Exercise 6.5.2

From equation (4), we have $(x_{k+1}-x^{\text{∗}})=x_k-x^{\text{∗}}-\frac{a_i{a}_i^T}{a_i^T{a}_i}(x_k-x^{\text{∗}})=(I-\frac{a_i{a}_i^T}{a_i^T{a}_i})(x_k-x^{\text{∗}})=P(x_k-x^{\text{∗}})$

So we have $P=(I-\frac{a_i{a}_i^T}{a_i^T{a}_i})$, it’s easy to see that $P^T=P$. Also

$P^2=(I-\frac{a_i{a}_i^T}{a_i^T{a}_i})(I-\frac{a_i{a}_i^T}{a_i^T{a}_i})=I-\frac{a_i{a}_i^T}{a_i^T{a}_i}-\frac{a_i{a}_i^T}{a_i^T{a}_i}+\frac{a_i{a}_i^T}{a_i^T{a}_i}\frac{a_i{a}_i^T}{a_i^T{a}_i}=I-\frac{a_i{a}_i^T}{a_i^T{a}_i}=P$

So $P$ is a projection matrix, and $x_{k+1}-x^{\text{∗}}=P(x_k-x^{\text{∗}})$

Note, this $P$ projects the error $x_k-x^{\text{∗}}$ onto plane $a_i^{T}x=b_i$, so $x^{\text{∗}}$ is on the plane, also $x_{k+1}$ is on the plane by design. The plane has a normal direction of $a_i$. So $a_i^T(x_{k+1}-x^{\text{∗}})=0$. And $(x_k-x^{\text{∗}})-P(x_k-x^{\text{∗}})=\frac{a_i{a}_i^T}{a_i^T{a}_i}(x_k-x^{\text{∗}})$ is the vector perpendicular to the plane $a_i^{T}x=b_i$. This is actually a projection (using another projection matrix) of $x_k-x^{\text{∗}}$ onto the line $a_i$ according to Problem VI.5.1. You see it has the same direction as $a_i$.