Let u and v be vectors in R^n, and let T be a linear operator on R^n. Prove that T(u) dot T(v) = u dot v

Question

Let u and v be vectors in R^n, and let T be a linear operator on R^n. Prove that T(u) dot T(v) = u dot v

IFF A^T = A^-1 where A is the standard matrix for T.

Answer 1

To prove the given statement, we need to show both directions of the equivalence.

First, we will assume that T(u) · T(v) = u · v, and then prove that A^T = A^(-1), where A is the standard matrix for T.

Let's suppose u = [u1, u2, ..., un]^T and v = [v1, v2, ..., vn]^T. The dot product of two vectors can be expressed as u · v = u1v1 + u2v2 + ... + unvn.

Using the standard matrix A for T, we can express T(u) as A * u and T(v) as A * v.

Now, let's compute the dot product T(u) · T(v):

T(u) · T(v) = (A * u) · (A * v)

By the dot product property, we know that (A * u) · (A * v) = (A * u)^T * (A * v)

Expanding the transpose, we have (A * u)^T * (A * v) = u^T * A^T * A * v

Since matrix multiplication is associative, we can simplify further:

u^T * A^T * A * v = (u^T * A^T) * (A * v)

Now, using the fact that T(u) · T(v) = u · v (as given):

(u^T * A^T) * (A * v) = u · v

Comparing the corresponding terms, we have:

u^T * A^T = u^T

Since this equation holds for any non-zero vector u, we can conclude that A^T = I, where I is the identity matrix. Therefore, A^T = A^(-1) (since A is square).

Now, we need to prove the converse. If A^T = A^(-1), then T(u) · T(v) = u · v.

Using the fact that T(u) = A * u and T(v) = A * v, we can compute the dot product:

T(u) · T(v) = (A * u) · (A * v)

Expanding the dot product, we have (A * u) · (A * v) = u^T * A^T * A * v

Since A^T = A^(-1), we can substitute it in the equation:

u^T * A^T * A * v = u^T * A^(-1) * A * v

Again, using the associativity of matrix multiplication:

u^T * A^(-1) * A * v = (u^T * A^(-1)) * (A * v)

Now, substituting T(u) = A * u and T(v) = A * v back in:

(u^T * A^(-1)) * (A * v) = (u^T * A^(-1)) * T(v)

Finally, using the fact that A^T = A^(-1):

(u^T * A^(-1)) * T(v) = (u^T * A^T) * T(v) = u · v

Since we have shown that both directions hold, we can conclude that T(u) · T(v) = u · v if and only if A^T = A^(-1), where A is the standard matrix for T.