1. ## Some inner product proofs help please

1) if a,b,c belong to an inner product space X, find all solutions to the equation <a,x>b=c

I know that I should set c=landa*b but i get stuck from there.

2) find solutions to equation x + <a,x>c=b

No idea how to do this one at all

$\displaystyle <a,x>b=c$ has solutions only if b and c are parallel, that is, if $\displaystyle c = \lambda b$ for some constant $\displaystyle \lambda$. So, we set $\displaystyle <a,x> = \lambda$. We can decompose x into a piece that's parallel to a and a piece that's perpendicular to a, i.e., $\displaystyle x = u + v$ with $\displaystyle <u,a> = 0$ and $\displaystyle v = \sigma a$ for some constant $\displaystyle \sigma$.

Since the dot product is linear, this gives $\displaystyle <a,x> = <a,u+v> = <a,u> + <a,v> = <a,v> = \sigma <a,a> = \lambda$. Thus, we see that u can be any vector we like as long as it's perpendicular to a, and we must have $\displaystyle \sigma = \frac{\lambda}{<a,a>}$. The set of solutions, then, is
$\displaystyle x = u + \frac{\lambda}{<a,a>}a$ for all u orthogonal to a.

3. wow thanks, that helps alot. Now im guessing the second part isnt as easy as that =s . Any suggestions how to do it ?

4. Second part is a little trickier, but doable. Think of this as an equation of the form $\displaystyle Lx=b$, where L is a linear operator. In this case we can write this operator $\displaystyle L = I + ca^*$. Here I is the identity, and $\displaystyle a^*$ is the conjugate transpose of a, so applying it to a vector gives $\displaystyle a^*x = <a,x>$.

So, the equation we want to solve is $\displaystyle Lx = (I + ca^*)x = x + <a,x>c = b$. In order to solve it, we need to find an inverse of the operator L, because then the solution is just $\displaystyle x = L^{-1}b$.

Now I can't remember if there's an easy way to see this, but there happens to be an inverse of the form $\displaystyle L^{-1}=I + ua^*$ for some vector u. Suppose this is true. Then we want $\displaystyle (I+ua^*)(I+ca^*) = I$. Now expand this out: $\displaystyle I + ua^* + ca^* + <a,c>ua^* = I$. All that's left is to solve for u, and apply the inverse operator you get to b to get your final result.