What's the geometric meaning of a matrix decomposed into its symmetric and skew-symmetric parts? For example, a skew-symmetric matrix on its own can be interpreted as an infinitesimal rotation. As an another example, the polar decomposition of a matrix A=UP=QU for a unitary matrix U and symmetric positive definite matrices P and Q means that we can interpret a matrix as a stretching (the positive definite matrix) followed by a rotation (the unitary matrix) or vice versa.
Basically, the decomposition of a matrix into its symmetric and skew symmetric parts $A=(1/2)(A+A^T)+(1/2)(A-A^T)$ is causing me problems because I don't know how to geometrically interpret the symmetric part (if it was positive definite it would be a stretch) nor the sum (if it was a product then one action would follow the other.)
Edit 1
The answer from @user_of_math helped me figure the rest of this out.
Assume that we have a matrix $B$ thats symmetric. Then, we know that it has a spectral decomposition $B=VDV^T$. What this says is that the action of $B$ on a vector $x$ occurs in three steps
- We determine how much we will project $x$ onto the columns of $V$, $V^T x$
- We scale this projection by the eigenvalues of $A$, $D V^T x$
- Then, we put everything back into the original coordinate axis, $V D V^T x$
In other words, a symmetric operator scales the space along the orthogonal axes defined by $V$. When $B$ is positive definite, the elements of $D$ are positive and this is a stretch. When $B$ is indefinite or negative definite, we can potentially flip or collapse an axis. In any case, provides one explanation of the geometric the meaning of a symmetric operator.
Now, let $C$ be a skew-symmetric operator. Then, $exp(C)$ is orthonormal matrix, which corresponds to a rotation.
At this point, we have an idea of what happens with a symmetric operator and a skew-symmetric operator, but there's a disconnect because our explanation assumed the skew-symmetric operator corresponds to something infinitesimal and the symmetric operator does not. As such, consider an infinitesimal shift of a vector x by a symmetric operator B
$$
\lim\limits_{\alpha\rightarrow 0} (I+\alpha B)x
$$
More specifically, we want to do an infinite number of these transformations, so we have
$$
\lim\limits_{n\rightarrow\infty} \left(I+\frac{1}{n} B\right)^n x = exp(B)x
$$
Now, since $B$ is symmetric, we have the spectral decomposition, $B=VDV^T$. Hence,
$$
exp(B)x=exp(VDV^T)x=Vexp(D)V^Tx,
$$
which is interesting because this says that $exp(B)$ is a positive definite operator. In other words, and infinite number of shifts by a symmetric operator eventually yields a single stretch by a positive definite operator.
In any case, let's look at the general situation now. Let $B=(A+A^T)/2$ be the symmetric part of $A$ and $B=(A-A^T)/2$ be the skew-symmetric part. Then, consider an infinite number of shifts by $A$,
$$
\lim\limits_{n\rightarrow\infty} \left(I+\frac{1}{n} A\right)^n = exp(A) = exp(B+C) = \lim\limits_{n\rightarrow \infty}(exp(B/n)exp(C/n))^n
$$
where the last equality comes from the Lie product formula. This says that the meaning of $exp(A)$ is an infinite, alternating application of infinitesimally small stretches, defined by $exp((A+A^T)/2)$, and rotations, defined by $exp((A-A^T)/2)$. In other words, the geometric meaning of exp(A), when $A$ is decomposed into its symmetric and skew-symmetric components, is an alternating application of stretches and rotations defined by these symmetric and skew-symmetric components, respectively.
Best Answer
You can use physical ideas to help construct a simple counterexample to show that unless additional conditions are imposed, like positive definiteness etc, the symmetric part can represent quite different things.
Consider (for simplicity) a 2D body (a unit square with one vertex at the origin) in a fixed cartesian basis, and subjected to the following homogeneous deformation.
$$ x = X + 2Y \\ y = X - Y $$
This is not change of basis, but an actual change in the position / shape of the body, with lowercase denoting the "deformed" configuration and uppercase the coordinates of a point in the body in its original configuration.
In matrix terms $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} $$
We'll call this matrix $F$ and write it as a sum of symmetric and skew-symmetric components $$ F = F_{s} + F_{skew} \\ F = \frac{1}{2}(F + F^T) + \frac{1}{2}(F - F^T) \\ = \begin{bmatrix} 1 & 3/2 \\ 3/2 & -1 \\ \end{bmatrix} + \begin{bmatrix} 0 & 1/2 \\ -1/2 & 0 \\ \end{bmatrix} $$
Now consider what the first (symmetric) matrix $F_s$ does to our unit square with vertices $a=(0,0), b=(1,0), c=(1,1) \; and \; d=(0,1)$ in that order. (We might as well start with symmetric F, but anyway).
The new coordinates of the unit square are (let's call them $a', b', c', d')$ under this symmetric map are \begin{bmatrix} a' & b' & c' & d' \\ 0 & 1 & 5/2 & 3/2 \\ 0 & 3/2 & 1/2 & -1 \\ \end{bmatrix}
Looks ok. However, if you sketch this figure you'll see that the handedness of the figure has flipped. In particular, its Jacobian is negative, so the area changes sign. What this implies is that no matter how you parametrize or decompose (say using products) the map $F_s$, at some stage, material in the square must interpenetrate to reach the final configuration. So in this instance, the symmetric matrix with no other restriction represents an impossible deformation (if you look at it physically). Mathematically, it represents a change of handedness by interchanging sides PLUS a stretch in this case.
Even weirder things can happen when the jacobian is 0. The symmetric map \begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}
results in collapsing our unit square into a straight line passing through (-1,1) and (1,-1) with obviously no area! So you see that symmetry by itself doesn't tell you what's going on geometrically unless you look at additional characteristics of the map.