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Let $A_{n\times n} = aI+bJ$, where $I$ is the identity matrix and $J$ is the matrix of all ones. Is it possible to find the expression of $A^{1/2}$ such that $A^{1/2}A^{1/2} = A$? In particular $A = I_{n\times n} - \frac{(1-\alpha)}{n+\alpha(2-n)}J_{n\times n}$, where $0<\alpha<1$.

Satya Prakash
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  • There have been a number of previous Questions about the eigenvalues and eigenvectors of $A$, which is evidently real and symmetric. Provided $a,b$ are such that $A $ is positive definite, it is possible to take a matrix "square root" but the result is not unique. – hardmath Jan 16 '18 at 21:45
  • I have given the particular form of $A$. Any suggestions? – Satya Prakash Jan 16 '18 at 22:17

2 Answers2

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Here is a quick and easy way to get a matrix $A^{1/2}$, provided one is willing to limit oneself to solutions of the form

$C = \alpha I + \beta J; \tag 1$

then

$C^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 J^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 n J = \alpha^2 I + (2 \alpha \beta + n \beta^2)J, \tag 2$

where we have used the fact that $J^2 = nJ$, which is easily verified; with

$A = a I + bJ \tag 3$

and

$C^2 = A, \tag 4$

since $I$ and $J$ are linearly independent over $\Bbb C$, we must take

$\alpha^2 = a \tag 5$

and

$2 \alpha \beta + n \beta^2 = b; \tag 6$

then

$\alpha = \pm \sqrt a, \tag 7$

and thus (6) becomes a quadratic equation in $\beta$ whose coefficients are known:

$n\beta^2 + 2 \alpha \beta - b = 0; \tag 8$

enter the quadratic formula:

$\beta = \dfrac{-2\alpha \pm \sqrt{4 \alpha^2 + 4nb}}{2n}; \tag 9$

using (5), we simplify

$\beta = \dfrac{-2\alpha \pm \sqrt{4 a + 4nb}}{2n} = \dfrac{-2\alpha \pm 2\sqrt{a + nb}}{2n} = \dfrac{-\alpha \pm I\sqrt{a + nb}}{n}; \tag{10}$

with $\alpha$ and $\beta$ given by (5) and (10) we take

$A^{1/2} = C. \tag{11}$

Note: It is not yet clear to me whether or not all solutions are of this form. Further investigation required. End of Note.

Robert Lewis
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    +1 [It is not yet clear to me whether or not all solutions are of this form. ]---> I think not as $A$ itself satisfies a quadratic polynomial. – Duchamp Gérard H. E. Jan 17 '18 at 01:49
  • $A^2 - 2a A + a^2 I = (A - aI)^2 = b^2 J^2 = n b^2 J = nb (bJ) = nb (A - aI) \Longrightarrow A^2 - (2a + nb) A + (a^2 + nab)I = 0$ – Robert Lewis Jan 17 '18 at 02:05
  • Right ! It factors as $(A-(a+nb))(A-a)$ which is not the same thing (neither stronger nor weaker) than saying $sp(A)=[a+nb]+(n-1)[a]$ (spectrum, see other answer). – Duchamp Gérard H. E. Jan 17 '18 at 07:11
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The $n\times n$ matrix $A$ has a real symmetric square root $A^{1/2}$ if and only if the eigenvalues of $A$ are nonnegative. The eigenvalues of $J$ have been found in a number of previous Questions, namely $n$ is an eigenvalue of multiplicity one and $0$ as an eigenvalue has multiplicity $n-1$.

Therefore $A = aI+bJ$ has eigenvalues $a+bn$ with multiplicity one and $a$ with multiplicity $n-1$.

So for $A$ to be positive semi-definite (i.e. have nonnegative eigenvalues) it is required that $a \ge 0$ and $a+bn \ge 0$. This is indeed true for the case "in particular" in the revised Question, where:

$$ a = 1\; , \; b = -\frac{(1-\alpha)}{n+\alpha(2-n)} $$

given that $0\lt \alpha \lt 1$. When we simplify:

$$ a+bn = \frac{2\alpha}{2\alpha + n(1-\alpha)} $$

Assuming this, we can deduce the existence of positive semi-definite symmetric $A^{1/2}$ and explicitly construct one by taking an orthonormal basis of eigenvectors for $A$ to diagonalize it:

$$ D = Q^T A Q $$

where $Q$ has columns that are the orthonormalized eigenvectors and $D = \operatorname{diag}(a+bn,a,\ldots,a)$, so that:

$$ A^{1/2} = Q D^{1/2} Q^T$$

Here we mean to take square roots of the diagonal entries of $D$ to produce $D^{1/2}$, but there are additional matrices $B$ that satisfy the relationship $B^2 = A$. For example, $B = -A^{1/2}$, but for $n\gt 1$ there are further possible matrix "square roots".

hardmath
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