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I have come across an observation about the determinant of a matrix, but I don't know how to prove it in general. Let me demonstrate it through an example.

$$ \begin{align} \left| \begin{matrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \\ \end{matrix} \right| \cdot \left| \begin{matrix} 6 & 7 \\ 10 & 11 \\ \end{matrix} \right| &= \left| \begin{matrix} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 9 & 10 & 11 \\ \end{matrix} \right| \cdot\left| \begin{matrix} 6 & 7 & 8 \\ 10 & 11 & 12 \\ 14 & 15 & 16 \\ \end{matrix} \right|\\ &- \left| \begin{matrix} 2 & 3 & 4 \\ 6 & 7 & 8 \\ 10 & 11 & 12 \\ \end{matrix} \right| \cdot \left| \begin{matrix} 5 & 6 & 7 \\ 9 & 10 & 11 \\ 13 & 14 & 15 \\ \end{matrix} \right| \end{align} $$

What this essentially says is that you can find the determinant of a larger matrix by breaking it down into this "ad-bc"-style of smaller determinants, however I can't work out why this should be true. I've expanded out the determinant for a 3x3 example in full generality and confirmed that it works, but I'd like to know if there's a more straightforward reason for this observation.

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3 Answers

up vote 2 down vote accepted

I agree that your formula works. Proving it is possible, but slightly messy. It is something that I may expand on in the future, but let me give you some assurance that my agreement has some validity.

Denote the matrix of cofactors as $$\left[ \begin{matrix} 1_c & 2_c & 3_c & 4_c \\ 5_c & 6_c & 7_c & 8_c \\ 9_c & 10_c & 11_c & 12_c \\ 13_c & 14_c & 15_c & 16_c \\ \end{matrix} \right]$$ The matrix of cofactors is the transpose of the inverse scaled by the determinant. In other words, each value is the determinant of the corresponding submatrix, hence the name matrix of cofactors. For example the regular Laplace expansion for the determinant (call it $\Delta$) along the last row is $$\Delta = 13\cdot 13_c + 14\cdot 14_c + 15\cdot 15_c+ 16\cdot 16_c$$ This notation allows us to write your formula more concisely as $$\Delta\cdot \left| \begin{matrix} 6 & 7 \\ 10 & 11 \\ \end{matrix} \right| = 16_c \cdot 1_c - 13_c \cdot 4_c$$

Using a formula derived in a similar manner as this answer of mine gives $$\left| \begin{matrix} 6 & 7 \\ 10 & 11 \\ \end{matrix} \right| =\frac{16_c\cdot 1_c - 4_c\cdot 13_c}{\Delta}$$ and you can see that indeed your formulation agrees with mine. Generally speaking the formula works for the higher dimensional matrix: $$\left(\matrix{a & \vec{m}^\top & b \\ \vec{g} & N & \vec{h} \\ c & \vec{k}^\top & d}\right)$$

let $d_c$ denote the corresponding determinant of the submatrix $$d_c = \left|\matrix{a & \vec{m}^\top \\ \vec{g} & N }\right|$$ and similarly for $a_c$, $b_c$, and $c_c$. Then the same formula is valid: $$\operatorname{det}N = \frac{a_cd_c - b_cc_c}{\Delta}$$

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Thank you, this looks to be essentially what I'm after. I haven't got the time to spend fully thinking it through right now, but hopefully comparing my question against the other question you linked to might help reveal the tools I need to prove my observation. –  Daniel Mar 27 '13 at 11:31
    
Thank you again for your reply Adam, I can see now why you have agreed with my initial observations. I've read through the linked answer you provided, but I'm having trouble reconstructing a proof in a similar manner as your linked derivation. Could you perhaps point me along the right direction with any more specifics and hopefully I can fill in the blanks? –  Daniel Apr 2 '13 at 11:25
    
Convert everything to be not in terms of inverse, but in terms of inverse times the determinant. Which would be the adjoint or the matrix of cofactors (the $a_c$, $b_c$ etc). You get the same simple rank one update as the linked answer. "Popping" one corner, then looking at the other corner element gives the determinant for the $N$. Or, later, if you ask it as a separate question can be more specific. –  adam W Apr 2 '13 at 12:56
    
Thanks, I've played around with your suggestion and adjusted your formulation slightly, but I keep running in circles and getting myself mixed up between the cofactors/matrices of cofactors. I think I've been looking at it so long I just can't see it another way, so I might just consolidate all of my information together and form up a new clearer question on this specific avenue. –  Daniel Apr 3 '13 at 10:09
    
New question can be found here –  Daniel Apr 3 '13 at 10:32
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Let's step back and ask what the determinant is and why it is useful. Then we'll get to this particular property.

