Consider for example a mass matrix $M$, $\lambda$ one eigenvalue and $X$ a corresponding eigenvector. Then $[M]=\text{mass}$ (the brackets indicate the "unit operator"), and $MX=\lambda X$ so $[M][X]=[\lambda][X]$, so $[\lambda]=\text{mass}$. That's why for example in oscillators, the pulsations $\omega$ are such that $[\omega^2]=[M^{-1}K]=\text{seconds}^{-2}$.

But what about the eigenvectors? I would tend to think that they are dimensionless, because during a change of basis $u=Pq$, $q$ and $u$ have the same units, while $P$ gathers the eigenvectors $X$; the vectors designate a change of basis but the vector space remains the same.

Is it so?


2 Answers 2


The units of the eigenvector can be anything you choose. Normally you want them to be dimensionless, but other choices can be sensible on occasion.

The reason that any dimension is valid is because if $X$ is an eigenvector, then $\mu X$ is also an eigenvector for any scalar $\mu$, which can in principle have any dimensions you want.

The main implication of that is that eigenvectors are only defined up to a multiplicative constant, and if you want to fix that then you should look at exactly what you'll be using them for. Because the constant can be dimensionful, this carries the dimensionality of the eigenvectors with it.

In general, though, what you want to do with your eigenvectors is to use them as a basis. In that case, it is usually best practice to make them dimensionless, so that their norms will be unity. If you don't care so much about this, then you can essentially do whatever you feel you need to.


Not just with eigenvectors, but with any vector.

I always wondered when you decompose any vector into magnitude and direction who gets the units. Is it $(1,0.1,0)\cdot3\mathrm{m}$ or $3\cdot(1\,\mathrm{m}, 10\,\mathrm{cm},0)$ ?

I think it is up to the user to interpret a decomposition any way they see fit.

With eigen vectors you have $$\hat{y}=A \hat{x} = \lambda \hat{x} $$

which is interpeted as $A$ and $\lambda$ carrying the units of $\hat{y}$, or $\hat{x}$ carrying the units of $\hat{y}$. Or it is both.

For mechanics to understand the units you need to phrase the problem in terms of forces

$$ K \hat{x} - \omega^2 M \hat{x} = 0$$

and then multiply by $M^{-1}$ giving $$M^{-1} K \hat{x} = A \hat{x} = \omega^2 \hat{x} $$

So in this case it would make sense to retain the units of distance for the eigenvectos $\hat{x}$. But it doesn't have to be like this. All it matters is that $\lambda \hat{x}$ is the same units as $A \hat{x}$.


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