Equations involving the inner product

Question

I've been reading chapter 2 of Ward Cheney's Analysis for Applied Mathematics, and he gives the following question:

Find all solutions to the equation $\langle x,a \rangle c = b $, assuming that $a$, $b$, and $c$ are given vectors in an inner-product space.

This problem looks like it will require some case work. For instance, if $b=0$ then any $x\in \{a\}^\perp$ is a solution. Also, there can be no solution if $b\neq0$ and either $a$ or $c$ is zero. Further, if $b\neq0$ and $c$ is orthogonal to $b$, then there can be no solution since we would have

$$\langle b,b \rangle = \langle \langle x,a \rangle c,b \rangle = 0,$$ which is a contradiction.

So then the most interesting case is when none of the three knowns is zero, and $\langle b,c \rangle \neq0$. However, I am not sure how to proceed.

Answer 1

If $a=0$ then $b=0$ and $c$ is arbitrary.

If $c=0$ then $b=0$ and $x$ can be anything.

The LHS is a scalar times $c$. So if $b$ is not a scalar times $c$ then there is no solution.

If $a\ne0$ and $c\ne0$ and $b$ is a scalar times $c$ then $b=\lambda c$, where $\lambda$ can be found since $b$ and $c$ are given. So $\langle x,a\rangle=\lambda$. Write $x=\alpha a+y$, where $y\perp a$; then we get $$\alpha=\frac{\lambda}{\langle a,a\rangle}\ .$$ So the solution in this case is $$x=\frac{\lambda}{\langle a,a\rangle}a+y$$ where $b=\lambda c$ and $y\perp a$.

Answer 2

Hint: If $b \neq \alpha c$ for some scalar $\alpha$ (and $c \neq 0$), consider taking the inner product of both sides with $c - \frac{\langle b,c \rangle}{|\langle b,b \rangle|} b$. Note that this non-zero vector is orthogonal to $b$ but not to $c$.