Let $M$ be a subspace of $\mathbb R^4$ which is spanned by the vectors $v_1 = (1,0,-1,1)$ , $v_2=(0,1,2,1)$. Find the orthogonal complement $M^T$ of $M$ and the orthogonal projections of the vector $v=(4,3,0,1)$ on the subspaces $M,M^T$.
My solution for the first part :
Let $A=[v_1 | v_2]$, $v_1,v_2$ are already in an opened-form and linearly independent and thus solving : $A\vec x=0$ yields : $\{x – z + w = 0, y + 2z + w = 0\}$ which gives you two linearly independent vectors : $v_3,v_4$ : $v_3 = (1,-2,0,1)$,$v_4=(0,-3,1,1)$ and we get that : $M^T = span(v_3,v_4) = \langle v_3,v_4\rangle=\langle (1,-2,0,1),(0,-3,1,1) \rangle.$
Now, I have trouble finding (I do not know in that case) how to work for the orthogonal projection for the second part. Does it have to do with the Gram-Schmidt procedure ? I would appreciate a hint or an explanation on how to work over finding the orthogonal projections generally or in that example. Thanks in advance.
Best Answer
If you have, say a $4$ by $4$ square matrix that's represendted by $A=[c_1,c_2,c_3,c_4]$ where $c_i$'s are colums, then you now that given a column vector $v$, $Av=\sum_i c_iv_i$ where $v_i$ are elements of $v$.
So by finding basese for $M^T$ and $M$ you know basis for whole space since $\mathbb{R}^4=M^T \oplus M$ and to find projection of some vector $v$ onto these spaces you need to represent $v$ in terms of these basis vectors. So you are trying to find $a_i$ such that $[v_1,v_2,v_3,v_4] a = v$ where $a$ is column vector with elements $a_i$ and $v_i$ are bases vectors in column form.
Since you know $v_i$'s are linearly independent you know $A=[v_i]$ is an invertible matrix, and inverse can easily be found using Gaussian elimination - so you can easily find $a$, and you are done