Unique solution of $A^3 = B$, where $A$, $B$ is a diagonal matrix given below.

Question

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 ?

Answer 1

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.

Answer 2

The main points about this situation are that (1) the matrix $B$ is diagonalisable over 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$~B$.