Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question

5 Answers 5

up vote 10 down vote accepted

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}$.

share|cite|improve this answer
There are two problems with this proof. First, you write that since $A=C^3$, every eigenvector of $A$ is clearly an eigenvector of $C$: with no further justification, this is false (for example the cube of the rotation of angle $2\pi/3$ is identity, for which every nonzero vector is an eigenvector). You also write that since $5$ is an eigenvalue of $A$ and $\sqrt[3]{5}$ is a cube root of $5$, $\sqrt[3]{5}$ must be an eigenvalue of $C$, and the same for $-1$: with no further justification, this is false (for one thing, what disqualifies the other cube roots?). – Did Mar 29 '11 at 5:54
@Didier: you are correct, I did not explain that these statements hold true given the type of the specific matrix we have (I thought it was trivial, because the matrix is diagonalizable, though I am now not sure if that's a sufficient condition). – InterestedGuest Mar 29 '11 at 5:59
That $A$ is diagonalizable is (true and) irrelevant. – Did Mar 29 '11 at 6:30
@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? – InterestedGuest Mar 29 '11 at 7:02

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 $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$$

share|cite|improve this answer
Is the answer unique over $\mathbb{R}$? – Fixee Mar 18 '11 at 7:25
@Fixee: I think it should be since eigenbasis is unique, and so are the corresponding eigenvalues. – InterestedGuest Mar 18 '11 at 7:29
Interesting, so Meyer's method above (and others) shows that there is always a cube root over $\mathbb{R}$ for any symmetric matrix. – Fixee Mar 18 '11 at 7:37
@Fixee: in fact any odd root, and perhaps even, too (though one has to be careful, not sure if that would be unique). – InterestedGuest Mar 18 '11 at 7:53
@Fixee: The answer isn't unique in general. E.g., $\begin{bmatrix}0&1\\-1&-1\end{bmatrix}^3=I$. A real symmetric cube root of a real symmetric matrix is unique. – Jonas Meyer Mar 18 '11 at 7:53

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.

share|cite|improve this answer
typo: you want $v$ not $u$ for the second vector – Fixee Mar 18 '11 at 7:19
@Fixee Thanks. Fixe(e)d. – Did Mar 18 '11 at 7:58

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 $\left[ \matrix{\lambda & 0 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$.

share|cite|improve this answer

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


share|cite|improve this answer
I can't comment on Jonas's post, so to rephrase: a symmetric positive definite matrix is always guaranteed to have a unique symmetric even root. Symmetric odd roots are always possible. – Juan Joder Mar 18 '11 at 8:07
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 $I$. – Jonas Meyer Mar 19 '11 at 6:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.