Quickly calculating an orthogonal matrix & Spectral Theorem
I was trying to find a quicker way to identify whether a set of vectors was an orthogonal set or not. I discovered something interesting, but I don't understand how to put this intuition into words.
Let S = a set of 3 vectors, which I combined into a $M_{33}$
$
S
=
begin{bmatrix}
3\1\1
end{bmatrix}
,
begin{bmatrix}
-1\2\1
end{bmatrix}
,
begin{bmatrix}
-.5\-2\3.5
end{bmatrix}
=
begin{bmatrix}
3&-1&-.5\1&2&-2\1&1&3.5
end{bmatrix}
\$
$S^T$*S = D, a diagonal matrix
D =
$begin{bmatrix}
11&0&0\0&6&0\0&0&16.5
end{bmatrix}$
I noticed that this is a pattern between all orthogonal square matrices. I also noticed that this may hold up for non-square matrices as well. It seems that this has something to do with eigenvalues and the spectral theorem, but I only briefly learned how to calculate eigenvalues and eigenvectors just yesterday yesterday, and we aren't going to touch this 'spectral theorem' thing until the end of the semester.
My questions are:
1) If you could sum up briefly, how would you describe what is happening?
2) Does my theory hold up for non-square matrices as well? My proofing skills are not strong enough to walk through this by myself.
linear-algebra
asked Nov 21 '18 at 20:02
Evan Kim
998
|
show 3 more comments
I was trying to find a quicker way to identify whether a set of vectors was an orthogonal set or not. I discovered something interesting, but I don't understand how to put this intuition into words.
Let S = a set of 3 vectors, which I combined into a $M_{33}$
$
S
=
begin{bmatrix}
3\1\1
end{bmatrix}
,
begin{bmatrix}
-1\2\1
end{bmatrix}
,
begin{bmatrix}
-.5\-2\3.5
end{bmatrix}
=
begin{bmatrix}
3&-1&-.5\1&2&-2\1&1&3.5
end{bmatrix}
\$
$S^T$*S = D, a diagonal matrix
D =
$begin{bmatrix}
11&0&0\0&6&0\0&0&16.5
end{bmatrix}$
I noticed that this is a pattern between all orthogonal square matrices. I also noticed that this may hold up for non-square matrices as well. It seems that this has something to do with eigenvalues and the spectral theorem, but I only briefly learned how to calculate eigenvalues and eigenvectors just yesterday yesterday, and we aren't going to touch this 'spectral theorem' thing until the end of the semester.
My questions are:
1) If you could sum up briefly, how would you describe what is happening?
2) Does my theory hold up for non-square matrices as well? My proofing skills are not strong enough to walk through this by myself.
linear-algebra
asked Nov 21 '18 at 20:02
Evan Kim
998
Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35
|
show 3 more comments
I was trying to find a quicker way to identify whether a set of vectors was an orthogonal set or not. I discovered something interesting, but I don't understand how to put this intuition into words.
Let S = a set of 3 vectors, which I combined into a $M_{33}$
$
S
=
begin{bmatrix}
3\1\1
end{bmatrix}
,
begin{bmatrix}
-1\2\1
end{bmatrix}
,
begin{bmatrix}
-.5\-2\3.5
end{bmatrix}
=
begin{bmatrix}
3&-1&-.5\1&2&-2\1&1&3.5
end{bmatrix}
\$
$S^T$*S = D, a diagonal matrix
D =
$begin{bmatrix}
11&0&0\0&6&0\0&0&16.5
end{bmatrix}$
I noticed that this is a pattern between all orthogonal square matrices. I also noticed that this may hold up for non-square matrices as well. It seems that this has something to do with eigenvalues and the spectral theorem, but I only briefly learned how to calculate eigenvalues and eigenvectors just yesterday yesterday, and we aren't going to touch this 'spectral theorem' thing until the end of the semester.
My questions are:
1) If you could sum up briefly, how would you describe what is happening?
