Jtdylktuy

Prove that there exists $delta > 0$ so that for all square matrices $Ain mathbb{R}^{ntimes n}$ with $|A-I| < delta$ (where $I$ denotes the identity matrix) there exists $Bin mathbb{R}^{ntimes n}$ so that $B^2=A$.

My attempt so far:

$$A-I= begin{bmatrix}
a_{11}-1 & a_{12} & ... & a_{1n} \
a_{21} & a_{22}-1 & ... & a_{2n} \
vdots \
a_{n1} & a_{n2} & ... & a_{nn}-1 \
end{bmatrix} $$
Taking $x=(1,0,...,0)in mathbb{R}^n$, we have that
$$|A-I|_{op}=underset{|x|=1}{sup}|(A-I)x|= sqrt{(a_{11}-1)^2+...+a_{n1}^2}<delta$$
Intuitively, this seems to suggest that for each basis vector of $mathbb{R}^n$ means that $A-I$ can be made close to the zero matrix, and hence, close to being a diagonal matrix. But I am having trouble going from here. Anyone have any hints?

Let

$I subset U subset Bbb R^{n times n}, tag 1$

where $U$ is open. Then we may define the matrix function

$F(B) = B^2:U to Bbb R^{n times n}, tag 2$

and we note that

$F(I) = I^2 = I, tag 3$

that is, $I$ is in the range of $F(B)$.

$F(B) = B^2$ is in fact differentiable; we have, for $Delta in Bbb R^{n times n}$,

$F(B + Delta) = (B + Delta)^2 = (B + Delta)(B + Delta) = B^2 + BDelta + Delta B + Delta^2, tag 4$

$F(B + Delta) - F(B) - (B Delta + Delta B) = (B + Delta)^2 - B^2 - (BDelta + Delta B) = Delta^2; tag 5$

$Vert F(B + Delta) - F(B) - (B Delta + Delta B) Vert = Vert (B + Delta)^2 - B^2 - (BDelta + Delta B) Vert = Vert Delta^2 Vert le Vert Delta Vert^2; tag 6$

since

$dfrac{Vert Delta Vert^2}{Vert Delta Vert} = Vert Delta Vert to 0 ; text{as} ; Delta to 0, tag 7$

we see that $F(B)$ is differentiable and that its derivative is the linear map

$DF(B)(Delta) = BDelta + Delta B; tag 8$

thus

$DF(I)(Delta) = I Delta + Delta I = 2 Delta, tag 9$

which is clearly non-singular. It follows from the inverse function theorem that there is some neighborhood $V$ of $I$ and a function

$R:V to U subset R^{n times n}, : R(I) = I, tag{10}$

such that for $A in V$

$F(R(A)) = (R(A))^2 = A; tag{11}$

if we now choose $delta > 0$ sufficiently small we may ensure that the set

$B(I, delta) = {A in Bbb R^{n times n}, ; Vert I - A Vert < delta } subset V, tag{12}$

and thus for $A in B(I, delta)$ we may set

$B = R(A), tag{13}$

and we have

$B^2 = (R(A))^2 = F(R(A)) = A, tag{14}$

as desired. $OEDelta$

Hint:

$$
sqrt{x}=sqrt{1-(1-x)}=sum_{k=0}^infty {1/2choose k} (x-1)^k
$$
By the binomial series. Can you use the above taking $x=A$ and proving it is a norm convergent series for a square root of $A$ with your bound?

$x^{T}Axgeq x^{T}x+x^{T}(A-I)x geq (1-delta) x^{T}x geq 0$. Hence $A$ (being non -negative definite) is digonalizable with non-negative diagonal entries. Replace diagonal entries by their square roots to get $B$.