Why does an orthogonal matrix have the property $Q^T Q = Q Q^T = I$?
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According to Wikipedia, an orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. It also says that this definition is equivalent to saying that an orthogonal matrix $Q$ is a matrix for which $Q^T Q = Q Q^T = I$, where $I$ is the identity matrix. Why are these definitions equivalent?
linear-algebra matrices
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
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add a comment |
$begingroup$
According to Wikipedia, an orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. It also says that this definition is equivalent to saying that an orthogonal matrix $Q$ is a matrix for which $Q^T Q = Q Q^T = I$, where $I$ is the identity matrix. Why are these definitions equivalent?
linear-algebra matrices
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
$endgroup$
3
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
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– Michael Burr
Dec 21 '18 at 14:07
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See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46
add a comment |
$begingroup$
According to Wikipedia, an orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. It also says that this definition is equivalent to saying that an orthogonal matrix $Q$ is a matrix for which $Q^T Q = Q Q^T = I$, where $I$ is the identity matrix. Why are these definitions equivalent?
linear-algebra matrices
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
$endgroup$
According to Wikipedia, an orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. It also says that this definition is equivalent to saying that an orthogonal matrix $Q$ is a matrix for which $Q^T Q = Q Q^T = I$, where $I$ is the identity matrix. Why are these definitions equivalent?
linear-algebra matrices
linear-algebra matrices
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
asked Dec 21 '18 at 14:04
K. ClaessonK. Claesson
17110
17110
3
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
$endgroup$
– Michael Burr
Dec 21 '18 at 14:07
$begingroup$
See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46
add a comment |
3
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
$endgroup$
– Michael Burr
Dec 21 '18 at 14:07
$begingroup$
See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46
3
3
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
$endgroup$
– Michael Burr
Dec 21 '18 at 14:07
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
$endgroup$
– Michael Burr
Dec 21 '18 at 14:07
$begingroup$
See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46
$begingroup$
See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46
add a comment |
2 Answers
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The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $ineq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
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add a comment |
$begingroup$
As pwerth noted, in the product $Q^top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
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2 Answers
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2 Answers
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$begingroup$
The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $ineq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $ineq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $ineq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
$endgroup$
The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $ineq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
edited Dec 21 '18 at 14:44
David C. Ullrich
61k43994
61k43994
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
answered Dec 21 '18 at 14:09
pwerthpwerth
3,243417
3,243417
add a comment |
add a comment |
$begingroup$
As pwerth noted, in the product $Q^top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
As pwerth noted, in the product $Q^top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
As pwerth noted, in the product $Q^top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
$endgroup$
As pwerth noted, in the product $Q^top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:27
SvanNSvanN
2,0661422
2,0661422
add a comment |
add a comment |
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3
$begingroup$
If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$.
$endgroup$
– Michael Burr
Dec 21 '18 at 14:07
$begingroup$
See math.stackexchange.com/questions/2028990/…
$endgroup$
– Widawensen
Dec 21 '18 at 14:46