# Conditions for Matrix to be Product of Near-Identity Matrices

For $\epsilon > 0$, let $M_{\epsilon}$ be the family of $n$ x $n$ real matrices A such that $||$A$-$I$_n|| < \epsilon$, where $|| \cdot ||$ is the standard operator norm. If $\epsilon$ is chosen sufficiently small, then all finite products of members of $M_{\epsilon}$ have positive determinant (i.e., they are orientation-preserving). Is this the only requirement for an $n$ x $n$ matrix to be expressible as such a product? If so, that would imply the result that any non-singular $n$ x $n$ matrix can be expressed as a product of $n$ x $n$ matrices that each change only one coordinate (as this is clearly the case for any matrix in $M_{\epsilon}$), which is what I'm trying to prove.

Let $A \in \mathbb{R}^{n \times n}$

Now if $\|A- I_{n} \| < \epsilon$ then I can express $A$ as product of two nearly orthogonal matrices. Right.

An orthogonal matrix is $QQ^{T} = Q^{T}Q = I_{n}$ now...each column of $Q$ is unit normal. So if we build an orthogonal matrix and alter slightly we manipulate the bounds on $\epsilon$ Like the following..

n= 3;
A = rand(n,n);
[Q,R] =qr(A);
I  = eye(n);
err = norm(Q*Q' - I);

now this is zero...for instance...

epsilon = 5;

Q1 = epsilon*Q(:,3);
Q1 = [Q(:,1),Q(:,2),Q1];
err1 = norm(Q1*Q1' -I)

err1 =

24.0000

From the matrix norms it slightly less than 25...like I expected. This comes from the matrix norm equality

$$\|AB \| \leq \|A\| \|B\|$$

and $$\| c A\| \leq |c| \| A \|$$

illustrating that this bounds it closer change epsilon to $1$

n= 3;
A = rand(n,n);
[Q,R] =qr(A);
I  = eye(n);
err = norm(Q*Q' - I);

epsilon = 1;

Q1 = epsilon*Q(:,3);
Q1 = [Q(:,1),Q(:,2),Q1];
err1 = norm(Q1*Q1' -I);

err1 =

5.1650e-16

Q1 = epsilon*Q(3,3);
Q2 = Q;
Q2(3,3) = Q1;
err1 = norm(Q2*Q2' - I)

Note that since 1 doesn't modify anything it will be close to machine precision or our $\epsilon$

In retrospect that was kind of dumb. We're going to create a matrix from the outer product of two other and subtract it from $I_{n}$

$$A =I_{n} - vu^{t}$$

let $vu^{T}_{ij} = 0 , vu^{T}_{i=j=1} = \epsilon ,$

So you have a zero matrix we create we subtract off epsilon from the identity.

$$\| A - I_{n} \| = \epsilon$$

$$\| I_{n} - vu^{t} -I_{n} \| = \| vu^{T} \| = \epsilon$$

we can demonstrate this like the following..

n=3;
I =eye(n);
Z = zeros(n);
epsilon = 1e-3;
Z(1,1) = epsilon;
A = I-Z;

error = norm(A-I);

error =

0.0010

So you simply create an $\epsilon$ and make it smaller.

• I'm afraid I don't see how I could use this to answer my question. If an orientation-preserving matrix A is really far away from the identity, how does its QR decomposition help me get it as a product of matrices close to the identity? – Davey Aug 6 '18 at 18:01
• I don't understand your concern. The QR decomposition is gram schmidt. I produced a matrix $Q$ which is orthogonal. I've demonstrated that with both code and math. Also there are two parts..one where I create a diagonal matrix like you state because what you're saying is confused. If you subtract $\epsilon$ from identity in the first spot then subtract that from $I_{n}$ you get $\epsilon$ for the norm. – Shogun Aug 7 '18 at 0:01
• your constraint is to make something called $\epsilon$ typically people it is very close it can be any number. further more determinant of orthogonal matrices is 1. – Shogun Aug 7 '18 at 0:15

I got the following solution from my algebra professor:

If $$H$$ is the group of $$n \times n$$ real matrices with positive determinant, we have that $$H$$ forms a connected set in the standard topology on $$\mathbb{R}^{n^2}$$. This is complicated to prove, but it essentially involves showing that any matrix in $$H$$ can continuously be moved to the identity by adding multiples of some rows to other rows and scaling rows by positive constants (most easily by induction on $$n$$). So given any matrix $$A$$ in $$H$$, there's a path $$C$$ in $$H$$ from $$I$$ to $$A$$.

We then fix a small $$\epsilon > 0$$ and let $$V$$ be the ball around $$I$$ in $$H$$ of radius $$\epsilon$$. We can choose a smaller open ball $$U$$ around $$I$$ such that for any matrices $$X, Y \in U$$, $$XY^{-1} \in V$$, that is, the multiplicative distance between any two matrices in $$U$$ is in $$V$$.

We then note that every matrix $$X$$ along $$C$$ yields a multiplicative translate $$XU$$ of $$U$$ that is still an open set (because $$X$$ is just a change of coordinates), and by compactness finitely many of these will cover $$C$$, so we have a chain $$U_1, ..., U_k$$ of open sets along $$C$$ linking $$I$$ to $$A$$, such that, by openness and the connectedness of the path, each $$U_i$$ overlaps with $$U_{i+1}$$. We can thus take a chain of points (matrices) $$X_1, ..., X_{k+1}$$ where $$X_1 = I$$, $$X_{k+1} = A$$, and for $$2 \leq i \leq k$$, $$X_i \in U_{i-1} \cap U_{i}$$. For each $$i$$, we have that $$X_{i+1}^{-1}X_i \in V$$, by the definition of $$U$$, so we can move from $$I$$ to $$A$$ via a product of $$k$$ matrices in $$V$$.