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I'm going through the article on orthogonal projections on Wikipedia and it reads

If $u_1,\dots,u_k$ is a basis and $A$ is matrix with these vectors as columns then the projection is $$P_A=A(A^TA)^{-1}A^T$$

What if $u_1,\dots,u_k$ are not linearly independent? Does this not still form a projection matrix onto the column space of $A$? Is it because then $(A^TA)^{-1}$ fails to be invertible?

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Note that it says "If $u_1, ..., u_k$ is a (not necessarily orthonormal) basis", meaning that $u_1, ..., u_k$ still must be a basis for this to hold. –  Suugaku Apr 16 '13 at 6:43
    
@Suugaku No I know, I just don't know why it has to be a basis –  crf Apr 16 '13 at 6:45
    
If it's not a basis, then $(A^T A)^{-1}$ doesn't make sense. Or does it? Check! Can $A^T A$ be invertible otherwise? –  Suugaku Apr 16 '13 at 6:54
    
@crf : it is not hard to show $A$ and $A^T A$ have the same null space. If $A$ has linearly independent columns, then by the rank-nullity theorem, the null space of $A$ is trivial, so $A^T A$ is nonsingular. If $A$ does not have linearly independent columns (say, $A$ is a matrix of zeros), then $A^T A$ will be singular. –  Stefan Smith Apr 16 '13 at 23:29

2 Answers 2

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Yes. If the columns of $A$ are not linealy independent then $\text{det} (A)=0$, thus $\text{det}(A\cdot A^T)=\text{det}(A)^2=0$ and so $A\cdot A^T$ is not invertible.

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At a crucial step in the derivation of the of the projection matrix onto the column space of $A$ we take the inverse of $AA^t$. But how do we know this matrix is invertible? It is invertible, because the columns of A, the vectors $u_{i}$ , were assumed to be linearly independent.

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