788 Sep 16 '14 at 09:47
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    Everything of interest about determinants can be proved without such horrible formulas. What's the motivation for delving into them? No real surprise you couldn't find this in any book. – Marc van Leeuwen Sep 17 '14 at 12:37
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    @MarcvanLeeuwen - the question is not whether or not there are more elegant proofs of this theorem (obviously there are). I want to understand this specific proof, and I'm almost there. I just can't understand one nuance - everything else is pretty straightforward, albeit looks bulky. – user175788 Sep 17 '14 at 12:50

1 Answers1

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Note that

$$\bar M_{j_0k_0}^{i\,\,1}=\bar M_{k_0j_0}^{i\,\,1}$$ Thus, you always have two different coefficients for the same minor that add up: $$j=j_0 , k=k_0 ~~ (j_0>k_0): \\ j>k\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}\\$$

$$ j=k_0, k=j_0:\\ j<k\colon\; (-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0} $$

Dehumanizer
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