Show that $\phi (x)\cdot \phi (y)=x\cdot y$ for all $x,y\in \mathbb{R}^3$

Question

Let $A$ be a matrix where the columns form an orthonormal basis and let $\phi (v)=Av$.

I want to show that $\phi (x)\cdot \phi (y)=x\cdot y$ for all $x,y\in \mathbb{R}^3$.

Since the columns of $A$ form an orthonormal basis that means that $A^TA=I$, or not?

So let $\phi (x)\cdot \phi (y)=B$ then we have the following: $$A\cdot x\cdot A\cdot y=B\Rightarrow A^TA\cdot x\cdot A\cdot y=A^TB\Rightarrow x\cdot A\cdot y=A^TB\Rightarrow x\cdot A\cdot y=A^TB$$ Is that the correct way? But how do we continue?

Answer 1

You should probably try to be very careful with notation and not write "$\cdot$" for a matrix product.

1. $x\cdot y = \sum x_i y_i = x^T y$. This expresses the dot product as a matrix product.
2. $(AB)^T = B^TA^T\!$. Proof: let $A=(A_{ij}),B=(B_{ij})$, and $C=AB$ so that the entries of $C$ are $C_{ij}=\sum_k A_{ik} B_{kj}$. Then $C^T=(C^T_{ij})$ has entries $$C^T_{ij}=C_{ji}=\sum_k A_{jk} B_{ki} =\sum_k B_{ki} A_{jk} = \sum_k B^T_{ik} A^T_{kj} = (B^TA^T)_{ij}. $$
3. $\phi(x)\cdot \phi(y) = Ax\cdot Ay\overset{\text{by 1.}}= (Ax)^T Ay \overset{\text{by 2.}}= x^T A^TAy.$
4. Now apply $A^TA=I$.

Your calculation is correct for the very different quantity, $A(x^TAx) = A(x\cdot Ax),$ which is a matrix (not a scalar.) Only here is it possible to cancel $A$ by multiplication on the left (actually $x\cdot Ax$ is a scalar, so multiplying on the right also works): $$ B=A(x\cdot Ax) \implies A^TB = x\cdot Ax$$ In contrast, $Ax\cdot Ay=(Ax)\cdot(Ay)$ is a dot product of vectors, and hence a scalar.

Answer 2

It's true that $A^TA=I$.
Now, note that $\phi(x).\phi(y)=(\phi(x))^T(\phi(y))=x^TA^TAy=x^Ty=x.y$

Note: Please note that $Ax$, where $x\in \mathbb R^3$, is linear combination of columns of matrix $A$ so $A.x$ as you have used in your post, does not make sense.

Answer 3

It's not clear what you mean by $A\cdot x$. At first sight it looks to me that you are confusing inner product with matrix multiplication.

In general, it's true that $Ax \cdot y = x\cdot A^ty$ for any matrix $A$. You can prove it by choosing $x$ and $y$ to be elements of an orthonormal basis, both sides of the equality will become appropriate matrix coefficients.

Furthermore, it is true that in your case $AA^t = A^tA = I$. So, $$Ax\cdot Ay = x\cdot A^tAy = x\cdot Iy = x\cdot y.$$