Unique Solution to Linear System of Equations

Question

I want to show that $x=\left(\frac { 1}{ 1+{ b }^{ T }b}\right)*b$ is a unique solution to the Linear System $({ I }_{ q }+b{ b }^{ T })x=b$ for

$b\ \epsilon { \Re }^{ q }$ (${ \Re }$ denoting the reals)

Since b is a vector with q-entries $b{ b }^{ T }$ becomes a q × q Matrix, so the term in the first brackets becomes some Matrix, lets call it A. In order to have a unique solution to the System we need to find conditions on A to be invertible, how can I do this, i feel like that I have too little information of A to argue this way. and How do Igenerally approach Problems of this kind?

Answer 1

Clearly, $$ (bb^T)b=b(b^Tb)=(b^Tb)b, $$ since $b^Tb$ is scalar. Hence $$ (I+bb^T)b=b+(bb^T)b=b+(b^Tb)b=(1+b^Tb)b, $$ and thus $$ (I+bb^T)\left(\frac{1}{1+b^Tb}b\right)=b, $$ and as $I+bb^T$ is positive definite and thus invertible, $$ x=\frac{1}{1+b^Tb}b $$ is the unique solution of $$ (I+bb^T)x=b. $$

Note. The symmetric matrix $I+bb^T$ is positive definite since, for every $x\in\mathbb R^n$ $$ \langle(1+bb^T)x,x\rangle=|x|^2+\langle bb^Tx,x\rangle=|x|^2+\langle b^Tx,b^Tx\rangle=|x|^2+|b^Tx|^2\ge |x|^2. $$

Answer 2

If we write the inner product $\langle x, y \rangle = y^{T}x$, then the linear system can be interpreted as: $$x + b\langle b,x \rangle = b$$ From which we get: $$x = b(1 - \langle b, x \rangle)$$ If we take an inner product of $b$ with both sides of the equation, we get: $$\langle b, x \rangle = \langle b, b \rangle (1 - \langle b, x \rangle)$$ Which easily simplifies to: $$\langle b, x \rangle = \frac{\langle b, b \rangle}{1 + \langle b, b \rangle}$$ Plugging this back into the second line yields the answer: $$x = \left(\frac{1}{1 + \langle b, b \rangle}\right)b = \left(\frac{1}{1 + b^{T}b}\right)b$$