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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...

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    $\begingroup$ 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. $\endgroup$ – Omnomnomnom Nov 27 '16 at 7:18
  • $\begingroup$ Also, you computed $A^TA$ incorrectly. $A^TA$ is always symmetric, and each entry is the dot product of two columns from $A$. $\endgroup$ – Omnomnomnom Nov 27 '16 at 7:22
  • $\begingroup$ 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? $\endgroup$ – slydez Nov 27 '16 at 7:26
  • $\begingroup$ @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 math.stackexchange.com/questions/3519005/… ) $\endgroup$ – justmyfault Jan 23 at 8:26
  • $\begingroup$ @justmyfault If you mean the Moore-Penrose generalized inverse, then yes. $\endgroup$ – Omnomnomnom Jan 23 at 8:28
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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.

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    $\begingroup$ 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? $\endgroup$ – slydez Nov 27 '16 at 8:18
  • $\begingroup$ Looks good to me $\endgroup$ – Omnomnomnom Nov 27 '16 at 12:34

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