# Let $T$ is a non-negative linear operator. Prove that there exists a unique non-negative linear operator $S$ such that $S^2 = T$.

Let V be a finite-dimensional inner product space, and $T$ is a non-negative linear operator. Prove that there exists a unique non-negative linear operator $S$ such that $S^2 = T$.

Proof given:

Let O be an orthonormal basis for V such that $[T]_O = diag(\lambda_1, ... \lambda_n)$ where $\lambda_1 \geq 0$ are the eigenvalues of T. Let $S \in \mathbb{L(V)}$ such that $[S]_O = diag(\sqrt{\lambda_1},...\sqrt{\lambda_n}).$ Then $S$ is non-negative and $S^2 = T$

Suppose that there exist non-negative $S_1, S_2 \in \mathbb{L(V)}$ such that $S^2_1 = S^2_1 = T$. Then there exist orthonormal bases $B_1, B_2$ for V such that $[S_1]_{B_1} = [S_2]_{B_2} = diag(\sqrt{\lambda_1},...\sqrt{\lambda_n})$ . Then $[T]_{B_1} = [S_2]_{B_2} = diag(\lambda_1, ... \lambda_n)$.

Let $P = (a_{ij})$ be the transition matrix from $B_1$ to $B_2$. Then $P[S_1]_{B_1}P^* = [S_1]_{B_2} \implies P[S^2_1]_{B_1}P^* = [S^2_1]_{B_2} \implies P[T]_{B_1}P^* = [T]_{B_2}$ So $P diag(\lambda_1, ... \lambda_n) = diag(\lambda_1, ... \lambda_n)P$. Then the proof goes on to use this result prove that $Pdiag(\sqrt{\lambda_1},...\sqrt{\lambda_n}) = diag(\sqrt{\lambda_1},...\sqrt{\lambda_n})P$. And therefore $S_1 =S_2$

My question is

1. Why $P[T]_{B_1}P^* = [T]_{B_2}$ gives $P diag(\lambda_1, ... \lambda_n) = diag(\lambda_1, ... \lambda_n)P$? Is it because $P^*P = I$? Is this always true $P^*P = I$?

2. To prove $Pdiag(\sqrt{\lambda_1},...\sqrt{\lambda_n}) = diag(\sqrt{\lambda_1},...\sqrt{\lambda_n})P$, it uses $P diag(\lambda_1, ... \lambda_n) = diag(\lambda_1, ... \lambda_n)P$. But this result is proved using $P[S_1]_{B_1}P^* = [S_1]_{B_2}$. Doesn't this make it a circular argument? Can't I just move the $P^*$ to the other side to get $Pdiag(\sqrt{\lambda_1},...\sqrt{\lambda_n}) = diag(\sqrt{\lambda_1},...\sqrt{\lambda_n})P$? Of course, this is assuming $P^*P = I$

To your first question: the transition matrix between orthonormal bases is always unitary. That is, if $B_1,B_2$ are orthonormal, then the transition matrix $P$ between them satisfies $P^*P = I$.
To your second: the fact that $P[S_1]_{B_1}P^* = [S_1]_{B_2}$ follows from the definition of a transition matrix and the fact that $P$ is unitary.
• 1. Can I know why transition matrix is always unitary? 2. So can I just straight away get $P[(S_1)]_{B_1} = [S_1]_{B_2}P$ without proving $P[T]_{B_1} = [T]_{B_2}P$. To me that part is redundant – macho Nov 30 '14 at 5:43
• For 1: this is generally given along with (or near) the definition of a unitary matrix; a matrix is unitary if and only if its columns form and orthonormal basis. For 2: the whole point is to exploit the fact that $S_1^2 = S_2^2$. So no, that part is not redundant. – Omnomnomnom Nov 30 '14 at 5:54