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Suppose $y$ is orthogonal to a subspace spanned by the columns of matrix $A$.

My question is, what is the projection of $y$ onto $A$?

I know that the projection of $y$ onto $A$ is the vector in $A$ such that the distance between that vector and $y$ is minimal.

But I think every vector in $A$ in this case attains same distance.

Hence the projection of $y$ onto $A$ is the set of vectors spanned by $A$.

Is my intuition correct?

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  • $\begingroup$ Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $\pmatrix{0 & 0 \\ 1 & 2}$, and $y = \pmatrix{1\\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $\pmatrix{0\\2}$. But the transformation represented by $B = \pmatrix{0 &0 \\ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $\pmatrix{0\\3}$. It just happens to not be an orthogonal projection. $\endgroup$ Jan 23, 2019 at 8:08

2 Answers 2

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The projection of $y$ onto $A$ will be the zero vector. For any other vector $v\in A$ you will have $||y-v|| = \sqrt{||y||^2+||v||^2} > ||y||.$

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Zero.

The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{\perp}$.

That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.

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