# Find the orthogonal projection of vector $x$ on the space of $\{a_1,a_2,a_3\}$

Find the orthogonal projection of vector $x$ on the space of $\{a_1,a_2,a_3\}$.

$x = [4,-1,-3,4]^T, a_3 = [1,0,0,3]^T , a_2 = [1,2,2,-1]^T , a_3 = [1,1,1,1]^T$

Well, $x,a_1,a_2,a_3$ are given in the question. I want to ask how can I approach this question.

1. I use that $P_{Col(A)}x = Ax$ and it is really straight forward.

2. I'm not sure how to implement this but, I do know that every space/subspace has a basis, I check if the given matrix $A$ columns are linearly independent. and if yes, I somehow need to continue from there. maybe Make this basis orthonormal and somehow use fourier coeffcients.

• Is there a mistake? The $a_1$, $a_2$ and $a_3$ are linealy dependent. – Fly by Night Mar 25 '16 at 12:40
• Fourier coefficients??? I don't see any exponential or trigonometric functions here... – Marc van Leeuwen Mar 25 '16 at 12:41
• @Marc van Leeuwen, Fourier coefficients refer to $\langle x, e_i\rangle$ where $\{e_i\}$ is orthonormed basis in Hilbert space. Trigonometric functions are just a particular case. – Ennar Mar 25 '16 at 12:44
• Note that $a_1+a_2=2a_3$: your vectors are linearly dependent. – Marc van Leeuwen Mar 25 '16 at 14:22

Deduce an orthonormal basis for the subspace first. Then use the standard inner product: $$p_{\langle a_1,a_2,a_3\rangle}(x)=\langle x, a_1\rangle a_1 +\langle x, a_2\rangle a_2 +\langle x, a_3\rangle a_3.$$
• Bernard, do $a_1,a_2,a_3$ need to form an orthonormal basis before I can use what you wrote? or just be linearly independent? (in other words, do $a_1$ has to be a normalized vector so I can use the formula u wrote – Ilan Aizelman WS Mar 25 '16 at 12:44
You are right, you can use Gram-Schmidt process on your vectors $\{a_1,a_2,a_3\}$ and then find coefficients $\langle x,e_i\rangle$ where $\{e_i\}_i$ is orthonormed basis for $\operatorname{span}\{a_1,a_2,a_3\}$.
• Ennar, what about $Ax = p$? will it work? – Ilan Aizelman WS Mar 25 '16 at 12:47
• Yes, by the uniqueness property of any basis. You will have explicit decomposition of $x = a + b$ where $a\in\operatorname{span}\{a_1,a_2,a_3\}$ and $b\in\operatorname{span}\{a_1,a_2,a_3\}^\perp$. – Ennar Mar 25 '16 at 12:51