Permit me to introduce a new product of vectors, called the wedge product. If $a, b$ are vectors, then their wedge product is $a \wedge b$. The result is not a vector, but instead, we interpret it geometrically as an oriented plane--exactly the plane perpendicular to $a \times b$, as a matter of fact, but we can continue wedging--for instance, forming $a \wedge b \wedge c$, which represents a volume.

Of course, in 3d space, there is only one unit volume (you could call it $\hat x \wedge \hat y \wedge \hat z$, but let me call it $i$ for short. All other volumes are scalar multiples of this volume, at least in terms of magnitude and orientation (you may be thinking "orientation?" I submit that a coordinate system following the right-hand rule is oppositedly oriented to one following a left-hand rule, and the unit volumes formed by wedging their unit vectors are oppositely oriented). It is for this reason that the volume elements are often called pseudoscalars, because they are only different from scalars by how they have orientations where scalars do not.

How does this relate to matrices? Well, matrices are used to represent linear operators on vectors, but these operators can act on wedge products of vectors or pseudoscalars too. The "matrices" you use to represent these extensions of the original operator are different from the original matrix. We define the relationships by a simple rule. If $\underline T$ is a linear operator on a vector, then we define $\underline T(a \wedge b) = \underline T(a) \wedge \underline T(b)$, and so on (note that the cross product does not have this nice property except under rotations).

But again, we said that there is only one unit pseudoscalar ($i$), so the action of a linear operator on $i$ must be some scalar multiple of $i$. That is, if $\alpha$ is a scalar,

$$\underline T(i) = \alpha i$$

We define this number $\alpha$ to be the determinant, telling us geometrically how the unit volume is shrunk or dilated (or changes orientation) under the action of a linear operator.

You can find the determinant of a linear operator by writing down the matrix representation and wedging the vectors that appear there. This is perfectly well-founded. You just need to know that $a \wedge b = - b \wedge a$--vectors anticommute under the wedge--and that the wedge is associative. Knowing these properties makes it possible to do computations with the wedge.

(If you're puzzled why wedging vectors gives the determinant, feel free to ask and I'll clarify this point.)

So let's take three vectors $f, g, h$ that appear in the matrix representation of a linear operator and wedge them to find the determinant. Let $f = f^x e_x + f^y e_y + f^z e_z$ and so on. We can then write out the following:

$$\begin{align*} f \wedge g \wedge h = f^x e_x \wedge (g \wedge h) + f^y e_y \wedge (g \wedge h) + f^z e_z \wedge (g \wedge h)\end{align*}$$

I've expanded the wedge product through linearity (the distributive property, which I failed to mention earlier but is also valid for the wedge). This is the foundation for the technique you'e come across, which is called Laplace expansion, or cofactor expansion, or expansion by minors. The antisymmetry of the wedge means that we can write $f^x e_x \wedge (g \wedge h)$ as

$$f^x e_x \wedge (g \wedge h) = f^x e_x \wedge (g^y e_y + g^z e_z) \wedge (h^y e_y + h^z e_z)$$

Why? Because $e_x \wedge e_x = 0$ always, so if any $e_x$ appeared in $g$ or $h$, they would be irrelevant. We can ignore them, and instead, we find the "determinant" of a linear operator on the $yz$-plane. Thus, the method of expansion by minors follows by recursively expanding a single vector through linearity to reduce the subsequent wedge products to a wedge product (a "determinant") you might already know. The method has its roots in the relationship between the determinant and volumes, how all the terms that make up the determinant must contain 1 and only 1 component of some vector from each of the coordinate directions--no more, no less.

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Going to read it, but +1 for the effort. –  dineshdileep Mar 26 '13 at 4:28
    
Thanks for your work, I'm pretty happy with the idea of determinants as oriented volumes and the idea of expanding via minor matrices. That's what's stumped me, because I'm used to expanding along a single row of a matrix, whereas in this case I have a collection of minors that overlap with each other in terms of the original matrix. It's a form that I can't see how it prove. –  Daniel Mar 26 '13 at 8:52
    
All right, perhaps I've jumped the gun here. Can you give more examples that fit the form of what you want to prove? To me, what you described in the question in words is exactly expansion by minors, but the example you've given doesn't seem to fit that very well at all. –  Muphrid Mar 26 '13 at 13:15
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This is a determinant identity for Schur complement of a matrix. It's also closely related to http://en.wikipedia.org/wiki/Dodgson_condensation. "It is named for its inventor Charles Dodgson (better known as Lewis Carroll)."

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