# 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

*A*

**x**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 (

*A*

**x,y**)−(

**x,**

*A*

**y**).

**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.

**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.**

*a*as the angular velocity.

_{ij}**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.**

^{1}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 \( e_i\).