I'm studying the oscillations of systems with more than one degree of freedom from Landau & Lifshitz's Mechanics Third Edition (for those who have the book, my question corresponds roughly to explaining the last paragraph before equation 23.9).

First I give a brief background to my problem to show what I understand. So far I understand that given the Lagrangian, the equation of motion is:

$${m_{ik}\ddot{x}_k + K_{ik}x_k = 0}\tag{23.5}$$

Using the usual ansatz $x_k = A_k e^{i \omega t}$ then, we get the set of equations:

$$(-\omega^2m_{ij} + k_{ik})A_k = 0\tag{23.7}$$

where it can be easily seen that the section in brackets must have determinant 0, leading us to the characteristic equation:

$$|k_{ik} - \omega^2m_{ik}| = 0\tag{23.8}$$

where furthermore it can be easily shown that for each solution to the characteristic equation $\omega^2_\alpha$, $\alpha = 1, \ldots, S$ where $S$ is the degrees of freedom of the system, $\omega_\alpha$ must be positive and real.

I now quote in full the paragraph I'm having difficulty with:

The frequencies $\omega_\alpha$ having been found, we substitute each of them in equations (23.7) and find the corresponding coefficients $A_k$. If all the roots $w_\alpha$ of the characteristic equation are different, the coefficients $A_k$ are proportional to the minors of the determinant (23.8) with $\omega = \omega_\alpha$. Let these minors be $\Delta_{k\alpha}$. A particular solution of the differential equations (23.5) is therefore $x_k = \Delta_{k\alpha}C_{\alpha}\exp{(i\omega_\alpha t)}$ where $C_\alpha$ is an arbitrary complex constant.

I think my main problem here is my poor knowledge of linear algebra. I don't understand why the roots all being different implies that the coefficients are proportional to the minors of the determinant. Furthermore I'm slightly confused by the wording "minors of the determinant" since to me, a determinant is a scalar value, it seems like they should be saying "minors of the matrix $(k_{ik} - \omega^2m_{ik})$" Can someone explain the linear algebra behind all this to me? I've taken a Linear Algebra course but I was quite poor at it.

  • 3
    $\begingroup$ Very well-written question! :-) $\endgroup$
    – David Z
    Jul 1, 2012 at 4:12

2 Answers 2


$\def\M{{\bf M}} \def\x{{\bf x}}$Here are two examples that may help you to understand what Landau means by "the coefficients $A_k$ are proportional to the minors of the determinant."

Two by two

Let $\M = \left(\begin{array}{cc}a&b\\ c&d\end{array}\right)$. Notice that if $\det \M = 0$, then $\M \x = 0$ is solved by $$\x = \left(\begin{array}{c} \det(b) \\ -\det(a) \end{array}\right) = \left(\begin{array}{c} b \\ -a \end{array}\right).$$

Three by three

Let $$\M = \left(\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array}\right).$$ Again, if $\det \M = 0$, then $\M \x = 0$ is solved by $$\x = \left(\begin{array}{c} \det\left(\begin{array}{cc}b&c\\ e&f\end{array}\right) \\ -\det\left(\begin{array}{cc}a&c\\ d&f\end{array}\right) \\ \det\left(\begin{array}{cc}a&b\\ d&e\end{array}\right) \end{array}\right).$$

If you sometimes find unwinding Landau difficult, you are not alone. It is usually well worth the effort.

Addendum: Notice the statement $\M\x=0$ is equivalent to the statement that $\det\M = 0$ and the determinant of a matrix with linearly dependent rows is zero.


The basic fact from Liner Algebra underlying this derivation is the Cramer's rule which expresses solution to a system of linear equations via determinants.

The problem expressed by equation (23.7) is homogeneous, i.e. the r.h.s. of the system of linear equations is a null-vector, therefore the determinants entering Cramer's rule (see here) become minors.

  • $\begingroup$ OK, I remember Cramer's rule (vaguely). So if I understand correctly you're saying that the problem is equivalent to the usual Cramer's rule problem of finding $x$ in $Ax = b$ where $A = (-\omega^2m_{ik} + k_{ik})$ $x = (A_k)$ and $b = 0$ Hmm...but since we're substituting in the solutions to the characteristic equation won't it be the case that the determinant of $A$ is 0, and therefore Cramer's rule won't be applicable since we have 0/0? $\endgroup$
    – Mark Allen
    Jul 1, 2012 at 15:06

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