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# Classroom Notes: Symmetric Matrices and (a Little) Work

Here are a few unpretentious observations that occurred to me several years ago after teaching a linear algebra course. I am not making any claim to their originality.

1. The elegant but also a bit antiseptic definition of a symmetric $$n × n$$ real matrix A as the one satisfying the identity

$(A{\bf x} ,{\bf y} ) - ({\bf x}, A{\bf y})=0$

$\textrm {for all} \quad {\bf x}, {\bf y} \in {\mathbb R}^n \qquad (1)$

has a physical interpretation: this identity is equivalent to saying that the work done by the linear force field $${\bf F} ({\bf x})=A {\bf x}$$ around the parallelogram generated by $${\bf x}$$ and $${\bf y}$$ vanishes (see Figure 1).

Figure 1. Vector field Ax and the closed parallelogram path.

Figure 2 illustrates proof of this equivalence. The average forces on each of the sides of the parallelogram are equal to the forces $${\bf F}_i$$ at the midpoints $$M_i$$. The total work $$W$$ around the parallelogram, grouping parallel sides together, is

$W=({\bf F}_1- {\bf F}_3,{\bf x})+ ({\bf F}_2- {\bf F}_4 , {\bf y} );$

$\textrm {and since} \quad {\bf F}_1- {\bf F}_3 = -A {\bf y} \quad \textrm {and} \quad {\bf F}_2- {\bf F}_4 = A{\bf x},$

$\textrm {this gives} \quad W= (A{\bf x}, {\bf y}) - ({\bf x}, A{\bf y}).$

In particular, $$(1)$$ expresses the conservativeness of the vector field $$A{\bf x}.$$

Figure 2. Physical meaning of (Ax,y)−(x,Ay).

2. Here is a physical reason why eigenvalues of a symmetric matrix are real. Assuming for a moment that they are not, consider the plane spanned by the real and the imaginary parts $${\bf u, v}$$ of the eigenvector $${\bf w} = {\bf u} + i {\bf v}$$. At each point $${\bf x}$$ in this plane the force $$A{\bf x}$$ lies in the plane (so that we can forget about the rest of $${\mathbb R} ^n).$$ And since the work done by $$A{\bf x}$$ around a circle in this plane—centered around the origin—is zero, the tangential component of $$A{\bf x}$$ changes sign at some point(s) $${\bf x_0}$$ on the circle, which is to say that $$A{\bf x_0}$$ is normal to the circle at $${\bf x_0}$$. Thus, $${\bf x_0}$$ is a (real) eigenvector.

Figure 3. Geometrical proof of orthogonality.
3. Orthogonality of the eigenvectors: a physical and geometrical proof. Let $${\bf u, v}$$ be two distinct eigenvectors of a symmetric $$n × n$$ matrix $$A$$ with the eigenvalues $$\lambda\not= \mu=0$$ (the latter assumption involves no loss of generality since we can take $$\mu=0$$ by replacing $$A$$ with $$A- \mu I$$). Figure 3 shows the force field $$A{\bf x}$$ of such a matrix. Consider the work of $$A{\bf x}$$ around the triangle $$OQP$$. The only contribution comes from $$PO$$ since $$A{\bf x}$$ vanishes along $$OQ$$ and is normal to $$QP$$. And if $$A{\bf x}$$ is conservative, then $$W_{PO} = 0$$ and hence $$P = O,$$ implying $${\bf u} \perp {\bf v}$$. This completes a “physical” proof of orthogonality of the eigenvectors of symmetric matrices.

Figure 4. aij as the angular velocity.
4. The entry $$a_{ij}, \enspace i\not= j$$ of a square matrix $$A = (a_{ij})$$ has a dynamical interpretation: it is the angular velocity, in the $$(ij)$$–plane, of $$e_i$$ moving with the vector field $$A{\bf x}$$.1 Indeed, $$a_{ij} = (A\textrm{e}_i, \textrm{e}_j)$$, the projection of the velocity $$A\textrm{e}_i$$ onto $$\textrm{e}_j,$$ Figure 4. And thus the symmetry condition $$a_{ij} = a_{ji},$$ illustrated in Figure 4, also amounts to stating that the 2D curl in every $$ij$$–plane vanishes. The symmetry for $$3 × 3$$ matrices is equivalent to curl $$A{\bf x}=0$$. In fact, decomposition of a general square matrix into its symmetric and antisymmetric parts amounts to decomposing the vector field $$A{\bf x}$$ into the sum of a curl-free and divergence-free field, a special case of Helmholtz’s theorem, itself a special case of the Hodge decomposition theorem.

And the diagonal entries $$a_{ii}$$ give the rate of elongation of $$\textrm{e}_i$$; this explains geometrically why the cube formed at $$t = 0$$ by $$\textrm{e}_i$$ and carried by the velocity field $$A{\bf x}$$ changes its volume at the rate $$\textrm {tr} A$$ (at $$t = 0$$). This also offers a geometrical explanation of the matrix identity det $$e^A = e^{tr \enspace A}.$$

1 to be more precise, we should be referring to the moving vector instantaneously aligned with $$e_i$$.

Mark Levi (levi@math.psu.edu) is a professor of mathematics at the Pennsylvania State University.