Let $A$ be a matrix. The orthogonal projection of a vector $x$ onto $\operatorname{Col}(A)$ is unique if and only if the columns of $A$ are linearly independent. True or False?
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http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/projections-onto-subspaces/MIT18_06SCF11_Ses2.2sum.pdf it is true because in this case matrix is ivertible and there exist unique representtaion – dato datuashvili Dec 13 '13 at 18:34
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http://math.stackexchange.com/questions/91278/projections-onto-ranges-subspaces – dato datuashvili Dec 13 '13 at 18:36
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Answer: False. The orthogonal projection onto the column space is unique.
For a given inner product, the orthogonal projection of a vector $x \in \mathbb{R}^{N}$ (or $\mathbb{C}^{N}$) onto a linear subspace $M \subseteq \mathbb{R}^{N}$ is unique, but it does depend on the inner-product you use. It doesn't matter whether $M$ is described as linear combinations of a linearly-independent set $S$ of vectors or as linear combinations of a linearly-dependent set $D$ of vectors.
In all cases, an orthogonal projection of $x$ onto the subspace $M$ is defined as a vector $m \in M$ such that $(x-m)\perp M$, which is to say that $(x-m,m')=0$ for all $m' \in M$. There can be at most one such $m$ because, if $m, m' \in M$ satisfy $(x-m)\perp M$ and $(x-m')\perp M$ then $(m-m') \perp M$ which means that $(m-m')\perp(m-m')$, or $\|m-m'\|^{2}=0$, which forces $m=m'$. So $m$ is unique if it exists. The existence of $m$ is a little trickier, but you didn't ask about that.
Disintegrating By Parts
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