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Show that $A^3 = B$ has exactly one REAL solution (which is easy to find), where: $$B = \begin{bmatrix}8& 0& 0\\ 0& -1& 0\\ 0& 0 &27\end{bmatrix}$$

Finding the solution is really easy. But how can we show that it's the only solution ? I know that the power of a diagonal matrix is a diagonal matrix, but is the opposite: the nth root of a diagonal matrix is a diagonal matrix, also true ? And if so, how to prove it ?

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    $\begingroup$ It is not true if you allow complex matrices. The question shold make clear what kind of matrices are considered. $\endgroup$ – Marc van Leeuwen Sep 29 '17 at 12:46
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    $\begingroup$ Also, it is not in general true that any square root of a diagonal matrix must be diagonal; for instance, a rotation by a quarter turn in a plane defined by two coordinate axes has as square a diagonal matrix, but it is not itself diagonal. $\endgroup$ – Marc van Leeuwen Sep 29 '17 at 12:50
  • $\begingroup$ Thank you for helping and answering the second question. The main question ask for real solution. $\endgroup$ – user3192711 Sep 29 '17 at 13:20
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The main points about this situation are that (1) the matrix $B$ is diagonalisable of the real numbers (it is even diagonal), and (2) its eigenspaces all have dimension$~1$ (because the diagonal entries are all distinct); also relevant is of course that cube roots of scalars always uniquely exist over the real numbers.

If $A$ is such that $A^3=B$, then clearly $A$ and $B$ must commute. Now if $v$ any eigenvector for $B$, then $Av$ is also an eigenvector for $B$, for the same eigenvalue: say $Bv=\lambda v$ then $B(Av)=A(Bv)=A(\lambda v)=\lambda(Av)$. But by property (2) above, this means $Av$ is a scalar multiple of $v$, in other words $v$ is also an eigenvector for$~A$. Then any basis of eigenvectors for$~B$ is also a basis of eigenvectors for$~A$, and $A$ will be diagonal when expressed on such a basis. In the example, the standard basis is such a basis of eigenvectors, in other words $A$ must, just like$~B$, be a diagonal matrix. As you already saw, it can only be the matrix obtained by taking cube roots of each of the diagonal entries of$~B$.

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The result follows from a direct computation using Buchberger's algorithm as follows. Write $A=(a_{ij})$ with $9$ parameters as entries. The Buchberger's algorithm immediately gives the following: the diagonal elements $a_{ii}$ have to satisfy one of the following equations: $$ (a_{11},a_{22},a_{33})=(2,-1,3), $$ or $$ a_{11}^2 + 2a_{11} + 4=0, $$ or $$ a_{22}^2 - a_{22} + 1=0, $$ or

$$ a_{33}^2 + 3a_{33} + 9=0. $$ The last three equations do not have real solutions. Then we easily see that $A={\rm diag}(2,-1,3)$ is the only real solution.

Edit: In this question

$A$ is a symmetric real matrix. Show that there is $B$ such that $B^3=A$

the claim is proved, too, since our matrix here is indeed symmetric.

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    $\begingroup$ Could you explain more about the Buchberger's algorithm ? Because I am studying pretty basic linear algebra and concepts, so a simple solution without needing too much knowledge/methods is expected and would be better. Thank you in advance $\endgroup$ – user3192711 Sep 29 '17 at 14:43
  • $\begingroup$ For the algorithm see here. It is a standard tool in computer algebra. Of course, one can argue whether or not one should use an algorithm, like for computing a determinant. However, you can do without it, using substitution and manipulating the polynomial equations. The same can be done computing a determinant by hand, if necessary. $\endgroup$ – Dietrich Burde Sep 29 '17 at 15:31
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Let $$Q = \begin{bmatrix}2& 0& 0\\ 0& -1& 0\\ 0& 0 &3\end{bmatrix}$$

We are given $$0=A^3-Q^3=(A-Q)(A^2+AQ+Q^2)=(A-Q)(A-\alpha Q)(A-\overline{\alpha}Q)$$ where $\alpha = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}.$

The solutions are

$$\begin{align} A &=Q, &(\textrm{real})\phantom{000}\\ A &= \alpha Q, &(\textrm{complex})\\ A &= \overline{\alpha}Q, &(\textrm{complex}) \end{align}$$

Notice $\{1, \alpha, \overline{\alpha} \}$ are the three cube roots of one.

UPDATE

These are not the only solutions. We could have taken $Q$ to be any of

$$Q = \begin{bmatrix} \mp 2& 0& 0\\ 0& -1& 0\\ 0& 0 &\mp 3\end{bmatrix}$$,

so it looks like there are $(4)\cdot (3) = 12$ solutions, not only $3$.

To address the last part of the question:

If $A^n = \Lambda$, Let $Q$ be an $n^\textrm{th}$ root of $\Lambda$ and $\alpha$ an $n^\textrm{th}$ root of unity. Then $A \in \{ Q, \alpha Q , \dots, \alpha^{n-1} Q \}$, are solutions. As Marc pointed out, the solutions need not be diagonal, for example, a rotation matrix that rotates by an angle of $\frac{2\pi}{n}$.

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  • $\begingroup$ Your factorization requires that $AQ=QA$, which is not clear to me. However, if you have this, then standard results give that $A$ and $Q$ are simultaneously diagonalizable and the problem reduces to 3 one dimensional problems. $\endgroup$ – Aaron 6 hours ago

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