[Math] Orthogonal Projection onto the kernel

Question

Ok this might be my last question on Linear Algebra hopefully, but how do I approach this problem? Find the orthogonal projection of $u = (3,2,1,4,5,6)$ onto the kernel of:

$$
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 & 1 & 1 \\
\end{pmatrix}
$$

I know that to find the orthogonal projection we have to have orthogonal vectors and then solve for $$ u^´= u – u^" $$ which I know how to do.

But I'm kinda stuck on how to begin, do I solve for $$Ax = 0$$ directly? Or do I solve for $$U^T = Ax = 0$$? (Imagine the T upside down as I don't know how to type the perpendicular sign)

Maybe you can create a plane too but somehow I am lost with this 6×2 matrix.

Answer 1

So you have problems to compute the kernel? I assume $A$ should be $$ A=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & -1 & -1 & 1 & 1 \end{pmatrix} $$ So you have to solve $Ax=0$ for $x=(x_1,x_2,x_3,x_4,x_5,x_6)^T$. First, you see $$ Ax=0 \Leftrightarrow \begin{pmatrix}1&1&0&0&1&1\\0&0&1&1&0&0\end{pmatrix}x=\begin{pmatrix}0\\0\end{pmatrix} $$ by using Gauß transformations. Now you have to choose $4$ variables free. Let be $a,b,c,d\in\mathbb R$ then $$ \begin{pmatrix}1&1&0&0&1&1\\0&0&1&1&0&0\end{pmatrix}x=\begin{pmatrix}0\\0\end{pmatrix} \Leftrightarrow\begin{pmatrix}1&1&0&0&1&1\\0&1&0&0&0&0\\0&0&1&1&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{pmatrix}x=\begin{pmatrix}0\\a\\0\\b\\c\\d\end{pmatrix} $$ and you get $$ x=\begin{pmatrix}-a-c-d\\a\\-b\\b\\c\\d\end{pmatrix}=a\begin{pmatrix}-1\\1\\0\\0\\0\\0\end{pmatrix}+b\begin{pmatrix}0\\0\\-1\\1\\0\\0\end{pmatrix}+c\begin{pmatrix}-1\\0\\0\\0\\1\\0\end{pmatrix}+d\begin{pmatrix}-1\\0\\0\\0\\0\\1\end{pmatrix} $$ Defining $$ v_1=\begin{pmatrix}-1\\1\\0\\0\\0\\0\end{pmatrix},~v_2=\begin{pmatrix}0\\0\\-1\\1\\0\\0\end{pmatrix},~v_3=\begin{pmatrix}-1\\0\\0\\0\\1\\0\end{pmatrix},~v_4=\begin{pmatrix}-1\\0\\0\\0\\0\\1\end{pmatrix} $$ You get $v_1,v_2,v_3,v_4$ as the base for the kernel of $A$ and you get the orthogonal projection by $$ P:\mathbb R^6\to \mathbb R^6, x\mapsto \sum_{i=1}^4 \langle x,v_i\rangle v_i $$

Finally you can check your orthogonal projection of $u$ into the kernel of $A$:

So you have just to compute $Pu=\langle u,v_1\rangle v_1+\langle u,v_2\rangle v_2+\langle u,v_3\rangle v_3+\langle u,v_4\rangle v_4$. Compute $$\langle u,v_1\rangle=-1,~\langle u,v_2\rangle=3,~\langle u,v_3\rangle=2,~\langle u,v_4\rangle=3$$ and you get $$Pu=(-1)\begin{pmatrix}-1\\1\\0\\0\\0\\0\end{pmatrix}+3\begin{pmatrix}0\\0\\-1\\1\\0\\0\end{pmatrix}+2\begin{pmatrix}-1\\0\\0\\0\\1\\0\end{pmatrix}+3\begin{pmatrix}-1\\0\\0\\0\\0\\1\end{pmatrix}=\begin{pmatrix}-4\\-1\\-3\\3\\2\\3\end{pmatrix}$$