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For a positive matrix $A$. Here, we assume that all positive matrices are self-adjoint. Show that \begin{align} A + A^{-1} \geq 2I. \end{align}

Here, $A≥0$ means that A is self-adjoint and for all $x∈ℂn,⟨Ax,x⟩≥0.$

DRich
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    What does $A\ge B$ mean for matrices? Is it entrywise? – user2345215 Apr 12 '14 at 21:34
  • It means that $a_{ij}\ge b_{ij }$ for all $i,j$ – Vladhagen Apr 12 '14 at 21:35
  • It probably means that $\forall i,j, \quad A_{i,j}=B_{i,j}$ – Jack Apr 12 '14 at 21:36
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    Assume positive means positive definite symmetric matrix, then $A + A^{-1} \ge 2I$ because $(A + A^{-1}) - 2I = (A^{1/2} - A^{-1/2})^2$ is the square of a symmetric matrix. – achille hui Apr 12 '14 at 21:37
  • $A \ge B$ means that $A - B$ is positive semi-definite, i.e. that $\langle x, (A - B)x \rangle \ge 0$ for all vectors $x$. – Robert Lewis Apr 12 '14 at 21:38
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    In my experience matrix positivy is defined entry wise. – Git Gud Apr 12 '14 at 21:39
  • Why do you assume that positive matrices are self adjoint? – Vladhagen Apr 12 '14 at 21:40
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    @GitGud That's far from standard, though: I've often seen $A>0$ for $A$ positive-definite. – user7530 Apr 12 '14 at 21:40
  • @user7530 I have, especially in TeX context, seen $$A \succ 0$$ for positive definity and $$A > 0$$ for entrywise positivity. – AlexR Apr 12 '14 at 21:41
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    In Roger Horn's book Matrix Analysis, he defines a positive matrix exactly as I have above: entrywise. He also defines the ordering exactly as I have: entrywise. That is all there is to it. – Vladhagen Apr 12 '14 at 21:42
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    @Vladhagen: all there is to it from Roger Horn's point of view! – Robert Lewis Apr 12 '14 at 21:43
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    The statement is false if $A > 0$ means its entries are positive. eg. take $A$ as $\begin{pmatrix}1&2\2&1\end{pmatrix}$. – achille hui Apr 12 '14 at 21:47
  • @achillehui: Right! – Robert Lewis Apr 12 '14 at 21:49
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    Wow. Let this get away from me. Thanks for all the hearty debate, but here, $A \geq 0$ means that $A$ is self-adjoint and for all $x \in \mathbb{C}^{n}, ; \langle Ax,x \rangle \geq 0$.$ – DRich Apr 12 '14 at 23:06
  • @DRich: Your last comment is a very clear statement of what you mean by $A \ge 0$. If you edited your question so that it was that clear, I suspect you might increase your chances of it being re-opened, which incidentally is something I voted for. And for the question itself. So best of luck with this; no promises however: I can't speak for/control my colleagues here on MSE. Regards, RKL. – Robert Lewis Apr 13 '14 at 21:30
  • It seems to have been effectively answered by @voldemort. I will, however edit it so it is easy to read as stated. – DRich Apr 15 '14 at 00:04

3 Answers3

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$A \geq 0$ means $\langle Ax,x \rangle \geq 0$. $A \geq B$ means $A-B \geq 0$ which translates to $\langle Ax,x \rangle \geq \langle Bx,x \rangle$.

Now for your problem we need to show $A+A^{-1} \geq 2I$.

So, we need to show $\langle A+A^{-1}x,x \rangle \geq 2 \langle x,x\rangle =2 \lVert x\rVert^2$.

Now, $A$ ,$A^{-1}$ are self adjoint, and commute, and hence can be simultaneously diagonalised.

So, let $A=diag(\lambda_1,...\lambda_n)$. Then, $A^{-1}=diag(1/\lambda_1,...1/\lambda_n)$. Now, $\lambda_i$ is postive, as $A$ is a positive matrix. For a postive number $a$ we always hve $a+1/a \geq 2$. Hence your result follows.

kahen
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voldemort
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Consider a diagonalization of $A$.

This is also a diagonalization of $A + A^{-1}$. You are then reduced to show that $$ a + \frac 1{a} \ge 2 $$for each (potential eigenvalue) $a>0$.

The function $$ f(a) = a + \frac 1{a} $$ is minimum when $a=1$ and then the minimum is $2$.

mookid
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$$A+A^{-1} -2 I =A^{-1}(A^2+ I -2A)= A^{-1}(A-I)^2 $$ and both $A^{-1}$ and $(A-I)^2$ are positive-definite.

themaker
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