Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be an inner product space generated over $\mathbb{C}$ and $B$ is a $n\times n$ normal complex matrix.

(1)I need to show that there exists a matrix $C$ such that $C^{2}=B$. I know that B is orthogonally diagonalizable by a theorem which was proved in class. Could I should that $C=BC^{-1}$?

(2) If the eigenvalues of $B$ are real, then $B$ is self-adjoint. Not sure where to start on this one.

share|cite|improve this question
Can you find a square root of a diagonal matrix with complex entries? – alex.jordan Nov 30 '12 at 20:25
up vote 2 down vote accepted

Hint: Notice that if you have a matrix of the form $B=PDP^{-1}$ then $$B^2 = (PDP^{-1})(PDP^{-1}) = PD^2P^{-1}$$ Can you think of a square root for a diagonal matrix?

share|cite|improve this answer
Wouldn't it just be the square roots of the eigenvalues of the eigenvectors that make up the basis for the complex vector space? – MathScratch Nov 30 '12 at 20:32
Yes, it would just be the square roots of the eigenvalues. – EuYu Nov 30 '12 at 20:36
How could I show that if the eigenvalues are real, then B is self-adjoint? Is that because the matrix is orthogonally diagonalizable? – MathScratch Nov 30 '12 at 20:40
For every diagonal matrix, $D^T = D$, now you can use the fact that the eigenvalues are real. – Stefan Nov 30 '12 at 20:42
I'm not following... – MathScratch Nov 30 '12 at 20:50

Hint: $(ACA^{-1})^2=AC^2A^{-1}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.