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Given the matrix $A^3= \begin{bmatrix}83&84\\42&41\end{bmatrix}$, find A

I only know how to use the basic method to find A, which is $A^3= \begin{bmatrix}a&b\\c&d\end{bmatrix}^3$, then solve a,b,c,d. Is there any other method?

Grace Cheng
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3 Answers3

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Hint : $A^3$ is diagonalizable and its eigenvalues are $5^3$ and $(-1)^3$.

TheSilverDoe
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Note that $B:= A^3$ has minimal polynomial $(x-125)(x+1)$ so to apply the function $x \mapsto \sqrt[3]{x}$ we have to find a polynomial $p(x) = ax+b$ such that $$p(125) = \sqrt[3]{125} = 5, \quad p(-1) = \sqrt[3]{-1} = -1.$$ We get that $p(x) = \frac1{21}(x-20)$ so $$\sqrt[3]{B} = p(B) = \frac1{21}(B-20I) = \begin{bmatrix}3 & 4 \\ 2 & 1 \end{bmatrix}.$$ This is certainly a cube root of $B$.

mechanodroid
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  • I think that is not immediately obvious what you are doing with your $p(x)$. So I'll explain: $A^3=PD^3P^{-1}$ and $A=PDP^{-1}$ so $p(A^3)=Pp(D^3)P^{-1}=PDP^{-1}=A$ and since $D,D^3$ are diagonal with only $2$ values,we can go from one to another with an affine polynomial. We would need a quadratic for $3\times 3$ matrix, and so on... – zwim Jan 26 '21 at 14:58
  • I'm not the downvoter though... – zwim Jan 26 '21 at 16:35
  • @zwim I was going for the application of the holomorphic functional calculus on the function $z \mapsto z^{1/3}$ (principal cube root) which is holomorphic on a neighbourhood of ${125,-1}$. Then we can recall how is this defined on the Jordan form and realize that we only need a polynomial which coincides with the cube root on the spectrum. – mechanodroid Jan 26 '21 at 16:43
  • Anyway, from this approach it is in no way clear that the cube root we got is actually equal to $A$ (which I never claimed), maybe that's what's bothering someone. – mechanodroid Jan 26 '21 at 16:45
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Since the roots of the characteristic polynomial are $-1,125$, two distinct roots, in dimension two, $A^3$ is diagonalizable. Find a basis of eigenvectors, $\{v_1,v_2\}$. Then conjugate by the matrix $P$ whose columns are those eigenvectors, to get $A=P\begin{pmatrix}-1&0\\0&5\end{pmatrix}P^{-1}$.