Proof that Every Positive Operator on V has a Unique Positive Square Root

Question

Suppose $V$ is a finite-dimensional, nonzero, inner-product space over $\Bbb{F}$, and $\Bbb{F}$ denotes $\Bbb{R}$ or $\Bbb{C}$.

My thought is : suppose $T$ is a positive operator; thus, $T$ is self-adjoint. Every self-adjoint operator on $V$ has a diagonal matrix with respect to some orthonormal basis of $V$.

But this doesn't tell me that $T$ has distinct eigenvalues. It tells me that $V$ has an orthonormal basis consisting of eigenvectors of $V$, of course, they are linear independent, but it doesn't tell me each vector from the basis has a unique eigenvalue.

It seems that, without distinct eigenvalues, I can't prove the uniqueness of positive square root.

Answer 1

To prove existence, we note that the normality of $T$ implies that $V$ has an orthonormal basis, $\beta = \{v_{1}, ..., v_{n} \}$, of eigenvectors of $T$ with corresponding eigenvalues $\lambda_1,...,\lambda_n$, permitting a unique decomposition: $V = \bigoplus_{i=1}^{m}E_{\lambda_{i}}$. Note that $n \ge m$ and so some of the $\lambda_i$'s may be equal.

That $T$ is self-adjoint implies that $\lambda_{i} \in \mathbb{R}$. That $T$ is positive definite implies that $\lambda_{i} \geq 0$. Now define $S(v_i) = \sqrt{\lambda_{i}}v_{i}$. Check that $S$ is a positive linear operator and that $S^{2}(v_i) = T(v_i)$. Since every $v \in V$ is written uniquely as $v = \sum_{i=1}^{n} a_i v_{i}$ then this will imply that $S^2=T$ on $V$.

To prove uniqueness, suppose that there exists another $S'$ such that $S'$ satisfies the same conditions as $S$, i.e., $S'^2=T$ on $V$. It suffices to show that $S$ and $S'$ agree on an arbitrary vector element in a basis.

Since $S'$ is positive, $V$ has a basis of orthonormal vectors consisting of eigen-vectors of $S'$. Let $u_{j}$ be any element in this basis, then $Tu_{j} =S'^2(u_j)=S'(S'(u_{j})) = \alpha_{j}^{2}u_{j}$. So that $u_{j}$ is an eigenvector of $T$ and by our construction it is also an eigenvector of $S$. Suppose then $Su_{j} = \gamma_{j} u_{j}$. Then $T(u_j)=S^{2}u_{j} = \gamma_{j}^{2}u_{j}$. Therefore, $\gamma_{j}^{2}u_{j} = \alpha_{j}^{2}u_{j} \implies \gamma_{j}^{2}=\alpha_{j}^{2} \implies \gamma_{j} = \alpha_{j}$ by non negativity of these eigenvalues. Therefore $S(u_{j}) = S'(u_{j})$ and hence they two operators agree on V and so they are identical.

Answer 2

Hint: first consider the case of the identity operator.

Answer 3

If $Te_i=\lambda_i e_i$, take $B(e_i)=\sqrt{\lambda_i}e_i$. Clearly $B^2=T$ and it is non-negative. Let's suppose there is another operator $\tilde{B}$ satisfying this. If $v\in E_{\lambda}$ (the $\lambda$-eigenspace), then $T\tilde{B}(v)=\tilde{B}^3(v)=\tilde{B}T(v)=\lambda_i\tilde{B}(v)$. In this case, the restriction $\tilde{B}|_{E_{\lambda}}:E_{\lambda}\rightarrow E_{\lambda}$ becomes diagonalizable (self-adjoint) with basis $\{v_i|i=1,...,n\}$. Therefore, since $\tilde{B}^2(v_i)=\gamma_i^2 v_i=T(v_i)=\lambda v_i$, we have no choice but taking$\gamma_i=\sqrt{\lambda}$, we obtain $\tilde{B}|_{E{\lambda}}=\sqrt{\lambda} I$. In particular, $\tilde{B}(e_i)=\sqrt{\lambda_i} e_i=B(e_i)$ if $e_i \in E_{\lambda}$. This holds for every $e_i$ in $E_\lambda$ and also in every eigenspace. Hence, they coincide in the entire vector space and $B=\tilde{B}$.