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This is a question I had on a test, we were told not to use brute-force and figure out a smart way to solve the problem.

We have a matrix $A$ $\begin{pmatrix} 2 & 3\\ 3 & 2 \end{pmatrix}$. Find a matrix $C$ such that $C^{3}$=$A$.

What is the 'smart not brute-force' way to solve this, without picking numbers, looking for patterns and so on?

it was in eigenvalues section" in the end?

5 Answers 5

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One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det $ (\begin{bmatrix} 2-\lambda & 3\\ 3 & 2-\lambda \end{bmatrix})$= $(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respective eigenspaces.

Basis $\xi_{-1}$ = ker$(A+I)$=ker $(\begin{bmatrix} 3 & 3\\\ 3 & 3 \end{bmatrix})$. Row-reducing, you get ker $(\begin{bmatrix} 1 & 1\\\ 0 & 0 \end{bmatrix})$, which is equal to the span of $\begin{bmatrix} -1\\ 1 \end{bmatrix}$. So a basis of $\xi_{-1}$ = $\begin{bmatrix} -1\\ 1 \end{bmatrix}$.

Same for $\xi_{5}$ -- ker $(\begin{bmatrix} -3 & 3\\\ 3 & -3 \end{bmatrix})$, row-reducing we get ker $(\begin{bmatrix} 1 & -1\\\ 0 & 0 \end{bmatrix})$, so ker = span $\begin{bmatrix} 1\\ 1 \end{bmatrix}$. So a basis of $\xi_{5}$ = $\begin{bmatrix} 1\\ 1 \end{bmatrix}$.

Now you know that $A$=$C^{3}$=$CCC$. So if $\vec{x}$ is an eigenvector of $A$, it is clearly an eigenvector of $C$, but perhaps with a different eigenvalue. You know that $A$ \begin{bmatrix} 1\\ 1 \end{bmatrix}$=$CCC$ \begin{bmatrix} 1\\ 1 \end{bmatrix}$=$\begin{bmatrix} 5\\ 5 \end{bmatrix}$. So, $\sqrt[3]{5}$ must be an eigenvalue of $C$.

Same goes for $A$\begin{bmatrix} -1\\ 1 \end{bmatrix}$=$CCC$\begin{bmatrix} -1\\ 1 \end{bmatrix}$=$\begin{bmatrix} 1\\ -1 \end{bmatrix}$. This would be possible only if -1 was an eigenvalue of $C$, since $(-1)^{3}$=$-1$.

So now you can construct a system of equations -- you know that $C$ is of some form $(\begin{bmatrix} a & b\\\ c & d \end{bmatrix})$, and based on the eigenvectors and eigenvalues of $C$ that we just found out, you can devise the following system:

$\left\{\begin{matrix} a & + & b & = & \sqrt[3]{5}\\ c & + & d & = & \sqrt[3]{5}\\ -a & + & b & = & 1\\ -c & + & d & = & -1 \end{matrix}\right.$

Now solve for the variables -- $b=a+1$, substitute into the other $a, b$ equation to get $2a+1=\sqrt[3]{5}$, so $a=\frac{\sqrt[3]{5}-1}{2}$. $b$, then, is equal to $\frac{\sqrt[3]{5}-1}{2}+1$, as obvious from the third equation. Same for $c$ and $d$ -- $c=d+1$, so substituting, we get $2d+1=\sqrt[3]{5}$, so $d=\frac{\sqrt[3]{5}-1}{2}$, and hence $c=\frac{\sqrt[3]{5}-1}{2}+1$.

This results into $C$=$ \begin{bmatrix} \frac{\sqrt[3]{5}-1}{2} & \frac{\sqrt[3]{5}-1}{2}+1 \\ \frac{\sqrt[3]{5}-1}{2}+1 & \frac{\sqrt[3]{5}-1}{2} \end{bmatrix}$.

