Can a 4x4 matrix with two vanishing eigenvalues have a square root?

In summary, for a 4x4 matrix with two vanishing eigenvalues, the existence of a square root depends on the Jordan blocks corresponding to zero. If these blocks can be arranged such that the matrix is diagonalizable, then a square root can be constructed. Otherwise, it may not be possible to find a square root for the matrix.
  • #1
PRB147
127
0
If one [tex]4\times 4[/tex] matrix have two vanishing eigenvalues, does the matrix have a square root?
 
Last edited:
Physics news on Phys.org
  • #2
What do you mean by vanishing eigenvalues?
 
  • #3
"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
i don't know what's 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
transgalactic said:
in order to make a squere root of a matrix you need to build first a digonizable form of the given matrix
That's not true. There are many non-diagonalizable matrices that have square roots, e.g.

[tex]\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.[/tex]

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:

What is the square root of a matrix?

The square root of a matrix is another matrix that, when multiplied by itself, results in the original matrix.

Is there always a square root for every matrix?

No, not all matrices have a square root. A matrix must have all real, non-negative eigenvalues to have a square root.

How is the square root of a matrix calculated?

The square root of a matrix can be calculated using diagonalization or the Jordan canonical form.

What is the difference between the principal square root and non-principal square root of a matrix?

The principal square root of a matrix is the unique matrix with all positive eigenvalues, while non-principal square roots may have negative eigenvalues.

What are some applications of the square root of a matrix?

The square root of a matrix is used in various fields such as physics, engineering, and computer science to solve systems of equations, simulate processes, and analyze data.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
764
  • Linear and Abstract Algebra
Replies
1
Views
487
Replies
6
Views
2K
  • Linear and Abstract Algebra
Replies
13
Views
1K
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
11
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
891
  • Linear and Abstract Algebra
Replies
2
Views
912
  • Linear and Abstract Algebra
Replies
12
Views
1K
Back
Top