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Let $A$ be a matrix in the special orthogonal group, $A \in \mathrm{SO}_n$. This means that $A$ is real, $n \times n$, $A^t A = I$ and $\det(A)=1$, that is, the column vectors of $A$ make a positively-oriented orthonormal basis for $\mathbb R^n$.

Decompose $A$ as a block matrix

$$A = \begin{pmatrix} B & C \\ D & E\end{pmatrix}$$

where $B$ is $k \times k$ and $E$ is $(n-k)\times (n-k)$.

I'm looking for a basic linear-algebra argument that $\det(B) = \det(E)$, ideally something that could be presented in a 2nd or 3rd year undergraduate course. So I do not want people to invoke anything like tensor products or differential forms.

C.F.G
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Ryan Budney
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  • Do you agree with my edits? – Rodrigo de Azevedo Jul 31 '20 at 14:02
  • @RodrigodeAzevedo: Partially. Not certain why you erased my tags. The new tags narrow the context too much, in my opinion. – Ryan Budney Aug 01 '20 at 06:40
  • It's your question. You can always revert my edit if it distorts the original spirit of the question. New tags were created over the past 8 years. Narrowing too much is a problem but it can make questions easier to find. I am surprised I have spent over 4 years on Math SE and had not yet found this question. For example, this question is now one of the top questions on block matrices. – Rodrigo de Azevedo Aug 01 '20 at 08:55
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    It's a cute question on block matrices, although the purpose of the question is to describe the geometric properties of the Hodge star on differential forms. Depending on which definition you are using, this could be viewed as the statement that it is invariant under orientation-preserving isometries. I suppose with these comments it will be searchable, as well. – Ryan Budney Aug 01 '20 at 18:47
  • I reverted my edits. – Rodrigo de Azevedo Aug 01 '20 at 18:51

1 Answers1

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Some googling brought up this, Hudson's book "Kummer's quartic surface", containing the following argument.

Given $A = \begin{pmatrix} B & C\\ D & E\end{pmatrix}$, orthgonality gives us that $A^{-1} = A^t = \begin{pmatrix} B^t & D^t\\C^t & E^t\end{pmatrix}$.

The block matrix multiplication gives $$I = A^t A = \begin{pmatrix} B^tB + D^tD & B^t C + D^tE\\C^tB + E^t D & C^tC + E^tE\end{pmatrix}$$

so $B^tB + D^tD = I$, $B^tC + D^tE = 0$, and $C^tC + E^tE = I$ (but the two $I$s probably have different size).

Now, using these relations, it's easy to see that $$\begin{pmatrix} B^t & D^t\\0&I\end{pmatrix} A = \begin{pmatrix} B^t & D^t\\0&I\end{pmatrix}\begin{pmatrix}B&C\\D&E\end{pmatrix} = \begin{pmatrix} I & 0\\D &E\end{pmatrix}$$

(but I confess to being at a loss of how to motivate where this came from - it's in the book).

From here, taking the determinant of both sides, using the fact that $\det(A) = 1$ and that $\det$ is a homomorphism, gives $$\det\begin{pmatrix} B^t & D^t\\0&I\end{pmatrix} = \det\begin{pmatrix} I&0\\D&E\end{pmatrix}$$

but the determinant of a block matrix with one block $0$ is the product of the determinants of the diagonal blocks, so we get $\det(B^t) = \det(E)$. Finally, note that $\det(B^t) = \det(B)$, so we're done.

Hans Lundmark
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Jason DeVito
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    Thanks Jason, that's a lovely argument. I was hopeful something like that would work, but I didn't notice the step you found tricky to motivate. I'll see if I can massage it into something more natural. – Ryan Budney Mar 21 '12 at 01:40
  • Here's an argument I noticed in one of Will Jagy's posts: $$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$ This assumes the matrix $\left(\begin{array}{cc} A & B \\\ C & D\end{array}\right)$ is orthogonal. – Ryan Budney May 24 '19 at 11:13
  • Can something similar be said about the invertibility of the bottom left block D? – gen Oct 16 '21 at 01:43
  • $D$ certainly doesn't need to be invertible, e.g., then the matrix is the identity. It *can* be invertible (for example, take a standard $2\times 2$ rotation matrix by angle $\neq 0,\pi$.). I'm not sure what more can be said. – Jason DeVito Oct 16 '21 at 02:16
  • @Robert Fridz Why have you deleted your question ? It's not a good habit. Moreover, I was about to give you another link [here](https://mathoverflow.net/q/134547) – Jean Marie Nov 18 '21 at 22:26