How to show that the dot product of $h(x)$ and $h(y)$ can be expressed as $h(x)^{tr} * h(y)$?

Question

In the book of Analysis on Manifolds by Munkres, at page 175, it is given that

Let $h(x) = A \cdot x$, where $A$ is orthogonal n-by-n matrix; we show $h$ is an isometry by showing that is preserves the dot product.Now the dot product of $h(x)$ and $h(y)$ can be expressed as

$$h(x)^{tr} * h(y)$$ [...]

But I do not understand how can $$h(x) \cdot h(y) = h(x)^{tr} * h(y)$$ ?

Answer 1

An example, $n=4$. If you have two $4 \times 1$ matrices, $$ \mathbf{u} = \begin{bmatrix} a_1\\a_2\\a_3\\a_4 \end{bmatrix} \qquad \mathbf{v}= \begin{bmatrix} b_1\\b_2\\b_3\\b_4 \end{bmatrix} $$ Then the transpose of $\mathbf{u}$ is a $1 \times 4$ matrix $$ \mathbf{u}^{\mathrm{tr}} = \begin{bmatrix} a_1\quad a_2\quad a_3\quad a_4 \end{bmatrix} $$ When you multiply matrices, you get a $1 \times 1$ matrix $$ \mathbf{u}^{\mathrm{tr}}\;\mathbf{v} = \left[a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4\right] $$ Then, by convention, we consider a $1 \times 1$ matrix to be the same thing as a scalar, so we have the dot product $\mathbf{u}\,\cdot\,\mathbf{v}$.

Answer 2

$h(x),\,h(y)$ are represented by $n\times 1$ matrices, so $\;{}^{\mathrm t}h(x)\cdot h(x)\;$ is the product of a $1\times n$ matrix by an $n\times 1$ matrix, i.e. a number.