Let u and v be vectors in R^{n}, and let T be a linear operator on R^{n}. Prove that T(n) * T(v) = n * v if and only if A^{T} = A^{-1} where A is the standard matrix for T.
Any help would be great.
i think it's a typo. i believe the OP is asking to show that every inner-product preserving linear operator has an orthogonal matrix in the standard orthonormal basis, and that an orthogonal matrix represents an inner-product preserving linear operator.
the whole of this proof rests on the fact that (for the standard inner product in R^{n}):
$\displaystyle \mathbf{u}\cdot \mathbf{v} = \mathbf{v}^T\mathbf{u}$
so if:
$\displaystyle T\mathbf{u} \cdot T\mathbf{v} = \mathbf{u}\cdot \mathbf{v}$
for all u, v, then:
$\displaystyle ((A\mathbf{v})^T)(A\mathbf{u}) = \mathbf{v}^T\mathbf{u}$
that is:
$\displaystyle \mathbf{v}^TA^TA\mathbf{u} = \mathbf{v}^T\mathbf{u}$
for all u, so
$\displaystyle \mathbf{v}^TA^TA = \mathbf{v}^T$
so that:
$\displaystyle (\mathbf{v}^TA^TA)^T = (\mathbf{v}^T)^T$
$\displaystyle A^T(A^T)^T\mathbf{v} = \mathbf{v} = I\mathbf{v}$
for all v, thus:
$\displaystyle A^TA = I$
$\displaystyle A^T = A^{-1}$
these steps are all reversible.