I'm self-studying through Mathematical Methods for Physics and Engineering and am stuck on how to evaluate some matrix quantities.

The authors provide the kinetic energy $T$ as $T=\dot{q}^{\text{T}}\text{A}\dot{q}$ and the potential energy $V$ as $V = q^{\text{T}}\text{B}q$, with $q$ being a generalized coordinate and A and B as symmetric, positive definite matrices.

They apply conservation of energy to the energies as defined: \begin{align*} \frac{d}{dt}\left(T+V\right) &= 0\\ \frac{d}{dt}\left(\dot{q}^{\text{T}}\text{A}\dot{q}+q^{\text{T}}\text{B}q\right) &= 0\\ \ddot{q}^{\text{T}}\text{A}\dot{q}+\dot{q}^{\text{T}}\text{A}\ddot{q}+\dot{q}^{\text{T}}\text{B}q+q^{T}\text{B}\dot{q}&=0 \end{align*} Up until this point, I follow along easily. However, they then go on to say:

"Using A=A$^{\text{T}}$ B=B$^{\text{T}}$ gives $2\dot{q}^{\text{T}}\left(A\ddot{q}+\text{B}q\right)=0"$

I don't know how to operate the $\ddot{q}^\text{T}$ matrix on the A matrix (or any of the other operations that were performed.)

To be clear: I don't want the work to be done for me, I'm just looking for a resource that explains the concepts clearly, as the chapter on matrices in this book is not helpful for this problem.


1 Answer 1



Using $(AB)^T=B^TA^T$, we have


  • $\begingroup$ That did it! Thank you. One more question: can the transpose always be applied as such (i.e., to only some terms in an equation) or are we able to do that in this example because the equation is set to zero? $\endgroup$
    – Luciano
    Apr 12, 2017 at 12:24
  • 1
    $\begingroup$ You're welcome. My pleasure. And we can always write $(AB)^T=B^TA^T$. -Mark $\endgroup$
    – Mark Viola
    Apr 12, 2017 at 12:38

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