anonymous
  • anonymous
When making a projection of vector b on a, you take a*aT/(aT*a) * b, but when in Gram-Schmidt, you take aT*b/(aT*a) *a, which are not the same, or are they? if they are, why?
MIT 18.06 Linear Algebra, Spring 2010
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
if a is column vector and b row vector, why is a*aT*b equal to aT*b*a
anonymous
  • anonymous
As you have guessed, they are the same. This can be seen by working from the properties of inner products: a*a^T*b = a = a = a*b^Ta Eh, 10 days ago. I doubt you'll see this or need it anymore. The reason why this holds is because of the symmetric property of the inner product. Specifically, spaces where you can define an inner product allow you to generate an orthogonal basis, which is all you're really doing in the Gram-Schmidt process, which is why they are equivalent.
anonymous
  • anonymous
thanks daniel, I don't need it anymore, I had figured out that they are the same but why, I couldn't understand. thanks for help :))

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anonymous
  • anonymous
A simple way to think about this is the inner product is a scalar and a scalar times a vector (or matrix) is the same as the vector (or matrix) times the scalar. You can put the scalar anywhere.

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