Thread: 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 ? 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.