Final answer:
To show the parallelogram law in an inner product space, we use the norm and inner product definitions to expand the sides and diagonals of the parallelogram formed by vectors u and v, and demonstrate that the sum of the squares of the lengths of the sides equals the sum of the squares of the lengths of the diagonals.
Step-by-step explanation:
The parallelogram law in an inner product space states that for any two vectors u and v, the sum of the squares of the lengths of the sides of the parallelogram equals the sum of the squares of the lengths of the diagonals. This can be written as:
\(||u+v||^2 + ||u-v||^2 = 2(||u||^2 + ||v||^2)\)
To show that this law holds, we first note that in an inner product space, the norm of a vector u is defined as \(||u|| = \sqrt{\langle u, u \rangle}\), where \(\langle u, v \rangle\) denotes the inner product of u and v. The left hand side of the equation involves expanding inner products:
\(||u+v||^2 = \langle u + v, u + v \rangle\)
\(||u-v||^2 = \langle u - v, u - v \rangle\)
After expansion and simplification using the properties of inner products (linearity and symmetry), we find that cross terms cancel out, and we are left with:
\(||u+v||^2 + ||u-v||^2 = \langle u, u \rangle + 2\langle u, v \rangle + \langle v, v \rangle + \langle u, u \rangle - 2\langle u, v \rangle + \langle v, v \rangle = 2(||u||^2 + ||v||^2)\)
Thus, demonstrating the parallelogram law.