For the orthogonal decomposition, don’t you need two scalars? E.g. x=ay+bz . For example, in R2 , let x=(2,2),y=(0,1),z=(1,0). Then x=2y+2z, and there’s no way to write x as y+az.
Ow. Yes, you do. This wasn’t a typo either, I remembered the result incorrectly. Thanks for pointing it out, and props for being attentive enough to catch it.
Or to be more precise, you only need one scalar, but the scalar is for y not z, because z isn’t given. The theorem says that, given x and y, there is a scalar a and a vector z such that x=ay+z and y is orthogonal to z.
For the orthogonal decomposition, don’t you need two scalars? E.g. x=ay+bz . For example, in R2 , let x=(2,2),y=(0,1),z=(1,0). Then x=2y+2z, and there’s no way to write x as y+az.
Ow. Yes, you do. This wasn’t a typo either, I remembered the result incorrectly. Thanks for pointing it out, and props for being attentive enough to catch it.
Or to be more precise, you only need one scalar, but the scalar is for y not z, because z isn’t given. The theorem says that, given x and y, there is a scalar a and a vector z such that x=ay+z and y is orthogonal to z.