# Showing that $\sqrt{A} = \sqrt{B} \implies A = B$, where $A$ and $B$ are real, symmetric, positive definite matrices

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

• 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 $f(X)\cdot f(X)$.
– user239203
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). 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. Oct 20 '19 at 21:17

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 $$C$$.

• Yes - I am asking how to ensure that my definition of $\sqrt{A}$ is unique. I am not sure how to verify this. I will edit my question Oct 20 '19 at 21:12
• @p3ngu1n See my edit. Oct 20 '19 at 21:19

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