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.

  • $\begingroup$ You didn't define $\mathbf{T}$, just some numbers $T_i$ (not components of a matrix $T_{ij}$). What's the definition of $\mathbf{T}$? $\endgroup$
    – ACuriousMind
    Commented Dec 22, 2021 at 13:39
  • $\begingroup$ I've copied the whole passage. That's all what the author said. $\endgroup$
    – Kashmiri
    Commented Dec 22, 2021 at 13:41
  • 3
    $\begingroup$ It is without loss of generality symmetric. $\endgroup$ Commented Dec 22, 2021 at 13:59
  • $\begingroup$ 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 $\boldsymbol{a}$). $\endgroup$
    – tueda
    Commented Dec 22, 2021 at 14:18
  • $\begingroup$ Related: physics.stackexchange.com/q/607437/2451 $\endgroup$
    – Qmechanic
    Commented Dec 22, 2021 at 14:25

2 Answers 2


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.

  • $\begingroup$ +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. $\endgroup$
    – 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.


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