Question

verify the equality: |u+v|2+|u−v|2=2|u|2+2|v|2, and. derive the theorem:the sum of the square so fthe diagonals of a parallelogram is equal to the sum of the square so the sides

Answer 1

Answer on Question #43658 – Math – Functional Analysis

Verify the equality $\| \pmb{u} + \pmb{v} \|^2 + \| \pmb{u} - \pmb{v} \|^2 = 2 \| \pmb{u} \|^2 + 2 \| \pmb{v} \|^2$, and derive the theorem: the sum of the square so the diagonals of a parallelogram is equal to the sum of the square so the sides.

Solution:

Using the properties of the inner product:

$$\| \boldsymbol {u} + \boldsymbol {v} \| ^ {2} = \langle \boldsymbol {u} + \boldsymbol {v}, \boldsymbol {u} + \boldsymbol {v} \rangle = \langle \boldsymbol {u}, \boldsymbol {u} \rangle + \langle \boldsymbol {u}, \boldsymbol {v} \rangle + \langle \boldsymbol {v}, \boldsymbol {u} \rangle + \langle \boldsymbol {v}, \boldsymbol {v} \rangle$$ $$\| \boldsymbol {u} - \boldsymbol {v} \| ^ {2} = \langle \boldsymbol {u} - \boldsymbol {v}, \boldsymbol {u} - \boldsymbol {v} \rangle = \langle \boldsymbol {u}, \boldsymbol {u} \rangle - \langle \boldsymbol {u}, \boldsymbol {v} \rangle - \langle \boldsymbol {v}, \boldsymbol {u} \rangle + \langle \boldsymbol {v}, \boldsymbol {v} \rangle$$ $$\| \boldsymbol {u} + \boldsymbol {v} \| ^ {2} + \| \boldsymbol {u} - \boldsymbol {v} \| ^ {2} = \langle \boldsymbol {u}, \boldsymbol {u} \rangle + \langle \boldsymbol {u}, \boldsymbol {v} \rangle + \langle \boldsymbol {v}, \boldsymbol {u} \rangle + \langle \boldsymbol {v}, \boldsymbol {v} \rangle + \langle \boldsymbol {u}, \boldsymbol {u} \rangle - \langle \boldsymbol {u}, \boldsymbol {v} \rangle - \langle \boldsymbol {v}, \boldsymbol {u} \rangle + \langle \boldsymbol {v}, \boldsymbol {v} \rangle = 2 \langle \boldsymbol {u}, \boldsymbol {u} \rangle + 2 \langle \boldsymbol {v}, \boldsymbol {v} \rangle = 2 \| \boldsymbol {u} \| ^ {2} + 2 \| \boldsymbol {v} \| ^ {2}$$

In a parallelogram if given two vectors $\pmb{u}$ and $\pmb{v}$ then the diagonals of a parallelogram are $\pmb{u} + \pmb{v}$ and $\pmb{u} - \pmb{v}$, so using $\| \pmb{u} + \pmb{v} \|^2 + \| \pmb{u} - \pmb{v} \|^2 = 2 \| \pmb{u} \|^2 + 2 \| \pmb{v} \|^2$ we derive the theorem: the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals

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