If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$, why?
$begingroup$
In the lecture notes we have a fact:
If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$
Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?
Thank you
linear-algebra norm orthogonal-matrices
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
$endgroup$
add a comment |
$begingroup$
In the lecture notes we have a fact:
If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$
Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?
Thank you
linear-algebra norm orthogonal-matrices
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
$endgroup$
add a comment |
$begingroup$
In the lecture notes we have a fact:
If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$
Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?
Thank you
linear-algebra norm orthogonal-matrices
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
$endgroup$
In the lecture notes we have a fact:
If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$
Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?
Thank you
linear-algebra norm orthogonal-matrices
linear-algebra norm orthogonal-matrices
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
asked Dec 19 '18 at 1:47
YohanRothYohanRoth
6471715
6471715
add a comment |
add a comment |
2 Answers
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$begingroup$
The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that
begin{align}
| Ax |^2 &= (Ax)^T Ax \
&= x^T A^T A x \
&= x^T x \
&= | x |^2.
end{align}
Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = sum_i x_i u_i$, then
begin{align}
|y|^2&= sum_i | x_i u_i |^2 qquad text{(by Pythagorean theorem)} \
&= sum_i x_i^2 | u_i |^2 \
&= sum_i x_i^2 \
&= | x |^2.
end{align}
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
$endgroup$
add a comment |
$begingroup$
Since $A$ has orthonormal columns, $A^TA=I$.
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that
begin{align}
| Ax |^2 &= (Ax)^T Ax \
&= x^T A^T A x \
&= x^T x \
&= | x |^2.
end{align}
Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = sum_i x_i u_i$, then
begin{align}
|y|^2&= sum_i | x_i u_i |^2 qquad text{(by Pythagorean theorem)} \
&= sum_i x_i^2 | u_i |^2 \
&= sum_i x_i^2 \
&= | x |^2.
end{align}
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
$endgroup$
add a comment |
$begingroup$
The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that
begin{align}
| Ax |^2 &= (Ax)^T Ax \
&= x^T A^T A x \
&= x^T x \
&= | x |^2.
end{align}
Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = sum_i x_i u_i$, then
begin{align}
|y|^2&= sum_i | x_i u_i |^2 qquad text{(by Pythagorean theorem)} \
&= sum_i x_i^2 | u_i |^2 \
&= sum_i x_i^2 \
&= | x |^2.
end{align}
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
$endgroup$
add a comment |
$begingroup$
The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that
begin{align}
| Ax |^2 &= (Ax)^T Ax \
&= x^T A^T A x \
&= x^T x \
&= | x |^2.
end{align}
Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = sum_i x_i u_i$, then
begin{align}
|y|^2&= sum_i | x_i u_i |^2 qquad text{(by Pythagorean theorem)} \
&= sum_i x_i^2 | u_i |^2 \
&= sum_i x_i^2 \
&= | x |^2.
end{align}
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
$endgroup$
The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that
begin{align}
| Ax |^2 &= (Ax)^T Ax \
&= x^T A^T A x \
&= x^T x \
&= | x |^2.
end{align}
Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = sum_i x_i u_i$, then
begin{align}
|y|^2&= sum_i | x_i u_i |^2 qquad text{(by Pythagorean theorem)} \
&= sum_i x_i^2 | u_i |^2 \
&= sum_i x_i^2 \
&= | x |^2.
end{align}
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
edited Dec 19 '18 at 2:06
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
answered Dec 19 '18 at 1:55
littleOlittleO
30.3k648111
30.3k648111
add a comment |
add a comment |
$begingroup$
Since $A$ has orthonormal columns, $A^TA=I$.
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
$endgroup$
add a comment |
$begingroup$
Since $A$ has orthonormal columns, $A^TA=I$.
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
$endgroup$
add a comment |
$begingroup$
Since $A$ has orthonormal columns, $A^TA=I$.
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
$endgroup$
Since $A$ has orthonormal columns, $A^TA=I$.
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
answered Dec 19 '18 at 1:51
C.DingC.Ding
1,4111422
1,4111422
add a comment |
add a comment |
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