# Finding The Orthogonal Projection of a Vector Onto a Subspace

Find the orthogonal projection of the vector $\vec{v}=16(1,0,0,0,0,1,1,1,1,1)$ onto the subspace spanned by the vectors $\vec{u1}=16(0,1,0,0,0,1,1,1,1,1)$ and $\vec{u2}=16(0,0,1,0,0,1,1,1,1,1)$.

I've done questions like this before in 3-D space, and would usually find a normal to the plane using the cross product, find the projection of $\vec{v}$ on to the plane and then subtract this component from $\vec{v}$ itself:

$$proj_{subspace}(\vec{v})=\vec{v}-proj_{\vec{n}}(\vec{v})$$

However, in 10-D space I have no idea how to proceed. I realise there are many different vectors normal to the subspace and I don't know if there's a specific one to use or if all are equally valid etc. I also noticed that all but the first three components of these vectors are identical, so I thought perhaps I could reduce the problem to 3 dimensions that way, but the projection for the first three components came out as zero, which is incorrect. Any help would be much appreciate, thanks in advance!

By the way, the given answer is: $$proj_{subspace}(\vec{v})=8(0,1,1,0,0,2,2,2,2,2)$$

• First, you find an orthonormal basis of your subspace. I will denote the basis vectors as $a_1,\ a_2$. Then you calculate the projection with $p(v)=<v,a_1>a_1-<v,a_2>a_2$, where $<>$ is the inner product Feb 2, 2016 at 3:24

Let $S=\mathrm{span}\{(0,1,0,0,0,1,1,1,1,1),(0,0,1,0,0,1,1,1,1,1)\}$.

Note that

$\begin{multline}(1,0,0,0,0,1,1,1,1,1)=(1,-\frac{5}{11},-\frac{5}{11},0,0,\frac{1}{11},\frac{1}{11},\frac{1}{11},\frac{1}{11},\frac{1}{11})\\+(0,\frac{5}{11},\frac{5}{11},0,0,\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11})\end{multline}$.

Let $u=(1,-\frac{5}{11},-\frac{5}{11},0,0,\frac{1}{11},\frac{1}{11},\frac{1}{11},\frac{1}{11},\frac{1}{11})$ and $v=(0,\frac{5}{11},\frac{5}{11},0,0,\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11})$.

Now, $\langle u,(0,1,0,0,0,1,1,1,1,1)\rangle=0$ and $\langle u,(0,0,1,0,0,1,1,1,1,1)\rangle$. Thus $u\in S^\perp$.

On the other hand, $\begin{multline}(0,\frac{5}{11},\frac{5}{11},0,0,\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11})=\frac{5}{11}(0,1,0,0,0,1,1,1,1,1)\\+\frac{5}{11}(0,0,1,0,0,1,1,1,1,1)\end{multline}$

Thus $(0,\frac{5}{11},\frac{5}{11},0,0,\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11})\in S$.

Then $P[16(1,0,0,0,0,1,1,1,1,1)]=16P(1,0,0,0,0,1,1,1,1,1)=16(0,\frac{5}{11},\frac{5}{11},0,0,\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11},\frac{10}{11})$

Let $\{v_1,\dotsc,v_m\}$ be a basis for a subspace $V$ of $\Bbb R^n$. The matrix of the orthogonal projection onto $V$ is $$P=A\left(A^\top A\right)^{-1}A^\top$$ where $A$ is the matrix whose columns are $\{v_1,\dotsc,v_m\}$. A nice discussion of this can be found in these notes.

In our case we have $$A= \left[\begin{array}{rr} 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 1 & 1 \\ 1 & 1 \\ 1 & 1 \\ 1 & 1 \\ 1 & 1 \end{array}\right]$$ Thus $$P = \left[\begin{array}{rrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{6}{11} & -\frac{5}{11} & 0 & 0 & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} \\ 0 & -\frac{5}{11} & \frac{6}{11} & 0 & 0 & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{11} & \frac{1}{11} & 0 & 0 & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} \\ 0 & \frac{1}{11} & \frac{1}{11} & 0 & 0 & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} \\ 0 & \frac{1}{11} & \frac{1}{11} & 0 & 0 & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} \\ 0 & \frac{1}{11} & \frac{1}{11} & 0 & 0 & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} \\ 0 & \frac{1}{11} & \frac{1}{11} & 0 & 0 & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} & \frac{2}{11} \end{array}\right]$$ Now, the projection of $v=16\cdot(1,0,0,0,0,1,1,1,1,1)$ is $$Pv = \begin{bmatrix} 0 \\ \frac{80}{11} \\ \frac{80}{11} \\ 0 \\ 0 \\ \frac{160}{11} \\ \frac{160}{11} \\ \frac{160}{11} \\ \frac{160}{11} \\ \frac{160}{11} \end{bmatrix}$$ Note that I'm getting a slightly different answer than you, so you may want to double check!