# Linear Algebra Proof

• Dec 30th 2012, 05:52 PM
rocker71
Linear Algebra Proof
Let u and v be vectors in Rn, and let T be a linear operator on Rn. Prove that T(n) * T(v) = n * v if and only if AT = A-1 where A is the standard matrix for T.

Any help would be great.
• Dec 30th 2012, 07:51 PM
zhandele
Re: Linear Algebra Proof
What does n mean? Does it mean a vector normal to v?

The transpose equaling the inverse means that the columns of A form an orthonormal basis.
• Dec 30th 2012, 08:50 PM
Deveno
Re: Linear Algebra Proof
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 Rn):

$\mathbf{u}\cdot \mathbf{v} = \mathbf{v}^T\mathbf{u}$

so if:

$T\mathbf{u} \cdot T\mathbf{v} = \mathbf{u}\cdot \mathbf{v}$

for all u, v, then:

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

that is:

$\mathbf{v}^TA^TA\mathbf{u} = \mathbf{v}^T\mathbf{u}$

for all u, so

$\mathbf{v}^TA^TA = \mathbf{v}^T$

so that:

$(\mathbf{v}^TA^TA)^T = (\mathbf{v}^T)^T$

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

for all v, thus:

$A^TA = I$

$A^T = A^{-1}$

these steps are all reversible.