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

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.

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.

• I think your third line is to read $$"...v\in V\;\;\text{is written uniquely as}\;\;v=\sum_{i=1}^n\color{red}{a_i}v_i\;\;,\;\;a_i\in\Bbb F\;"$$ or else to put $$v=\sum_{i=1}^n w_i\;\;,\;\;w_i\in E_{\lambda_i}$$ and changing also definition of $\;S\;$ .
– user177692
Jun 10, 2015 at 9:26

Hint: first consider the case of the identity operator.

• you mean that identity operator (not having distinc eigenvalues) still enable to have unique square root ? But how do I prove that in general? Aug 21, 2014 at 6:31
• Can I just simply say: suppose S is a positive square root of T. If Tv = av for some scalar a, then $S=\sqrt{T}$, therefore, the operator S is just multiplication by $\sqrt{a}$. So S is unique. Aug 21, 2014 at 6:43
• the uniqueness comes from the square root. I avoid the distinct eigenvalue problem Aug 21, 2014 at 6:45
• The proof is not that trivial. Aug 21, 2014 at 6:51
• do u mind posting a complete proof ? thanks so much Aug 21, 2014 at 6:52

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 $$B=\tilde{B}$$.