I've been trying to look for an intuitive understanding for the Laplace expansion of the determinant. I first tried looking for the proof but let's just say it was way to complicated for my understanding.
So, I've devised a way to look at the Laplace expansion. A slight problem in it is that the cofactor formula turns to have a $i-j$ in the power instead of $i+j$. I'd like to know if my way of looking at the Laplace expansion is valid. Here's my intuition:
Consider a $3\times 3$ matrix:
$$M = \begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\a_{2,1}&a_{2,2}&a_{2,3}\\a_{3,1}&a_{3,2}&a_{3,3}\end{bmatrix}$$
If you ignore signs for just a second, here's how to think of the determinant:
Iterate through each column. During this iteration, choose an element that is in the $i^{th}$ row of this column such that the $i^{th}$ row hasn't been chosen in any of the previous iterations. So, for example, one of the products will be $a_{1,1}a_{3,2}a_{2,3}$. Think of the determinant as doing all the valid iterations starting from $a_{1,1}$ then starting from $a_{2,1}$ and then finally starting from $a_{3,1}$. This will look like the following:
$$det(M) = iterations(a_{1,1}) + iterations(a_{2,1}) + iterations(a_{3,1})$$
Here, $iterations(a_{1,1}) = a_{1,1}a_{2,2}a_{3,3} + a_{1,1}a_{3,2}a_{2,3}$. But, you can think of $iterations(a_{1,1}) = a_{1,1}(a_{2,2}a_{3,3} + a_{3,2}a_{2,3}) = a_{1,1}det(T_{1,1})$ where $T_{1,1}$ is the minor of $a_{1,1}$. In this way, if you completely ignore signs, you get the following if $M$ has a size $n\times n$
$$det(M) = a_{1,1}det(T_{1,1}) + a_{2,1}det(T_{2,1}) + ... + a_{n,1}det(T_{n,1})$$
Now, the trouble comes when trying to account for sign. Consider the $2\times 2$ case first: $$A = \begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\end{bmatrix}$$
Ignoring sign, the determinant would be $a_{1,1}a_{2,2} + a_{2,1}a_{1,2}$. The sign on each summand depends on the number of switches needed to bring its first indices in ascending order. For the first summand, the first indices are (1,2). So, no switches needed. But for the second summand, the first indices are (2,1). Therefore, you need to make one switch. If the number of switches is even, you will have a +1 as the coefficient. Else, you'll have -1.
Now, come back to the $3\times 3$ case. Consider $a_{1,1}det(T_{1,1})$. The triplet of first indices will be (1,_,_). The blank indices can be either 2 (for $a_{2,2}$) or 3 (for $a_{3,2}$). Now, we can make the following transformation $(1,\_,\_)\xrightarrow{k_1}(1,2,3)$. Whether $k_1$ is even or odd will determine the sign in front of $a_{1,1}det(T_{1,1})$. $(\_,\_)\xrightarrow{k_2}(2,3)$ represents the number of switches needed to bring the indices of any of the summands in $det(T_{1,1})$ to the order of (2,3). For example, $a_{2,2}a_{3,3}$ has a $k_2 = 0$ while $a_{3,2}a_{2,3}$ has a $k_2 = 1$.
So, to summarize, if you consider the starting point $a_{1,1}$ the number of switches needed to bring its products indices (the products being $a_{1,1}a_{2,2}a_{3,3}$ and $a_{1,1}a_{3,2}a_{2,3}$) to 1,2,3 is denoted by the following:
$$(1,\_,\_)\xrightarrow{k_1}(1,2,3)$$
Further, the number switches needed to bring the indices of any of the summands in $det(T_{1,1})$ to the order of (2,3) is the following:
$$(\_,\_)\xrightarrow{k_2}(2,3)$$
It's easy to see that $k_1 = k_2$. This implies that whatever the sign is for the summands of $det(T_{1,1})$, the same sign will carry forward for $a_{1,1}det(T_{1,1})$. However, now consider starting from $a_{3,1}$. The two equations for it will be the following: $$(3,\_,\_)\xrightarrow{k_1}(1,2,3)$$ $$(\_,\_)\xrightarrow{k_2}(1,2)$$
$k_2 \leq k_1$. So, if you do the $k_2$ of the $k_1$ switches, the LHS for the first equation will be $(3,1,2)$. Now, you need to do two more switches to get this to $(1,2,3)$. Therefore, $k_1 = k_2 + 2$. Essentially, if a certain summand in $det(T_{3,1})$ required a certain number of switches, the corresponding summand in $a_{3,1}det(T_{3,1})$ will require two more switches. So, if $k_2$ is even, $k_1$ will also be even. The same is true with the odd case. This implies that there will be a plus in front of $a_{3,1}det(T_{3,1})$. You can use the same logic to show that there will be a minus in front of $a_{2,1}det(T_{2,1})$.
Now, let me try to generalize. Let's say I start from $a_{i,1}$ and that $M$ is an $n\times n$ sized matrix. The first equation will be the following: $$(i,\_,\_,...,\_)\xrightarrow{k_1}(1,2,...,n)$$
The second equation will be: $$(\_,\_,...,\_)\xrightarrow{k_2}(1,...,i-1,i+1,...,n)$$
If you do the $k_2$ of the $k_1$ switches, you will get:
$$(i,1,...,i-1,i+1,...,n)$$
Now, you need to make an additional $i-1$ switches to turn this into $(1,...,i-1,i,i+1,...n)$. If $i-1$ is even, your sign will be positive. Else, it will be negative. Therefore, I can write the determinant formula as the following:
$$det(M) = (-1)^{1-1}a_{1,1}det(T_{1,1})+...+(-1)^{n-1}a_{i,1}det(T_{n,1})$$ $$det(M) = \sum_{i=1}^n(-1)^{i-1}a_{i,1}det(T_{i,1})$$
Although the equation works, the traditional way to express the Laplace expansion is to write it as the following:
$$det(M) = \sum_{i=1}^n(-1)^{i+1}a_{i,1}det(T_{i,1})$$
I have two questions:
- Is my intuition valid? If so, how can I write a more formal proof based on it?
- Is the discrepancy of i-1 and i+1 a flaw in my intuition?
