# Matrices Transformation

• May 28th 2014, 03:56 AM
mlg
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 uTv.
(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)
• May 28th 2014, 04:13 AM
HallsofIvy
Re: Matrices Transformation
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 $\displaystyle u^T$? How are such things multiplied- i.e. how is $\displaystyle u^Tv$ defined? How is "orthogonal" defined?
• May 28th 2014, 04:32 AM
mlg
Re: Matrices Transformation
The only information given is as follows:
Matrix A is orthogonal if A^t = A^-1, where A^t is the transpose of A.
• May 28th 2014, 07:30 AM
Deveno
Re: Matrices Transformation
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?
• May 28th 2014, 09:07 AM
mlg
Re: Matrices Transformation
Thank you Deveno.
I appreciate very much the big effort you put into your response.
I will now study this in more detail.