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Find the (projection) matrix which projects every vector in $R_{3}$ onto the subspace given by S = span {(2, 4, 6),(3, 6, 9),(1, 1, −1)}.

I have been looking at this problem for hours. So far, from my understanding, the projection matrix is:

$P=A(A^{T}A)^{-1}A^{T}$

So what I did was letting A equal:

A=$\begin{bmatrix}2 & 3 &1\\4 & 6 & 1\\6 & 9 & -1\end{bmatrix}$

Then I solved for $A^{T}A=\begin{bmatrix}56 & 84 &12\\84 & 126 & 18\\0 & 0 & 3\end{bmatrix}$

But, then I got, $(A^{T}A)^{-1}=0$

Does this mean that the projection matrix is just zero? The given vectors are not linearly independent right? So its impossible to even do this with my steps because A is not linearly independent. Is there another way to solve this, with the nonindependent vectors? I feel that I did something wrong given the simple answer...

slydez
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    You computed the inverse incorrectly; dividing by zero does not give you zero. Also, this formula only works if $A^TA$ is invertible, which only happens when $A$ has linearly independent columns. – Ben Grossmann Nov 27 '16 at 07:18
  • Also, you computed $A^TA$ incorrectly. $A^TA$ is always symmetric, and each entry is the dot product of two columns from $A$. – Ben Grossmann Nov 27 '16 at 07:22
  • Oh its dividing. Wait...now I have a bigger problem. How do I go about the 1/0 of the determinant? Is it correct that A does not have linearly independent columns by rref having only 2 pivots? – slydez Nov 27 '16 at 07:26
  • @Omnomnomnom In the case in which $A^T A$ is not invertible, does using the generalized inverse still give a projector on the subspace? I'm having a hard time understanding this. (Yesterday I asked a related question https://math.stackexchange.com/questions/3519005/projector-on-subspace-spanned-by-linearly-dependent-vectors ) – justmyfault Jan 23 '20 at 08:26
  • @justmyfault If you mean the Moore-Penrose generalized inverse, then yes. – Ben Grossmann Jan 23 '20 at 08:28
  • @Omnomnomnom yes I actually meant the Moore-Penrose pseudoinverse, thanks a lot! Then I guess this holds also for general Hilbert spaces (when the matrix $A^T A$ is replaced by the Gram matrix, see my other question if you have some spare time). By the way, thanks a lot for your contribution to this site. I'm a physicist who has some holes in his linear algebra, and I found many of your answers very useful in the past years :) – justmyfault Jan 23 '20 at 08:31
  • @justmyfault Thanks for saying so, that's nice to hear. – Ben Grossmann Jan 23 '20 at 08:32

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Hint: Find a basis for $S$, and make the vectors of your basis the columns of the matrix $A$. Then your formula will work correctly.

Ben Grossmann
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    Okay. So I found the rref and got the basis to be $\begin{bmatrix}2 & 1\4 & 1\6 & -1\end{bmatrix}$. Is this basis correct?

    If so, I then found $A^{T}A$ to equal $\begin{bmatrix}56 & 0\0 & 3\end{bmatrix}$. Then got the inverse of that matrix to be 1/168$\begin{bmatrix}3 & 0\0 & 56\end{bmatrix}$. After that, I plugged and chugged using the projection matrix formula. Are these good?

     – slydez Nov 27 '16 at 08:18
  • Looks good to me – Ben Grossmann Nov 27 '16 at 12:34