Square root of transpose and transpose of square root of a matrix.
$B^{\frac{1}{2}^T}=B^T{^{\frac{1}{2}}} $ (This is not true in general.)
where $B$ is symmetric.
But what are the instances where this would hold?
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But if $B$ is symmetric, then $B^T=B$. – TonyK Jun 20 '19 at 10:09
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I don't know how exactly you define the square root of $B$, but if $A$ is a square root of $B$, then $A^2=B$. Transposing that equation gives $(A^T)^2 = B^T$, thus $A^T$ is a square root of $B^T$. Whether it is the square root of $B^T$ depends on how you select the square root among all possible square roots of $B^T$.
celtschk
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"Whether it is the square root of $B^T$ depends on how you select the square root among all possible square roots of B." This is it. Thank You – Abhiram V P Jun 20 '19 at 11:53
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@JohnHughes: Actually it was demonstrably not perfect, since it had a typo (twice; now fixed) that even made it incorrect. But thank you. – celtschk Jun 20 '19 at 12:44
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Let $A\in M_n(\mathbb{C})$ which has no eigenvalues in $]\infty ,0]$; in particular, $A$ is invertible. Let $\log$ be the principal value of the logarithm, that is,
$\log: re^{i\theta}\in\mathbb{C}\setminus (]-\infty,0])\mapsto \log(r)+i\theta$, where $\theta\in ]-\pi,\pi[$.
We extend the definition as follows (cf. [1], Matrix of functions, Higham):
if the Jordan form of $A$ is $A=PUP^{-1}$ with $U=diag(\lambda_kI_{i_k}+J_{i_k})$ ($J_k$ being the nilpotent Jordan form of dimension $k$), then
$\log(A)=Pdiag(\log(\lambda_kI_{i_k}+J_{i_k}))P^{-1}$ with $\log(\lambda_kI_{i_k}+J_{i_k}))=\log(\lambda_k)+\log(I_{i_k}+1/\lambda_kJ_{i_k})$.
Finally $A^{1/2}=e^{1/2\log(A)}$ is the unique square root whose spectrum lies in the open right plane.
$\textbf{Proposition}$. With the above defintion, ${A^{1/2}}^T={A^T}^{1/2}$.
$\textbf{Proof}$. One has $A^{1/2}=\dfrac{2}{\pi} A\int_0^{+\infty}(t^2I+A)^{-1}dt$ ([1] p. 133). Then ${A^{1/2}}^T=\dfrac{2}{\pi} \int_0^{+\infty}(t^2I+A^T)^{-1}A^Tdt={A^T}^{1/2}$ because the integral is a polynomial in $A^T$. $\square$
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This is for A positive definite? For positive definite A, the square root is unique and is symmetric. So this is that proof? Thank you – Abhiram V P Jun 21 '19 at 04:24
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According to the definition, the result is valid for any complex matrices that have no eigenvalues in $]-\infty,0]$. If $A$ is real, then $A^{1/2}$ is real. On the contrary, $diag(-1,1)$ admits no real square roots. – Jun 21 '19 at 11:20
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