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Show that for any $B\in Mat(n; \mathbb{R})$ symmetric and positive definite, there exists a unique symmetric positive definite matrix $A\in Mat(n; \mathbb{R})$ such that $A^2 = B^3$ .

$\textbf{Hint}$: any symmetric matrix is diagonalisable.

V. Krumov
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  • Hint: Write eigenvalue decomposition of $B$, and compute $B^3$. Try to find a square root for that. – user25004 Mar 01 '13 at 21:55
  • The matrix $B^3$ is also symmetric positive definite. So it has a unique square root which is symmetric positive definite. That's $A$. – Julien Mar 01 '13 at 21:56
  • @julien? Sorry. What do you mean by unique square root? I do not think any matrix has unique square root. – user25004 Mar 01 '13 at 21:58
  • @user25004, why don't you think so? – Inquest Mar 01 '13 at 21:59
  • @user25004 A symmetric positive definite (resp semidefinite) matrix has a unique symmetric positive definite (resp semidefinite) square root. That's what is meant in my comment. – Julien Mar 01 '13 at 22:00
  • Because if $C$ is a square root, at least $-C$ would be another. – user25004 Mar 01 '13 at 22:00
  • Now, I see what you are saying. The PSD one can be unique. – user25004 Mar 01 '13 at 22:01
  • @user25004 Yes, $2$ and $-2$ are two square roots of $4$... – Julien Mar 01 '13 at 22:01
  • @user25004 If $A$ is a symmetric square root of $B$, it commutes with $B$ and you can diagonalize then simulataneously in an orthonormal basis. Then adding the condition positive forces the choice of the eigenvalues of $A$. My english was note very good in my first comment. I should have said: there is a unique PSD square root. – Julien Mar 01 '13 at 22:05

2 Answers2

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Rough Structure of Proof: \begin{align} B^3&=QS^{3}Q^T\\&=QS^{3/2}S^{3/2}Q^T\\&=QS^{3/2}Q^TQS^{3/2}Q^T\\&=A^2 \end{align}

Inquest
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  • Would this show uniqueness? – EuYu Mar 01 '13 at 21:56
  • I think so because every positive definite matrix has one and only one square root called the principal square root. – Inquest Mar 01 '13 at 21:57
  • Well, I'm aware of the principal square root, but I'm wondering whether we can take this fact for granted here when the question asks for a proof. – EuYu Mar 01 '13 at 22:00
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$(i)$ If $\rm B$ is a scalar in $\mathbb R^{\ast}_+$, then the polynomial $\rm X^2 - \rm B^3$ has a unique non negative real root, call it $\rm A$ ;

$(ii)$ If $\rm B$ is a positive definite diagonal matrix $\rm B = \text{ Diag }(B_1, ... , B_n)$, then follow $(i)$ for every $\rm B_i$ and set $\rm A = \text{ Diag }(\rm A_1,...,\rm A_n)$ ;

$(iii)$ if $\rm B$ is a symmetric definite positive matrix, then I can diagonalize it, so there exists a positive definite diagonal matrix $\rm D$ and an invertible matrix $\rm P$ such that $\rm B = P D P^{-1}$. Then apply $(ii)$ to $\rm D$ which gives me a diagonal matrix $\rm E$. Then set $\rm A = P E P^{-1}$.

Damien L
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