# Video: Inner Products of Vectors - The Polarization Identity

Given that 𝐮 and 𝐯 ∈ ℝ^(𝑛) and that <𝐮, 𝐯> = 𝑐[|𝐮 + 𝐯|² − |𝐮 − 𝐯|²], find the value of 𝑐.

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### Video Transcript

Given that 𝐮 and 𝐯 are in ℝ 𝑛 and that the inner product of 𝐮 and 𝐯 is equal to 𝑐 times the modulus of 𝐮 plus 𝐯 all squared minus the modulus of 𝐮 minus 𝐯 all squared, find the value of 𝑐.

An easy way for tackling any equations such as these is to directly evaluate both sides of the equation in terms of series. Since 𝐮 and 𝐯 are both in ℝ 𝑛, their tuples look something like this: 𝑢 one, 𝑢 two, all the way to 𝑢 𝑛, and likewise 𝑣 one, 𝑣 two, all the way to 𝑣 𝑛. We can use these tuples to directly evaluate both sides of the equation in terms of a series.

Let’s start with the left-hand side. In ℝ 𝑛, the inner product of 𝐮 and 𝐯 is just 𝑢 one times 𝑣 one plus 𝑢 two times 𝑣 two, et cetera, all the way to 𝑢 𝑛 times 𝑣 𝑛. We can reexpress this as a series as the sum from 𝑖 equals one to 𝑛 of 𝑢 𝑖 times 𝑣 𝑖.

Now let’s look at the terms on the right-hand side. The modulus of 𝐮 plus 𝐯 all squared is just the components of 𝐮 and 𝐯 added together and squared separately. So we have 𝑢 one plus 𝑣 one all squared plus 𝑢 two plus 𝑣 two all squared plus et cetera, all the way to 𝑢 𝑛 plus 𝑣 𝑛 all squared. We can once again express this as a series as the sum from 𝑖 equals one to 𝑛 of 𝑢 𝑖 plus 𝑣 𝑖 all squared.

Similarly, for the second term on the right-hand side, we have 𝐮 minus 𝐯 all squared equals 𝑢 one minus 𝑣 one all squared plus 𝑢 two minus 𝑣 two all squared plus et cetera, all the way to 𝑢 𝑛 minus 𝑣 𝑛 all squared. And again, we can reexpress this as a series as the sum from 𝑖 equals one to 𝑛 of 𝑢 𝑖 minus 𝑣 𝑖 all squared. So again, ignoring the 𝑐 for now, we can reexpress the right-hand side as the sum from 𝑖 equals one to 𝑛 of 𝑢 𝑖 plus 𝑣 𝑖 all squared minus 𝑢 𝑖 minus 𝑣 𝑖 all squared.

Expanding out the parentheses in the general term, we get 𝑢 𝑖 squared plus two 𝑢 𝑖 𝑣 𝑖 plus 𝑣 𝑖 squared minus 𝑢 𝑖 squared plus two 𝑢 𝑖 𝑣 𝑖 minus 𝑣 𝑖 squared. These positive and negative squares will cancel each other out. And this leaves us with the sum from 𝑖 equals one to 𝑛 of four 𝑢 𝑖 𝑣 𝑖.

Now, substituting these series into the original equation, we get the sum from 𝑖 equals one to 𝑛 of 𝑢 𝑖 𝑣 𝑖 is equal to 𝑐 times the sum from 𝑖 equals one to 𝑛 of four 𝑢 𝑖 𝑣 𝑖. We can take this constant of four outside of the series summation. And by direct comparison, we can see that 𝑐 is equal to one-quarter.