# Projection matrix onto a subspace

#### cheertcc101

Alright so I am trying to find the projection matrix for the subspace spanned by the vectors

[1] and [2]
[-1] [0]
[1] [1]

I actually have the solution to the problem, it is ...

P = [ 5 1 2 ]
(1/6) [1 5 -2]
[2 -2 2]

Every formula I have found is not for two vectors but a vector and a matrix or something else. I found the formula P = A(ATA)-1AT but that makes no sense for what A could be since I'm given two vectors and not a matrix. I feel like I'm missing something simple.

Thanks

Related Linear and Abstract Algebra News on Phys.org

#### HallsofIvy

Homework Helper
The vectors <1, -1, 1> and <2, 0, 1> are independent so there exist a third vector, independent of both and all 3 form a basis for R3. It should not be hard for you to find such a third vector. Then you want a matrix that maps <1, -1, 1> and <2, 0, 1> to themselves and the third vector to <0, 0, 0>. that gives you 9 equations to solve for the 9 numbers in the matrix. Actually they separate into 3 sets of 3 equations.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving