Hello, how can I proof the next theorem?

I have a linear transformation T(X) that can be express as T(X)=AX
and A is an orthogonal matrix, then ||T (X)||=||X|| , I was doing this:
||T (X)||=sqrt(<AX,AX>)
But I don't know what to do with the orthogonal matrix..
Please help

To prove the theorem ||T(X)|| = ||X||, where T(X) = AX and A is an orthogonal matrix, we can use the properties of orthogonal matrices.

First, let's start with the expression ||T(X)|| = sqrt(<AX, AX>).

Next, we need to use the property of orthogonal matrices that A^T * A = I, where A^T represents the transpose of A and I represents the identity matrix.

Substituting AX back into the expression, we have ||T(X)|| = sqrt(<A^T * AX, AX>).

Now, we can simplify the expression using the properties of the dot product. Recall that the dot product of two vectors X and Y is defined as <X, Y> = X^T * Y, where ^T represents the transpose of a vector.

Therefore, we have ||T(X)|| = sqrt((A^T * AX)^T * AX).

Using the property (XY)^T = Y^T * X^T and the fact that A^T * A = I, we can rewrite the expression as ||T(X)|| = sqrt(X^T * (A^T * A) * X).

Since A is an orthogonal matrix, A^T * A = I, and the expression simplifies to ||T(X)|| = sqrt(X^T * I * X).

Since I is the identity matrix, I * X = X, and we get ||T(X)|| = sqrt(X^T * X), which is equal to ||X||.

Therefore, we have proven that ||T(X)|| = ||X|| when T(X) = AX and A is an orthogonal matrix.