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Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$

I think I am doing something wrong here, so let me describe my attempt:

1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$

2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$

3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!

Have I perhaps overlooked something here or miscalculated the solution for $B$?

Edit: As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$

Rodrigo de Azevedo
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jsonnet
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6 Answers6

7

$$A=\begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}=PDP^{-1}$$

Where $$P=\begin{bmatrix}1& 1\\-1 & 2\end{bmatrix}$$is the matrix of eigenvectors

and $$D=\begin{bmatrix}-8& 0\\0 & 1\end{bmatrix}$$ is the matrix of eigenvalues.

Thus $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$

Mohammad Riazi-Kermani
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How in the world would the product of three real matrices turn out to have non-real entries? Something must have gone wrong. Try rewriting $\sqrt[3]{-8}=-2,$ and see if that makes a difference. It may be that your calculator chose one of the two non-real cube roots of $-8,$ instead, or perhaps you accidentally entered some even root of $-8$.

Cameron Buie
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3

For a $2\times 2$ matrix $\{A\}$ with distinct eigenvalues $\{\lambda_1, \lambda_2\}$, any function $\{f(A)\}$ can be evaluated as a linear polynomial, whose coefficients are determined solely by the eigenvalues $$\eqalign{ c_1 &= \tfrac{f(\lambda_1)-f(\lambda_2)}{\lambda_1-\lambda_2}\cr c_0 &= f(\lambda_1) - c_1\lambda_1 \cr f(A) &= c_1A + c_0I \cr\cr }$$ For the current problem, $B=f(A)=A^{1/3}$ and $(\lambda_1,\lambda_2)=(-8,1)$, therefore $$\eqalign{ f(A) &= \tfrac{1}{3}A + \tfrac{2}{3}I\cr }$$

greg
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  • Are there any elementary (studying operator algebras is a bit too much) links/refs that elaborate on these ideas? / (+1) for 'slickness'. – CopyPasteIt Apr 06 '20 at 04:20
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    @CopyPasteIt As a consequence of Cayley-Hamilton, any analytic function of an $(n\times n)$ matrix can be written as a polynomial of degree $\leq n.;$ Also, any functional relationship which is true for the whole matrix is true for each eigenvalue, e.g.
    $$\eqalign{ \exp(A)=f(A) &= \sum_{k=0}^{n-1} c_k A^k \quad\implies\quad \exp(\lambda_j) &= \sum_{k=0}^{n-1} c_k \lambda_j^k \ }$$ Given the spectrum of $A$, you can solve the EV system for the $c_k$ coefficients which can then be used to evaluate $f(A)$. An EV of multiplicity $=(\mu+1)$ must evaluate derivatives through $(d^\mu f)$.     – greg Apr 06 '20 at 15:47
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    @CopyPasteIt I've always called this technique Sylvester-Hermite Interpolation, but I can't locate any references to that name. – greg Apr 06 '20 at 15:49
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The characteristic polynomial of $A$ is \begin{align} p(\lambda)&= (-5-\lambda)(-2-\lambda)-18\\ &= \lambda^2+7\lambda-8 \\ &= (\lambda +8)(\lambda-1). \end{align} So $(A+8I)(A-I)=0$. That means \begin{align} A(A+8I)&=(A+8I) \\ A(A-I) &= -8(A-I) \\ I &= \frac{1}{9}((A+8I)-(A-I)) \\ A^{1/3}&=\frac{1}{9}((A+8I)+2(A-I)) \\ &= \frac{1}{9}(3A+6I) = \frac{1}{3}(A+2I) \\ &= \frac{1}{3}\begin{pmatrix}-3 & 3 \\ 6 & 0\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 2 & 0 \end{pmatrix}. \end{align}

Disintegrating By Parts
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0

Note: My goal here was to work the problem using triangular matrices.

Let $ A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}. $

The eigenvalues for the matrix $A$ are $1$ and $-8$.

Set

$ \quad R = \begin{bmatrix}-2 & 1 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad S = \begin{bmatrix}-8 & 3 \\ 0 & 1 \end{bmatrix} $

Observe that

$\tag 1 R^3 = S$

Set

$ \quad T = \begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix} $

Observe that

$\tag 2 T^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$

and that

$\tag 3 T\,A\,T^{-1} = S$

Set

$\quad B = T^{-1} \, R \, T$

Using algebra it is easy to see that $B^3 = A$.

Calculating

$ \quad B = \begin{bmatrix}-1 & 1 \\ 2 & 0 \end{bmatrix} $

CopyPasteIt
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-1

Standard computer methods (e.g., Mathematica Solve) gives the answer directly:

myB = {{b11, b12}, {b21, b22}}; 
myB /. Solve[myB.myB.myB == {{-5, 3}, {6, -2}}, {b11, b12, b21, b22}][[1]]

(*

{{-1, 1}, {2, 0}}

*)

$${\bf B} = {-1, 1 \choose 2 , 0}$$.

David G. Stork
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