If a vector is orthogonal to a subspace, what is its projection?
$begingroup$
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?
linear-algebra
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
$endgroup$
add a comment |
$begingroup$
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?
linear-algebra
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
$endgroup$
$begingroup$
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.
$endgroup$
– John Hughes
Jan 23 at 8:08
add a comment |
$begingroup$
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?
linear-algebra
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
$endgroup$
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?
linear-algebra
linear-algebra
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:37
The man of your dreamThe man of your dream
23110
23110
$begingroup$
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.
$endgroup$
– John Hughes
Jan 23 at 8:08
add a comment |
$begingroup$
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.
$endgroup$
– John Hughes
Jan 23 at 8:08
$begingroup$
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.
$endgroup$
– John Hughes
Jan 23 at 8:08
$begingroup$
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.
$endgroup$
– John Hughes
Jan 23 at 8:08
add a comment |
2 Answers
2
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$begingroup$
The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
$endgroup$
add a comment |
$begingroup$
The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
$endgroup$
add a comment |
$begingroup$
The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
$endgroup$
The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
answered Jan 23 at 6:42
Reiner MartinReiner Martin
3,434414
3,434414
add a comment |
add a comment |
$begingroup$
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.
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
$endgroup$
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.
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
edited Jan 23 at 7:47
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
answered Jan 23 at 7:42
Chris CusterChris Custer
14k3827
14k3827
add a comment |
add a comment |
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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.
$endgroup$
– John Hughes
Jan 23 at 8:08