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    @Didier: Yeah there are projection matrices that are diagonalizable yet I would not be able to apply the same reasoning that I applied in my answer. What would be a sufficient condition? Perhaps diagonalizable and does not have an eigenvalue of 0 (so isn't a projection) would be sufficient?2011-03-29
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This would be easy for a diagonal matrix, because $\begin{bmatrix}a & 0\\ 0 & b\end{bmatrix}^3=\begin{bmatrix}a^3 & 0\\ 0 & b^3\end{bmatrix}$, which means that you could just take the cube root of each diagonal entry to solve the problem. While $A$ is not diagonal, it is symmetric and therefore diagonalizable. If you're comfortable with diagonalizing, find $S$ such that $SAS^{-1}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}$. You know how to find a matrix whose cube gives the right hand side. Then notice how conjugation behaves with cubing: (S^{-1}C'S)^3=S^{-1}C'^3S. Therefore, you can take $C=S^{-1}\begin{bmatrix}\sqrt[3]{a} & 0 \\ 0 & \sqrt[3]{b} \end{bmatrix}S$.

(I was going to write more involving eigenvectors, but then other answers were posted covering this.)

An alternative approach using polynomial interpolation will work for all diagonalizable matrices having eigenvalues $-1$ and $5$, and does not require finding eigenvectors. For more on this and generalizations, see Chapter 1 of Higham's Functions of matrices.

In this case, the Lagrange interpolating polynomial of the cube root function on the spectrum $\{5,-1\}$ of $A$ is $p(t)=\sqrt[3]{5}\cdot\frac{t+1}{5+1}+\sqrt[3]{-1}\cdot\frac{t-5}{-1-5}=\frac{\sqrt[3]{5}+1}{6}\cdot t +\frac{\sqrt[3]{5}-5}{6},$ so that a matrix cube root for $A$ can be obtained as $p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$

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    @Fixee: The eigenvalues in that case are $2a+1$ and $-1$, with respective eigenspaces $x=y$ and $x=-y$, and $S$ can always be taken to be \frac{1}{\sqrt 2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}, in which case $S=S^{-1}$, and the cube root found in the above manner is \frac{1}{2}\begin{bmatrix}\sqrt[3]{2a+1}-1&\sqrt[3]{2a+1}+1\\\sqrt[3]{2a+1}+1&\sqrt[3]{2a+1}-1\end{bmatrix}. So, no.2011-03-21
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Remember that:

Matrices act on vectors.

Here the linearly independent vectors $u=\begin{pmatrix} 1\\ 1\end{pmatrix}$ and $v=\begin{pmatrix} 1\\ -1\end{pmatrix}$ are such that $Au=5u$ and $Av=-v$.

Hence a suitable $C=\begin{pmatrix} a & b\\ c& d\end{pmatrix}$ could be defined by the conditions that $Cu=5^{1/3}u$ and $Cv=-v$. This gives you a linear system of two equations for $(a,b)$ and another linear system of two equations for $(c,d)$, and the matrix $C$ follows.

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    @Fixee Thanks. Fixe(e)d.2011-03-18
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In fact, the only 2 x 2 matrices that do not have cube roots (over the complex numbers) are those with Jordan canonical form $\left[ \matrix{0 & 1\cr 0 & 0\cr}\right]$.
The 3 x 3 matrices with no cube root are those with Jordan form $\left[ \matrix{0 & 1 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$ or $\left[ \matrix{\lambda & 0 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$.

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Eigendecompose $\mathbf A$ (easily done since you have a symmetric matrix), take the cube root of the eigenvalues, and multiply back the matrix of eigenvectors appropriately.

I get

$\frac12\begin{pmatrix}\sqrt[3]{5}-1&\sqrt[3]{5}+1\\\sqrt[3]{5}+1&\sqrt[3]{5}-1\end{pmatrix}$

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    You mean "guaranteed to have a unique symmetric *positive* even root." Just being symmetric doesn't make it unique. For example, \begin{bmatrix}-1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&0\\0&-1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\-1&0\end{bmatrix} and $-I$ are among the square roots of $I$.2011-03-19