Is M unique given A=M^2 where A and M are real matrices?

profesorluissp 2022-09-13 Answered
Is M unique given A = M 2 where A and M are real matrices? My guessing is they are unique as I tried to diagonalize A to P D P 1 and no matter how I order the eigenvalues in D, it still gives the same M = P D 1 2 P 1 . But I am not sure this is true in general, since the diagonalization is too specific.
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Answers (1)

Jaden Mason
Answered 2022-09-14 Author has 15 answers
No, if M is any n × n matrix satisfying A = M 2 (we then say that M is a square root of A) then −M is also a square root of A. For this kind of problem it helps to first think about the case where A and M are scalars then think about whether or not the same argument holds for matrices.
Even if A is diagonalizable (which doesn't always hold), your argument using diagonalization doesn't really work here since there are many choices for D 1 / 2 (Note that D 1 / 2 is by definition any matrix satisfying D 1 / 2 D 1 / 2 = D). For instance, in the special case where A has n distinct nonzero eigenvalues, there are 2 n choices for this matrix. Explicitly, if D = diag ( λ 1 , , λ n ) then all square roots are of the form diag ( μ 1 , , μ n ) where μi is any number (real or complex) satisfying μ i 2 = λ i (there are exactly two of them for each i).

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