While self studying Linear Algebra from Hoffman and Kunze, I have a question in Theorem 14 in the section on unitary operators from Chapter 8.
Here are the relevant images.
My question is in highlighted line of the image. I am unable to get what reasoning is behind the line that $M_{2}^{-1} $, $ M_{1}M_{2}^{-1}, $ and $ {(M_{1} M_{2})}^{-1}$ are all in $T^{+}(n).$
Edit: I am adding a related question on normal operators here.
Suppose $U$ is any normal operator on $V$ and $\alpha$ is a vector in $V.$ Why should $||U \alpha || = || U^{*} \alpha ||$ always hold true?
Best Answer
Essentially, your first question is answered by the fact that $T^+(n)$ is a multiplicative group. By hypothesis, we have that $M_1$ and $M_2$ are both elements of $T^+(n),$ hence their inverses $M_1^{-1}$ and $M_2^{-1}$ and any product of these four matrices are also in $T^+(n),$ i.e., $M_2^{-1},$ $M_1 M_2^{-1},$ and $(M_1 M_2)^{-1}$ are all in $T^+(n).$
Perhaps a better question is to understand why $T^+(n)$ is a multiplicative group. Observe that the $n \times n$ identity matrix $I$ is an element of $T^+(n),$ and it also functions as the identity element of $T^+(n).$ Further, associativity of matrix multiplication holds, so we need only establish that $T^+(n)$ is closed under multiplication and that the inverse of a matrix in $T^+(n)$ is also in $T^+(n).$ For the latter, I suggest you check out this answer; the former is a matter of routine computation. By writing down the matrices $A$ and $B$ of $T^+(n)$ in terms of their respective elements $a_{ij}$ and $b_{ij}$ in the $i$th row and $j$th column, we have that $$[AB]_{ij} = \sum_{k = 1}^n a_{ik} b_{kj}.$$ Use the fact that $a_{ik} = 0$ whenever $k \geq i + 1$ and $b_{kj} = 0$ whenever $j \geq k + 1$ to see that $[AB]_{ij} = 0$ whenever $j \geq i + 1.$ Likewise, we have that $(AB)_{ii} > 0.$ Both of these together say that $AB \in T^+(n).$
On the matter of your second question, observe that for any vector $\alpha$ and any normal operator $U,$ $$||U \alpha|| = \langle U \alpha, U \alpha \rangle = \langle \alpha, U^* U \alpha \rangle = \langle \alpha, UU^* \alpha \rangle = \langle U^* \alpha, U^* \alpha \rangle = ||U^* \alpha||.$$