Letuandvbe vectors in R^{n}, and letTbe a linear operator on R^{n}. Prove that T(n) * T(v) =n*vif and only if A^{T}= A^{-1}where A is the standard matrix forT.

Any help would be great.

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- Dec 30th 2012, 04:52 PMrocker71Linear Algebra Proof
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. - Dec 30th 2012, 06:51 PMzhandeleRe: 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, 07:50 PMDevenoRe: 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 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.