# The square root of Matrix!

1. Mar 29, 2008

### PRB147

If one $$4\times 4$$ matrix have two vanishing eigenvalues, does the matrix have a square root?

Last edited: Mar 29, 2008
2. Mar 29, 2008

### morphism

What do you mean by vanishing eigenvalues?

3. Mar 29, 2008

### HallsofIvy

Staff Emeritus
"Vanishing" is pretty standard for "0" so I'll assume you mean that the matrix has one or more eigenvalues equal to 0. Uh- what about the 0 matrix? Isn't its square root pretty easy? And it has plenty of "vanishing" eigenvalues.

(I'm assuming that you meant "does the matrix have a square root?")

4. Mar 30, 2008

### transgalactic

i dont know whats vanishing eigen values but

in order to make a squere root of a matrix you need to build first a digonizable form of the given matrix

then you have the formula
P^-1 *f(D) *P=f(A)
in this case your "f" is a squere root function
you find the P and P^-1 then make the root operation on each of the numbers in the
diagonal matrix then take these three matrices and their multiplication
is your resolt

5. Mar 30, 2008

### morphism

That's not true. There are many non-diagonalizable matrices that have square roots, e.g.

$$\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right) = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right)^2.$$

If we're working over an algebraically closed field (like the complex numbers), then we can get a nice set of necessary and sufficient conditions for the existence of square roots of nxn matrices by putting everything into Jordan form. Then it becomes evident that the problem really lies in the Jordan blocks corresponding to zero. For 4x4 matrices with "two vanishing eigenvalues" the situation is relatively straightforward; the example above is such a matrix.

Last edited: Mar 30, 2008
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