Long story short, I've been asked to define a square root operator on the set of real, symmetric, positive definite matrices. I am not sure how to show my operator is 1-1.
Let $A$ be such a matrix. This means $A$ can be diagonalized into $Q \Lambda Q^{-1}$, where $Q$ is an orthogonal matrix whose columns are the eigenvectors of A, and $\Lambda$ is a diagonal matrix whose entries are the eigenvalues of $A$, $(\lambda_1,\lambda_2,\dots,\lambda_n)$.
I've defined $\sqrt{A}$ = $Q \Lambda^{1/2} Q^{-1}$, where $\Lambda^{1/2}$ is a diagonal matrix whose entries are
$$\left( \sqrt{\lambda_1},\sqrt{\lambda_2},\dots,\sqrt{\lambda_n} \right)$$
Intuitively, this should be unique, but I am not sure how to verify/prove this. I don't know what other conditions I need to include to get from $\sqrt{A} = \sqrt{B} \implies A = B$.
Best Answer
Hint If $\sqrt{A}=\sqrt{B}$ then $$A= (\sqrt{A})^2=(\sqrt{B})^2=B$$
The part where you need to be carefull is not the 1-1, you ahve to make sure your operator is well defined, i.e. for each $A$ your definition of $\sqrt{A}$ is unique.
Edit To show that this is unique.
If $B,C$ are symmetric, positive definite square roots of $A$, then prove that $B,C$ have the same eigenvalues, and the same eigenspaces.
You can do this by arguing that if $(\lambda,u)$ are eigenvalue/eigenvector for $B$ then $(\lambda^2, u)$ are eigenvalue/eigenvector for $A$. Also since $B$ is symmetric, the sum of dimensions of eigenspaces is $n$, the size of the matrix.
Do then exactly the same thing for $C$.