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I've the following exercise which I can't solve:

Prove that:

$$ AA^* + I $$

is invertible for all Matrix $ A $ in finite-dimensional field $V$ with inner product. $ A^* $ is the adjoint operator.

Any help will be appreciated.

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2  
But it is.${{}}$ –  Git Gud Jul 7 at 13:21
    
Oops, my mistake. It should is invertible, I fixed the title. –  danny11 Jul 7 at 13:25
1  
How is the adjoint defined for finite fields, where there isn't really an inner product? –  Nishant Jul 7 at 13:26
    
V is inner product space. –  danny11 Jul 7 at 13:29
    
Simple question always gets many answers LoL –  Mr.T Jul 7 at 13:46

4 Answers 4

up vote 3 down vote accepted

$<AA^*v+v,v>=<AA^*v,v>+<v,v>=<A^*v,A^*v>+<v,v>$

this is true from the property of inner product and definition of $A^*$.

Let's call $A^*v$ in another name, let's call it the vector $\alpha$.

So

$<AA^*v+v,v>=<A^*v,A^*v>+<v,v>=<\alpha,\alpha>+<v,v>=\|\alpha\|+\|v\|$

if $v \neq 0$, then this term is larger than zero. Which means $AA^*+I$ is positive definite, which means it only has positive eigenvalues, which means 0 isn't an eigenvalue, which means it is invertible.

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Assume that there's a $v\in V$ such that $AA^*v=-v$. Can you use the definition of the adjoint to conclude that $v=0$?

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If $ A^*A + I $ is not invertible then $ det(A^*A + I)=0 $ then $-1$ is an eigenvalue of $ A^*A $ then there exist nonzero $v$ such that $A^*Av=-v$ therefore $$ v^*A^*Av=-v^*v$$ which means nonsense because norm is not negative.

Then $ A^*A + I $ is always invertible.

Why does $v$ always imply column-vector and $v^*$ always imply row-vector?

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Assuming that there is an inner product in $V$. Then for every $x \in V, x \neq 0$, we have $$x^*(AA^*+I)x = \underbrace{(A^*x)^*(A^*x)}_{\geq 0}+\underbrace{x^*x}_{>0}$$ So $AA^*+I$ is positive definite and thus invertible.

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