# Why is this matrix symmetric?

Goldstein 3rd Ed, pg 339

In large classes of problems, it happens that $$L_{2}$$ is a quadratic function of the generalized velocities and $$L_{1}$$ is a linear function of the same variables with the following specific functional dependencies: $$L\left(q_{i}, \dot{q}_{i}, t\right)=L_{0}(q, t)+\dot{q}_{i} a_{i}(q, t)+\dot{q}_{i}^{2} T_{i}(q, t)\tag{8.22}$$ where the $$a_{i}^{\prime} s$$ and the $$T_{i}$$ 's are functions of the $$q$$ 's and $$t$$.

Under the given assumptions the Lagrangian can be written as$$L(q, \dot{q}, t)=L_{0}(q, t)+\tilde{\dot{\mathbf{q}}} \mathbf{a}+\frac{1}{2} \tilde {\dot{\mathbf{q}}} \mathbf{T} \dot{\mathbf{q}},\tag{8.23}$$

It's then said that $$\mathbf{T}$$ is symmetric. Why is it symmetric? Please give me a hint or an answer.

• You didn't define $\mathbf{T}$, just some numbers $T_i$ (not components of a matrix $T_{ij}$). What's the definition of $\mathbf{T}$? Commented Dec 22, 2021 at 13:39
• I've copied the whole passage. That's all what the author said. Commented Dec 22, 2021 at 13:41
• It is without loss of generality symmetric. Commented Dec 22, 2021 at 13:59
• Even if you include the antisymmetric part, $\boldsymbol{T} = \boldsymbol{T}_S + \boldsymbol{T}_A$, it never contributes ($\bar{\boldsymbol{a}} \boldsymbol{T}_A \boldsymbol{a} = 0$ for any vector $\boldsymbol{a}$). Commented Dec 22, 2021 at 14:18
• Commented Dec 22, 2021 at 14:25

Let us be general and suppose that you have a real column vector $$x$$ and a real matrix $$M$$. Define $$Q_M(x)=x^T Mx\tag{1}.$$

Now as is well-known you can split $$M = S+A$$ where $$S$$ is symmetric, $$S^T=S$$ and $$A$$ is anti-symmetric $$A^T=-A$$. One just has to define $$S = \dfrac{M+M^T}{2},\quad A=\dfrac{M-M^T}{2}\tag{2}.$$

Now observe that $$Q_M(x)=x^T Sx+x^T Ax\tag{3}.$$

It is easy, though, to see that the second term is zero. Indeed $$x^T A x$$ is a number, and hence is its own transpose. Therefore $$x^T Ax=(x^T Ax)^T= x^T A^T (x^T)^T=x^T(-A)x=-x^T A x\tag{4}$$

Equation (4) means that $$x^TAx=0$$ as claimed. Therefore $$Q_M(x)=x^T S x$$.

In other words: $$Q_M(x)$$ depends just on the symmetric part of $$M$$. Since the anti-symmetric part doesn't matter, we can without loss of generality assume that $$M$$ is symmetric, meaning that we have discarded the non-contributing anti-symmetric part.

• +1. If anyone's worried about the "transpose" of a number, note that while $x\cdot Mx$ is a number $x^TMx$ is a $1\times1$ matrix. See also here.
– J.G.
Commented Dec 22, 2021 at 14:59

You're supposed to define $$T$$ as the diagonal matrix with entries $$T_i(q,t)$$ along the diagonal. So in this fixed basis for the $$\dot{q}$$ we have $$\dot{q}Tq = \dot{q}_i^2 T_i$$, and $$T$$ is obviously symmetric. In some other basis, you get a general $$T$$ as the transform of the diagonal matrix. In particular, $$T$$ is therefore in general a diagonalizable real matrix with real eigenvalues $$T_i$$ and orthonormal eigenspaces (spanned by the initial choice of $$\dot{q}_i$$). Now, a real matrix is symmetric if and only if its eigenspaces are orthogonal and span the full vector space (see e.g. this math.SE question and its linked questions), therefore $$T$$ is symmetric.