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What is the geometric intuition behind the fact that only matrices that are similar to a symmetric matrix are diagonizable?

So e.g. why is it important that the multiplier of the the first component of the last basis vector be the same as the multiplier of the the last component of the first basis vector(i.e. that in an $n*n $ matrix $(n,1)$ be the same as $(1,n)$)?

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  • $\begingroup$ It seems that what you should really be after is "why are symmetric matrices diagonalizable"? You should look into proofs of the spectral theorem. $\endgroup$ Commented May 17, 2016 at 13:47

2 Answers 2

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When you were first learning about null spaces in linear algebra, your guess for the null space -- assuming you had some reasonable geometric intuition into the field -- was that the null space was orthogonal to the column space. After all, that makes sense. If your singular transformation collapses/projects $\mathbb{R}^2$ into a line, then the vectors that get mapped to the origin are the ones perpendicular to the column space.

Is the column space perpendicular to the row space?

Or at least, so it seems -- in reality, though, the projection doesn't need to be so nice and orthogonal. You could, for instance, rotate all vectors in the space by some angle and then collapse it onto a line.

It turns out the null space isn't perpendicular to the column space, but in fact to the row space instead -- these two spaces are only identical for matrices which do not perform a rotation.

This is a very important observation, because it tells you something about the character of matrices -- asymmetry in a matrix is a measure of how rotation-ish it is. Specifically, an antisymmetric matrix is the result of 90-degree rotations (like imaginary numbers) and a symmetric matrix is the result of scaling and skews (like real numbers).

$$A = \underbrace {\frac{1}{2}(A + {A^T})}_{\scriptstyle{\rm{symmetric }}\atop\scriptstyle{\rm{part}}} + \underbrace {\frac{1}{2}(A - {A^T})}_{\scriptstyle{\rm{antisymmetric }}\atop\scriptstyle{\rm{part}}}$$

All matrices can bet written as the sum of these two kinds -- a symmetric part and an anti-symmetric part -- much like all complex numbers can be written as the sum of a real part and an imaginary part. And this is fundamentally why symmetric matrices are "special" -- for the same reason that real numbers are special.


Notes:

(1) Scaling and skews are actually essentially the same thing, which is why it makes sense to include skews in the group of things that are "essentially real numbers", even though you can't really represent skews with any complex number -- real or otherwise. Skews are just scaling across a different set of axes, called "eigenvectors" (this is also why symmetric matrices have eigenvectors).

(2) My explanation of the analogy (between matrices and complex numbers) is oversimplified -- antisymmetric matrices actually represent 90 degree rotations only, and these rotations can actually be spirals, which means they do scaling too. But the analogy still holds, because this applies to imaginary numbers too (e.g. the complex number $8i$ is a rotation by 90 degrees followed by a scaling by 8).

(3) A more accurate way to phrase the analogy is "the antisymmetric part of the matrix operates in a sub-space orthogonal to the vector being transformed while the symmetric part operates in the direction of the vector itself, so their sum spans all possible vectors of the target space". In other words, the analogy is to the Cartesian form of complex numbers -- you get to represent transformations as linear combinations of the vector itself and vectors orthogonal to it.

(4) It is possible to deal with at least some matrices in a way that corresponds to the polar forms of complex numbers -- this is done by representing matrices as products of symmetric matrices and orthogonal matrices, much like $re^{i\theta}$ represents complex numbers as products of real numbers and unit complex numbers.

(5) Edit: Not sure why I was so long-winded in these notes. The point is:

  • symmetric matrices : real numbers :: antisymmetric matrices : imaginary numbers (Cartesian form)
  • positive-definite matrices : real numbers :: orthogonal matrices : unit complex numbers (Polar form)
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  • $\begingroup$ Thank you for this fantastic answer! $\endgroup$
    – kanso37
    Commented Aug 9, 2020 at 3:05
  • $\begingroup$ Great explanation! Any more references for this interpretation? $\endgroup$ Commented Sep 16, 2021 at 13:17
  • $\begingroup$ @JamesWhite I wrote some blog posts on it earlier, although I don't remember if there's anything additional in those articles related to this besides what's already included in the answer. $\endgroup$ Commented Sep 17, 2021 at 14:36
  • $\begingroup$ I'm so pleased I found this answer! Thank you! $\endgroup$
    – mattb
    Commented May 10, 2022 at 22:14
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only matrices that are similar to a symmetric matrix are diagonizable

This statement is true, but quite useless. I prefer the statement

only matrices that are similar to a diagonal matrix are diagonizable

which is, hum... the definition itself of diagonalizability. The geometrical intuition of diagonalizability is that you can decompose a transformation in homotheties, which are the simplest geometrical transformations you could imagine.

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