2) Does my theory hold up for non-square matrices as well? My proofing skills are not strong enough to walk through this by myself.
linear-algebra
asked Nov 21 '18 at 20:02
Evan Kim
998
I was trying to find a quicker way to identify whether a set of vectors was an orthogonal set or not. I discovered something interesting, but I don't understand how to put this intuition into words.
Let S = a set of 3 vectors, which I combined into a $M_{33}$
$
S
=
begin{bmatrix}
3\1\1
end{bmatrix}
,
begin{bmatrix}
-1\2\1
end{bmatrix}
,
begin{bmatrix}
-.5\-2\3.5
end{bmatrix}
=
begin{bmatrix}
3&-1&-.5\1&2&-2\1&1&3.5
end{bmatrix}
\$
$S^T$*S = D, a diagonal matrix
D =
$begin{bmatrix}
11&0&0\0&6&0\0&0&16.5
end{bmatrix}$
I noticed that this is a pattern between all orthogonal square matrices. I also noticed that this may hold up for non-square matrices as well. It seems that this has something to do with eigenvalues and the spectral theorem, but I only briefly learned how to calculate eigenvalues and eigenvectors just yesterday yesterday, and we aren't going to touch this 'spectral theorem' thing until the end of the semester.
My questions are:
1) If you could sum up briefly, how would you describe what is happening?
2) Does my theory hold up for non-square matrices as well? My proofing skills are not strong enough to walk through this by myself.
linear-algebra
linear-algebra
asked Nov 21 '18 at 20:02
Evan Kim
998
asked Nov 21 '18 at 20:02
Evan Kim
998
asked Nov 21 '18 at 20:02
Evan Kim
998
asked Nov 21 '18 at 20:02
Evan Kim
998
asked Nov 21 '18 at 20:02
Evan Kim
998
998
Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35
|
show 3 more comments
Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35
Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35
|
show 3 more comments
1 Answer
1
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oldest
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Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.
Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.
answered Nov 21 '18 at 21:48
amd
29.2k21050
add a comment |
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Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.
Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.
answered Nov 21 '18 at 21:48
amd
29.2k21050
add a comment |
Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.
Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.
answered Nov 21 '18 at 21:48
amd
29.2k21050
add a comment |
Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.
Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.
answered Nov 21 '18 at 21:48
amd
29.2k21050
Let’s go back to the definition of matrix multiplication. The elements of the product $C=AB$ are given by the formula $c_{ij}=sum_k a_{ik}b_{kj}$. That is, the $ij$-th element of the product of two matrices is the dot product of the $i$th row of the first matrix with the $j$th row of the second.
Now, the rows of $S^T$ are the columns of $S$, so $S^TS$ consists of all of the pairwise dot products of the columns of $S$. Thus, its diagonal elements are the squares of the norms of those columns, and the off-diagonal elements are zero iff the corresponding pair of columns is orthogonal. None of this requires that $S$ be square, but the product $S^TS$ will always be a square matrix.
answered Nov 21 '18 at 21:48
amd
29.2k21050
answered Nov 21 '18 at 21:48
amd
29.2k21050
answered Nov 21 '18 at 21:48
amd
29.2k21050
answered Nov 21 '18 at 21:48
amd
29.2k21050
29.2k21050
add a comment |
add a comment |
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Is your claim that, if you construct a matrix $S$ out of orthogonal vectors as above, you'll find that $S^T S$ is a diagonal matrix? (If so, you might want to state that explicitly.)
– aghostinthefigures
Nov 21 '18 at 20:35
Be careful with your terminology: in usual usage, an “orthogonal” matrix $M$ is one for which $M^TM=MM^T=I$, which is a stronger condition than simply having orthogonal columns.
– amd
Nov 21 '18 at 20:57
Think about what the elements of $S^TS$ are in terms of the columns of $S$.
– amd
Nov 21 '18 at 20:58
@amd That would be "orthonormal", wouldn't it?
– user58697
Nov 21 '18 at 21:34
@user58697 No. It’s an unfortunate bit of terminology. The rows/columns of an orthogonal matrix form an orthonormal set of vectors.
– amd
Nov 21 '18 at 21:35