11

"If $V$ is a complex inner product space and $T\in \mathcal{L}(V)$. Then $V$ has an orthonormal basis Consisting of eigenvectors of T if and only if $T$ is normal".

  I know that the set of orthonormal vectors is called the "spectrum" and I guess that's where the name of the theorem. But what is the reason for naming it?

Wuschelbeutel Kartoffelhuhn
  • 1,125
Hiperion
  • 1,763
  • 7
    Rather: the set of eigenvalues of a linear map is what is called spectrum.In the spectral theorem, you decompose the linear map in (very simple!) pieces, each piece coming from one element of the spectrum. – Mariano Suárez-Álvarez Oct 22 '12 at 06:33
  • @Mariano, yes! My mistake, thanks. – Hiperion Oct 22 '12 at 06:36
  • 2
    I'm teaching a Linear Algebra course at the moment and if things go according to schedule, I'll be discussing the Spectral Theorem on Halloween's day. :) – Andrea Mori Oct 22 '12 at 10:36
  • 1
    It is now the year 2019, I'll be teaching a Linear Algebra course starting next week, and if things go according to schedule, I'll be discussing the Spectral Theorem on Halloween's day! I am glad to maintain this important tradition. :-) – Zach Teitler Aug 25 '19 at 06:22

2 Answers2

23

The name is provided by Hilbert in a paper published sometime in 1900-1910 investigating integral equations in infinite-dimensional spaces.

Since the theory is about eigenvalues of linear operators, and Heisenberg and other physicists related the spectral lines seen with prisms or gratings to eigenvalues of certain linear operators in quantum mechanics, it seems logical to explain the name as inspired by relevance of the theory in atomic physics. Not so; it is merely a fortunate coincidence.

Recommended reading: "Highlights in the History of Spectral Theory" by L. A. Steen, American Mathematical Monthly 80 (1973) pp350-381

DarenW
  • 753
  • 1
    For more details and additional references, see Jeff Miller's "Earliest Known Uses of Some of the Words of Mathematics", http://jeff560.tripod.com/s.html (scroll down to "SPECTRUM"). – Zach Teitler Aug 25 '19 at 06:26
4

I think the top voted answer is very helpful:

Since the theory is about eigenvalues of linear operators, and Heisenberg and other physicists related the spectral lines seen with prisms

It might be worth explaining a bit more precisely how the theory is related to optics, especially for people not familiar with how a prism works (like me)

x

As shown in the figure, a prism automatically decomposes a light into a series of monochromatic lights and refracts each of them by a different angle.(cf Cauchy's transmission equation)

We could consider the following analogies with the Spectrum Theorem:

  • the Matrix Operator $A$ as a prism

  • the input vector $v$ as a polychromatic light

  • each eigenvector of $A$ as monochromatic light, and the associated eigenvalue as the refractive index of the monochromatic light.

With spectral theorem, we know that we can decompose symmetric matrices(I stay in $\mathbb{R}$ for simplicity) into $A = V^{-1}\Lambda V = \sum{\lambda_i \vec{v_i} \cdot \vec{v_i}^t }$.

And $A \vec{x}=\sum{\lambda_i \vec{v_i} \cdot \vec{v_i}^t } \vec{x}= \sum{\lambda_i \vec{v_i} x_i }$

This is very similar to what a prism does with light: decompose, refract, and output. So, the spectrum theorem tells us we could find a spectrum of eigenvectors of a given symmetric matrix S.

References: Spectrum of a matrix

Rafael
  • 153