# Orthogonal projection of the vector $p=(1,0,0,0)$ onto the subspace $W=[(1,-3,0,1),(1,5,2,3),(0,4,1,1),(1,-2,0,4)]$

I got a subspace $$W=[(1,-3,0,1),(1,5,2,3),(0,4,1,1),(1,-2,0,4)]$$

and I want to make an orthogonal projection of a vector $$p=(1,0,0,0)$$ onto $$W$$ and onto the orhhogonal complement of $$W$$.

Well, $$W$$ is $$[w_1,w_2,w_3]=[(1,-3,0,1),(0,4,1,1),(1,-2,0,4)]$$ for example and the orthogonal complement of $$W$$ is $$W' = [(-10,-3,11,1)]$$

So, we can project $$p$$ onto $$W$$ by the formula: $$p_{\| W} = \sum_{i =1}^k \frac{\langle p, w_i\rangle}{\langle w_i, w_i \rangle} w_i = \frac{\langle p , w_1 \rangle}{\langle w_1 , w_1 \rangle} w_1 + \dots + \frac{\langle p , w_k \rangle}{\langle w_k , w_k \rangle} w_k.$$

But I am somehow confused. Do the vectors $$w_i$$ have to be orthogonal to each other before I do the projection with the vector $$p$$? Because $$W$$ is not orthogonal itself. And examples I found contained the orthogonal subspaces already (so no Gramm-Schmidt process was needed).

• Yes, the basis of $W$ must be orthogonalized first. You could alternatively find the projection of $p$ onto $W_\perp$ and subtract that from $p$ to get its projection onto $W$. This should be faster since $W_\perp$ is unidimensional. – Shubham Johri May 17 at 14:55
• Thanks. And do I make the projection onto $W$ and $W'$ separately and then sum them? Or no summation? So the result would be two projected vectors? – Leif May 17 at 15:06
• I saw a similar example with the direct sum of the subspaces so this confuses me too. – Leif May 17 at 15:07
• $$p=p_{||W}+p_{||W_\perp}$$where $p_{||W_\perp}=\dfrac{\langle p, w\rangle}{\langle w, w\rangle} w$ and $w$ is the basis vector of $W_\perp$. – Shubham Johri May 17 at 15:46
• Thank you for making it clearer for me! – Leif May 17 at 16:07

Our space of interest $$W$$ is the column space of the matrix $$A$$ given by $$A=\left[\begin{array}{rrrr} 1 & 1 & 0 & 1 \\ -3 & 5 & 4 & -2 \\ 0 & 2 & 1 & 0 \\ 1 & 3 & 1 & 4 \end{array}\right]$$ A basis of $$W=\operatorname{Col}(A)$$ is given by the columns of $$A$$ corresponding to pivot columns in $$\operatorname{rref}(A)$$. The reduced row echelon form of $$A$$ is $$\operatorname{rref}(A)=\left[\begin{array}{rrrr} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$$ This shows that $$A$$ has rank three and that the vectors \begin{align*} \left\langle1,\,-3,\,0,\,1\right\rangle && \left\langle1,\,5,\,2,\,3\right\rangle && \left\langle1,\,-2,\,0,\,4\right\rangle \end{align*} form a basis of $$W$$.
Now, put these basis vectors into the columns of a matrix $$X = \left[\begin{array}{rrr} 1 & 1 & 1 \\ -3 & 5 & -2 \\ 0 & 2 & 0 \\ 1 & 3 & 4 \end{array}\right]$$ The projection matrix onto $$W$$ is $$P=X(X^\top X)^{-1}X^\top$$ In our case, we end up with $$P=\left[\begin{array}{rrrr} \frac{131}{231} & -\frac{10}{77} & \frac{10}{21} & \frac{10}{231} \\ -\frac{10}{77} & \frac{74}{77} & \frac{1}{7} & \frac{1}{77} \\ \frac{10}{21} & \frac{1}{7} & \frac{10}{21} & -\frac{1}{21} \\ \frac{10}{231} & \frac{1}{77} & -\frac{1}{21} & \frac{230}{231} \end{array}\right]$$ We can now project any vector $$v\in\Bbb R^4$$ onto $$W$$ by computing $$Pv$$.
If you're dead set on using orthonormal bases, then you could take these three basis vectors and apply the Gram-Schmidt algorithm. This gives the new basis \begin{align*} q_1 &= \left\langle\frac{1}{\sqrt{11}},\,-\frac{3}{\sqrt{11}},\,0,\,\frac{1}{\sqrt{11}}\right\rangle & q_2 &= \left\langle\frac{1}{\sqrt{7}},\,\frac{1}{\sqrt{7}},\,\frac{1}{\sqrt{7}},\,\frac{2}{\sqrt{7}}\right\rangle & q_3 &= \left\langle-\frac{1}{\sqrt{3}},\,0,\,-\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle \end{align*} We can then project any $$v\in\Bbb R^4$$ with the formula $$Pv=\langle q_1, v\rangle q_1 + \langle q_2, v\rangle q_2 + \langle q_3, v\rangle q_3$$
• @Leif The sum of the projections of $p$ onto $W$ and $W_\perp$ will just be $p$, as I mentioned. Why do you want to add them? The question only asks for the projections. – Shubham Johri May 17 at 19:01