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Assume that $A\in\mathbb{R}^{n\times n}$ is an upper triangular matrix which all the entries in the triangular part are positive. Let $n$ be a positive integer number. Is the following definition true? $$A^{\frac{1}{n}}:=\left[ \begin{array}{ccccc} \sqrt[n]{a_{11}}&\sqrt[n]{a_{12}}&\cdots& &\sqrt[n]{a_{1n}}\\ 0&\sqrt[n]{a_{22}}&\cdots& &\sqrt[n]{a_{2n}}\\ 0& 0& \sqrt[n]{a_{33}}& &\vdots\\ \vdots&\vdots& &\ddots & \\ 0& 0&\cdots& 0 &\sqrt[n]{a_{nn}}\\ \end{array} \right].$$ Moreover, can we say that the eigenvalues of $A^{\frac{1}{n}}$ are $\sqrt[n]{a_{11}},\ldots,\sqrt[n]{a_{nn}}$?

David Mayar
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    Did you try an example? Say for $n=2$? – quid Feb 11 '15 at 22:06
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    Definitions are always "true", by definition. The question is whether your definition of $A^{\frac{1}{n}}$ satisfies the presumably required property that $(A^{\frac{1}{n}})^n = A$. – Yuval Filmus Feb 11 '15 at 22:07
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    Since no has done it yet, I will. You really shouldn't be using an equal sign without explanation. This can actually make a definition meaningless. – Git Gud Feb 11 '15 at 22:08
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    The diagonal entries are correct, and the eigenvalues are correct (though you need not know the multiplicities). However, the off diagonal entries are not correct in general. I assume the $n$-th roots are the real positive ones when the $a_{ij}$ are positive. – Geoff Robinson Feb 11 '15 at 22:08
  • Yes, your answer is true. So, how can we define A^{\frac{1}{n}}? – David Mayar Feb 11 '15 at 22:10
  • This link to this question (http://math.stackexchange.com/questions/241764/eigenvalues-and-power-of-a-matrix) is useful for your second part, where you want to know how the eigenvalues of a matrix are exponentiated in the same way as the original matrix: – Xoque55 Jul 09 '15 at 21:05

2 Answers2

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In general, no, assuming $\left(A^\frac1n\right)^n = A$. Counterexample:

$$B \stackrel{\textrm{def}}{=} \begin{pmatrix} \sqrt{2} & \sqrt{3} \\ 0 & \sqrt{5} \end{pmatrix} \implies B^2 = \begin{pmatrix} 2 & \sqrt{2}\sqrt{3}+\sqrt{3}\sqrt{5} \\ 0 & 5\end{pmatrix}.$$

However, it is worth noting that the diagonal entries of a triangular matrix are the multiset of its eigenvalues.

Emily
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The answer to the second question is yes because the characteristic polynomial of $A$ is $$(\lambda-a_{11}^{1/n})...(\lambda-a_{nn}^{1/n})$$

The answer to the first question is no, consider $$A=\pmatrix{\sqrt{2}&\sqrt{3}\\0&\sqrt{5}}$$. We have $$A^2\ne \pmatrix{2&3\\0&5}$$

Peter
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