1

I would like to see a proof of the following statement:

A positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root.

I think the "proof" found in the wikipedia page is flawed; can somebody provide one, or maybe an hint?

Basically, I wanted to prove that the only orthogonal matrix that is symmetric and positive definite is the identity; although I am also interested in the theorem for its own sake :)

Ant
  • 21,098

1 Answers1

2

For your first question, could point out precisely which part of the proof bugs you?

To answer your second question:

Orthogonal symmetric positive definite matrix is an identity matrix. It is diagonalisable, hence it has an orthonormal basis of eigenvectors.

In this basis the matrix is a diagonal one. The norm of the matrix is $1$, therefore all eigenvalues are of the form $\pm 1$.

Finally, it is positive definite, therefore all eigenvalues are equal to $1$.

We deduce now that in this basis the matrix is an identity matrix, which allows us to conclude.

TZakrevskiy
  • 22,980
  • +1 Nice.. Pretty much what I did except I didn't connect the dots at the end (namely that the matrix must be diagonal so you don't need the more powerful theorem above). Anyhow I am more interested in the proof of the first theorem :) – Ant Dec 11 '14 at 21:43
  • the proof reported on the wiki page only works if there are $n$ distinct eigenvalues (and why the same argument does not work when there are two equal eigenvalues is not specified, so it is not clear at all) :-) – Ant Dec 11 '14 at 21:46
  • @Ant If the eigenvalues are not strictly positive, then we can not guarantee the existence of the square root. For example, the matrix $\begin{pmatrix}0&1\0&0\end{pmatrix}$ does not have a square root. – TZakrevskiy Dec 11 '14 at 21:49
  • @Ant as for the condition of having distinct eigenvalues - I don't see where they state it (and on the first glance they don't use it). – TZakrevskiy Dec 11 '14 at 21:52
  • Ah okay.. So if I understand correctly the correct statement is "A diagonalizable positive-semidefinite matrix has precisely one positive-semidefinite square root", correct? But again how to prove that? – Ant Dec 11 '14 at 21:52
  • They state the condition (although they don't use it) but without it the theorem does not hold; for example the 2x2 identity $I$ has infinite square roots, not only $2^2 = 4$ – Ant Dec 11 '14 at 21:53
  • Try to compute the square root of $\begin{pmatrix}1&1\0&1\end{pmatrix}$. It is not diagonalisable, yet it has a principal square root. – TZakrevskiy Dec 11 '14 at 21:54
  • I don't see how a $2\times 2 $ identity matrix can have more than $4$ different square roots. – TZakrevskiy Dec 11 '14 at 21:55
  • As stated at the beginning (although I have not checked) every matrix like $\frac 1t\begin{pmatrix}\mp s& \mp r\\mp r&\pm s\end{pmatrix}$ for every $r, s, t : r^2 + s^2 = t^2$ – Ant Dec 11 '14 at 21:57
  • Hm, nice, I overlooked that. To sum up: if eigenvalues are distinct, then there're $2^n$ square roots. If eigenvalues are not distinct, some weird stuff can happen: no roots, infinity of roots, etc. If the eigenvalues are strictly positive, then the square roots exists, and the principal square root is unique. – TZakrevskiy Dec 11 '14 at 22:03
  • This seems to be the case. I have find this answer interesting (http://math.stackexchange.com/questions/562566/are-there-sometimes-only-finitely-many-square-roots-of-a-positive-matrix) but I am still a little confused! :) – Ant Dec 11 '14 at 22:09