Video: Inner Products of Vectors - The Polarization Identity

Given that ๐ฎ and ๐ฏ โˆˆ โ„^(๐‘›) and that <๐ฎ, ๐ฏ> = ๐‘[|๐ฎ + ๐ฏ|ยฒ โˆ’ |๐ฎ โˆ’ ๐ฏ|ยฒ], find the value of ๐‘.

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Video Transcript

Given that ๐ฎ and ๐ฏ are in โ„ ๐‘› and that the inner product of ๐ฎ and ๐ฏ is equal to ๐‘ times the modulus of ๐ฎ plus ๐ฏ all squared minus the modulus of ๐ฎ minus ๐ฏ all squared, find the value of ๐‘.

An easy way for tackling any equations such as these is to directly evaluate both sides of the equation in terms of series. Since ๐ฎ and ๐ฏ are both in โ„ ๐‘›, their tuples look something like this: ๐‘ข one, ๐‘ข two, all the way to ๐‘ข ๐‘›, and likewise ๐‘ฃ one, ๐‘ฃ two, all the way to ๐‘ฃ ๐‘›. We can use these tuples to directly evaluate both sides of the equation in terms of a series.

Letโ€™s start with the left-hand side. In โ„ ๐‘›, the inner product of ๐ฎ and ๐ฏ is just ๐‘ข one times ๐‘ฃ one plus ๐‘ข two times ๐‘ฃ two, et cetera, all the way to ๐‘ข ๐‘› times ๐‘ฃ ๐‘›. We can reexpress this as a series as the sum from ๐‘– equals one to ๐‘› of ๐‘ข ๐‘– times ๐‘ฃ ๐‘–.

Now letโ€™s look at the terms on the right-hand side. The modulus of ๐ฎ plus ๐ฏ all squared is just the components of ๐ฎ and ๐ฏ added together and squared separately. So we have ๐‘ข one plus ๐‘ฃ one all squared plus ๐‘ข two plus ๐‘ฃ two all squared plus et cetera, all the way to ๐‘ข ๐‘› plus ๐‘ฃ ๐‘› all squared. We can once again express this as a series as the sum from ๐‘– equals one to ๐‘› of ๐‘ข ๐‘– plus ๐‘ฃ ๐‘– all squared.

Similarly, for the second term on the right-hand side, we have ๐ฎ minus ๐ฏ all squared equals ๐‘ข one minus ๐‘ฃ one all squared plus ๐‘ข two minus ๐‘ฃ two all squared plus et cetera, all the way to ๐‘ข ๐‘› minus ๐‘ฃ ๐‘› all squared. And again, we can reexpress this as a series as the sum from ๐‘– equals one to ๐‘› of ๐‘ข ๐‘– minus ๐‘ฃ ๐‘– all squared. So again, ignoring the ๐‘ for now, we can reexpress the right-hand side as the sum from ๐‘– equals one to ๐‘› of ๐‘ข ๐‘– plus ๐‘ฃ ๐‘– all squared minus ๐‘ข ๐‘– minus ๐‘ฃ ๐‘– all squared.

Expanding out the parentheses in the general term, we get ๐‘ข ๐‘– squared plus two ๐‘ข ๐‘– ๐‘ฃ ๐‘– plus ๐‘ฃ ๐‘– squared minus ๐‘ข ๐‘– squared plus two ๐‘ข ๐‘– ๐‘ฃ ๐‘– minus ๐‘ฃ ๐‘– squared. These positive and negative squares will cancel each other out. And this leaves us with the sum from ๐‘– equals one to ๐‘› of four ๐‘ข ๐‘– ๐‘ฃ ๐‘–.

Now, substituting these series into the original equation, we get the sum from ๐‘– equals one to ๐‘› of ๐‘ข ๐‘– ๐‘ฃ ๐‘– is equal to ๐‘ times the sum from ๐‘– equals one to ๐‘› of four ๐‘ข ๐‘– ๐‘ฃ ๐‘–. We can take this constant of four outside of the series summation. And by direct comparison, we can see that ๐‘ is equal to one-quarter.

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