Thread: Matrices Transformation

I would very much appreciate help with the following question:

(i) Prove that the dot product of any two vectors u,v ∈ Rn is u^Tv.
(ii) Given a Matrix A and vectors u,v ∈ Rn so that Au . Av = u . v , prove that A is orthogonal
(for part (ii)we can assume that (AB)^t= B^t A^t)

What definitions of all these things are you given? In particular, you title this "Matrices transformation". Are you given the vectors as matrices? Given vector u, what is vector ? How are such things multiplied- i.e. how is defined? How is "orthogonal" defined?

Thanks for the quick reply.
The only information given is as follows:
Matrix A is orthogonal if A^t = A^-1, where A^t is the transpose of A.

For (i):

If we have:

$\mathbf{u} = (u_1,\dots,u_n), \mathbf{v} = (v_1,\dots,v_n)$ then typically we have the definition of the dot product as:

$\displaystyle \mathbf{u}\cdot\mathbf{v} = u_1v_1 +\cdots +u_nv_n = \sum_{k = 1}^n u_kv_k$

However, we can also view vectors as nx1 matrices, in which case, for:

$\mathbf{u} = \begin{bmatrix}u_1\\ \vdots\\u_n\end{bmatrix}; \mathbf{v} = \begin{bmatrix}v_1\\ \vdots\\v_n\end{bmatrix}$

then the matrix product:

$\mathbf{u}^T\mathbf{v} = \begin{bmatrix}u_1& \dots&u_n\end{bmatrix} \begin{bmatrix}v_1\\ \vdots\\v_n\end{bmatrix}$

compute this last matrix.

For (ii):

By part (i) we have:

$\mathbf{Au}\cdot\mathbf{Av} = (\mathbf{Au})^T\mathbf{Av} = \mathbf{u}^T(\mathbf{A}^T\mathbf{A})\mathbf{v}$

Since we are told (by assumption) this is equal to $\mathbf{u}\cdot\mathbf{v} = \mathbf{u}^T\mathbf{v}$,

we have that:

$\mathbf{u}^T(\mathbf{A}^T\mathbf{A})\mathbf{v} - \mathbf{u}^T\mathbf{v} = 0$

that is:

$\mathbf{u}^T(\mathbf{A}^T\mathbf{A} - \mathbf{I})\mathbf{v} = 0$

for any pair $\mathbf{u},\mathbf{v}$.

In particular, we can choose $\mathbf{u} = \mathbf{e}_i, \mathbf{v} = \mathbf{e}_j$

in which case if we write $\mathbf{A}^T\mathbf{A} - \mathbf{I} = \mathbf{B} = (b_{ij})$,

then $\mathbf{u}^T(\mathbf{A}^T\mathbf{A} - \mathbf{I})\mathbf{v} = (\mathbf{e}_i)^T\mathbf{B}\mathbf{e}_j = b_{ij}$

which tells us that $b_{ij} = 0$ for all $i,j = 1,\dots,n$, that is $\mathbf{B}, = \mathbf{0}$.

So if $\mathbf{B} = \mathbf{A}^T\mathbf{A} - \mathbf{I} = \mathbf{0}$, what can you conclude from that?

Thank you Deveno.
I appreciate very much the big effort you put into your response.
I will now study this in more detail.