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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 $\sqrt{A} = \sqrt{B} \implies A = B$.

Rodrigo de Azevedo
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p3ngu1n
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  • If $f(X)$ is a matrix such that $f(X)\cdot f(X)=X$, then of course $f(X)=f(Y)$ implies $X=Y$! They're both $f(X)\cdot f(X)$. –  Oct 20 '19 at 21:06
  • Are you confusing yourself with the converse? A square root of a matrix is not necessarily unique. However, for a positive definite hermitian matrix, it has a unique positive definite Hermitian square root (but it may have non-positive definite Hermitian square root). – Batominovski Oct 20 '19 at 21:08
  • Here is a previously answered Question about the uniqueness of a positive definite (symmetric) square root of a positive definite (symmetric) real matrix. The title and body of your Question state a converse implication to what you really want to show a square root operator is defined on positive definite (symmetric) real matrices. – hardmath Oct 20 '19 at 21:17

2 Answers2

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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 $C$.

N. S.
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Take advantage of the fact that your eigenvalues are real and non-negative then the one-to-one property follows

Mohammad Riazi-Kermani
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