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

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

Since the dot product is linear, this gives $ = = + = = \sigma = \lambda$. Thus, we see that u can be any vector we like as long as it's perpendicular to a, and we must have $\sigma = \frac{\lambda}{}$. The set of solutions, then, is
$x = u + \frac{\lambda}{}a$ 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 $Lx=b$, where L is a linear operator. In this case we can write this operator $L = I + ca^*$. Here I is the identity, and $a^*$ is the conjugate transpose of a, so applying it to a vector gives $a^*x = $.

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

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