What conditions do I need on a matrix $A$ in order to know an $n^{th}$ root exists. In other words there is a matrix $B$ such that $B^n=A$ for $n \in \mathbb{Z}^+$.
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1One sufficient condition is if $A$ can be written as $A=\exp(M)$. – Berci Jan 22 '14 at 00:35
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@Berci So is there an easy test to apply to $A$ to determine if it can be written $\exp(M)$? Obviously if we can diagonalize $A$ we can take any root we want, but can we weaken this condition? – Wintermute Jan 22 '14 at 00:38
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Continue with your diagonalization idea. If the matrix is diagonalizable can you say something about its roots? – Betty Mock Jan 22 '14 at 00:46
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@BettyMock Sure if the matrix is diagonalizable the existence of the root will depend on the sign of the eigenvalues(depending on the parity of $n$) and we can use the diagonalization to get the root. I guess what I'm getting at is can a non-diagonalizable matrix have an $n^{th}$ root? – Wintermute Jan 22 '14 at 00:50
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If $A=\exp(M)$, then $A$ is invertible. However, I don't know if the exponential map is surjective – M Turgeon Jan 22 '14 at 01:01
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@MTurgeon Matrix exponential map is surjective to invertible matrices. – Sungjin Kim Jan 22 '14 at 01:14
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@mtiano Why would the existence of the root depend on the sign of the eigenvalues? Are we not admitting complex numbers? Also, if the matrix is not invertible the nth roots are still there even though those entries are zero; and the matrix is not expressible in exponential form. As per answer below, I was moving towards Jordan form -- if you can find the roots of a diagonalizable matrix, so too can you find the roots of the matrix in Jordan form. – Betty Mock Jan 23 '14 at 03:27
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If your matrix is invertable and you can get it into Jordan Normal form you can take the logarithm of a matrix using the techniques described here, and then exponentiate, taking advantage of the power rule for exponentials in the usual way. (If A is your matrix, then $\exp(\frac{1}{n}\log(A)) = A^\frac{1}{n}$)
Chris Bonnell
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If $A\in \textrm{M}_{2\times 2}(\mathbb{C})$ is a nonzero nilpotent matrix, then a square-root of the matrix $A$ is not defined.
To see this,
Suppose that $A=\pmatrix{{0}&{1}\\{0}&{0}}=B^2$, then $B$ is nilpotent as well.
Then $B$ is similar to either $A$ or $O$(zero-matrix), by considering Jordan block.
But, in either case, $B^2=O$, which is not possible.
Sungjin Kim
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But what can we say about the general case, not just the square root. – Wintermute Jan 22 '14 at 01:22
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1For this particular $A$, and for any $n\geq 2$, $A=B^n$ is impossible. – Sungjin Kim Jan 22 '14 at 01:23
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You've written $A=B^2$ here, which really leads one to want to claim that $B = \sqrt{A}$ for all practical purposes. – Chris Bonnell Jan 22 '14 at 01:54
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1This is "Proof by contradiction". I am not claiming that $B=\sqrt{A}$. – Sungjin Kim Jan 22 '14 at 01:58
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1If the dimension is not large, such as $3$, then there are two similarity classes of nilpotent matrices, that has square root. However, if dimension gets larger, then it would be difficult. – Sungjin Kim Jan 22 '14 at 02:27
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This answer seems to be false, as Sungjin Kim points out $$ \begin{pmatrix}0&0&0\1&0&0\0&1&0\end{pmatrix} \in \sqrt{\begin{pmatrix}0&0&0\0&0&0\1&0&0\end{pmatrix}}$$ is a counterexample. This answer hints at a more general treatment, but could use more links to further reading. – MRule Jul 09 '24 at 10:59