Hello everyone!

I was trying to look up the proof for the following:

$\displaystyle AX=\text{proj}_W (b) \Leftrightarrow \, A^t AX=A^t b$

Thanks!

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- Aug 7th 2010, 02:43 AMrebghbLeast Squares solution
Hello everyone!

I was trying to look up the proof for the following:

$\displaystyle AX=\text{proj}_W (b) \Leftrightarrow \, A^t AX=A^t b$

Thanks! - Aug 7th 2010, 05:07 AMHallsofIvy
A is a linear transformation from vector space U to vector space V, each with an inner product.

Use the fact that <Au, v>= <u, A^tv> where the first inner product is in V and the second in U.

Since Ax is equal to the projection of b on W, b- Ax is perpendicular to W: <b- Ax, w>= 0 for any vector w in W. But since A is projection onto W, w= Ay for some y. that is, <b- Ax, w>= <b- Ax, Ay> = 0. Use the property of the transpose, above, to write that as <A^t(b- Ax), y>= 0. this is now an inner product in U and now, y can be any vector in U. In order that <A^t(b- Ax, y>= 0 for**any**y in U, we must have A^t(b- Ax)= A^tb- A^tAx= 0 or A^tAx= A^tb.