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.
 A: 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. 
A: 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 $V$.
