Prove that a positive operator is invertible Help guys, I need to prove this:
Let $(V,\langle~,~ \rangle)$ a finite n-dimensional euclidean space.Let $T$ be a linear operator defined positive (There exists a non singular operator $S$ such that $T=S^*S$) on $V$, prove that $T$ is invertible
I tried this:
We know by hypothesis that $S$ is ivertible,then there exists $S^{-1}$ such that $SS^{-1}=I$ $S$ then I need to prove that $T=T^*$
But that is trivial, since $T=S^*S=(S^*S)^*=S^*(S^*)^*=S^*S=T$
Is that correct? I need help!
 A: Choose a non-zero eigenvector $v$ for $T$ with eigenvalue $\lambda$. It suffices to show that $\lambda > 0$ (note $0$ being an eigenvalue is equivalent to $T$ not being invertible). Now
$$
\langle Sv,Sv \rangle = \langle S^*Sv, v \rangle = \langle Tv,v \rangle = \langle \lambda v,v \rangle = \lambda \langle v, v \rangle.
$$
Since $S$ is non-singular $\langle Sv,Sv \rangle >0$, so we can conclude that $\lambda > 0$.
A: You should do like this:
$$\langle S(v),S(v) \rangle = \langle S^*S(v),v \rangle$$
$$=\langle T(v),v \rangle$$
$$=\langle \lambda v,v \rangle$$
$$=\lambda \langle v,v \rangle$$
where $v$ is non-zero eigen vector for $T$ and $\lambda$ is eigen value.
But $\langle v,v \rangle$ and $\langle S(v),S(v) \rangle$ are positive ;Therefore, $\lambda$ is positive. And you are done.
A: We want to show that the operator $T = S^*S$ is invertible. To show that this holds, it suffices to note that $(AB)^{-1} = B^{-1}A^{-1}$ is generally true whenever $A$ and $B$ are invertible.  Thus, because $T = S^*S$, we have
$$
T^{-1} = [S^*S]^{-1} = S^{-1} (S^*)^{-1} = S^{-1}(S^{-1})^*.
$$
So, $T$ has an inverse and is therefore invertible.
A: $T$ is invertible if $Tx=0$ implies $x=0$. In this case, $T=S^*S$ where $S$ is non-singular. Therefore, if $Tx=0$, it follows that
$$
     0 = \langle Tx,x\rangle=\langle S^*Sx,x\rangle=\langle Sx,Sx\rangle=\|Sx\|^2 \implies Sx=0 \implies x=0.
$$
So $T$ is invertible.
A: In addition to what others have said, you could just use a determinant argument
$$\det T = \det S^* \cdot \det S = \overline{\det S} \det S = |\det S|^2 > 0$$
Though as others have said, technically this is a positive-definite operator.
