# The square root of Matrix!

## Main Question or Discussion Point

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

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morphism
Homework Helper
What do you mean by vanishing eigenvalues?

HallsofIvy
Homework Helper
"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?")

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

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