Let $A$ be a $2times2$ matrix such that $A^TA = I_2$. [closed]
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How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.
linear-algebra matrices
closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53
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up vote
-1
down vote
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How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.
linear-algebra matrices
closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
1
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.
linear-algebra matrices
How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.
linear-algebra matrices
linear-algebra matrices
edited Nov 23 at 10:34
Yadati Kiran
1,444518
1,444518
asked Nov 23 at 10:23
DwightD
42
42
closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
1
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55
add a comment |
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
1
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
1
1
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55
add a comment |
3 Answers
3
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oldest
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up vote
1
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Well, since A is a $2times 2$ matrix, one way to do this is brute force:
$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$
$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$
From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
add a comment |
up vote
1
down vote
Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are
$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
add a comment |
up vote
1
down vote
If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:
- The first column has norm $1$;
- The second column has norm $1$;
- The columns ar orthogonal.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Well, since A is a $2times 2$ matrix, one way to do this is brute force:
$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$
$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$
From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
add a comment |
up vote
1
down vote
Well, since A is a $2times 2$ matrix, one way to do this is brute force:
$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$
$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$
From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
add a comment |
up vote
1
down vote
up vote
1
down vote
Well, since A is a $2times 2$ matrix, one way to do this is brute force:
$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$
$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$
From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$
Well, since A is a $2times 2$ matrix, one way to do this is brute force:
$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$
$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$
From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$
answered Nov 23 at 10:31
F.Carette
1,20112
1,20112
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
add a comment |
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
Actually you have written $AA^T$.
– Jean-Claude Arbaut
Nov 24 at 9:50
add a comment |
up vote
1
down vote
Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are
$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
add a comment |
up vote
1
down vote
Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are
$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
add a comment |
up vote
1
down vote
up vote
1
down vote
Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are
$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.
Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are
$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.
edited Nov 23 at 10:31
answered Nov 23 at 10:28
Yves Daoust
123k668219
123k668219
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
add a comment |
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
– Jean-Claude Arbaut
Nov 23 at 10:30
add a comment |
up vote
1
down vote
If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:
- The first column has norm $1$;
- The second column has norm $1$;
- The columns ar orthogonal.
add a comment |
up vote
1
down vote
If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:
- The first column has norm $1$;
- The second column has norm $1$;
- The columns ar orthogonal.
add a comment |
up vote
1
down vote
up vote
1
down vote
If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:
- The first column has norm $1$;
- The second column has norm $1$;
- The columns ar orthogonal.
If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:
- The first column has norm $1$;
- The second column has norm $1$;
- The columns ar orthogonal.
edited Nov 23 at 10:46
answered Nov 23 at 10:29
José Carlos Santos
147k22117218
147k22117218
add a comment |
add a comment |
that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27
Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31
See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46
1
Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55