Given Information
Given that $B$ is an $n \times n$ symmetric matrix such that $B ^ { 2 } = B$ . Also given that. $\mathbf { y }$ in $\mathbb { R } ^ { n } ,$ let $\hat { \mathbf { y } } = B \mathbf { y }$ and $\mathbf { z } = \mathbf { y } - \hat { \mathbf { y } }$ . We have to answer some questions based on that.

Step-1:
We have to prove that z is orthogonal to $\hat { \mathbf { y } }$. Find the product of z and $\hat { \mathbf { y } }$ \[\begin{array}{l}{\bf{z}}\widehat {\bf{y}} = ({\bf{y}} - \widehat {\bf{y}})(B{\bf{y}})\\ = {\bf{y}}(B{\bf{y}}) - (B{\bf{y}})(B{\bf{y}})\\ = {\bf{y}}(B{\bf{y}}) - B({\bf{y}}B){\bf{y}}\\ = {\bf{y}}(B{\bf{y}}) - {\bf{y}}(BB){\bf{y}}\\ = {\bf{y}}(B{\bf{y}}) - {\bf{y}}(B{\bf{y}})\\ = 0\end{array}\]Therefore,

z is orthogonal to $\hat { \mathbf { y } }$



Step-2:
If $W$ is the column space of $B$ . We have to show that $\mathbf { y }$ is the sum of a vector in $W$ and a vector in $W ^ { \perp } .$ \[\begin{aligned} ( \mathbf { y } - \hat { \mathbf { y } } ) \cdot ( B \mathbf { u } ) & = [ B ( \mathbf { y } - \hat { \mathbf { y } } ) ] \cdot \mathbf { u } \\ & = [ B \mathbf { y } - B B \mathbf { y } ] \cdot u \\ & = 0 \end{aligned}\]So, $\mathbf { y } - \hat { \mathbf { y } }$ is in $W ^ { \perp }$ Hence,. $\mathbf { y } = \hat { \mathbf { y } } + ( \mathbf { y } - \hat { \mathbf { y } } )$ means that, sum of a vector in $\mathrm { W }$ and a vector in $\mathrm { W } ^ { 1 }$ is equals to $\mathrm { y }$. Therefore,

The projection of yonto the column space of $B$