Definition of scalar product in relation to projections
$begingroup$
So if $x in X$ is an element in a vector space $X$, then $forall x in X$:
$$x = e_ix^i$$
where $e_i$ is a basis for $X$. However, I encountered that this is equivalent to:
$$x = e_i (e_i, x)$$
where $(e_i, x)$ is the scalar product between $e_i$ and $x$. Why is the following true? $$x^i = (e_i, x)$$
My lecturer mentioned something about projections, but I don't understand the relation.
linear-algebra vector-spaces vectors linear-transformations
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So if $x in X$ is an element in a vector space $X$, then $forall x in X$:
$$x = e_ix^i$$
where $e_i$ is a basis for $X$. However, I encountered that this is equivalent to:
$$x = e_i (e_i, x)$$
where $(e_i, x)$ is the scalar product between $e_i$ and $x$. Why is the following true? $$x^i = (e_i, x)$$
My lecturer mentioned something about projections, but I don't understand the relation.
linear-algebra vector-spaces vectors linear-transformations
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So if $x in X$ is an element in a vector space $X$, then $forall x in X$:
$$x = e_ix^i$$
where $e_i$ is a basis for $X$. However, I encountered that this is equivalent to:
$$x = e_i (e_i, x)$$
where $(e_i, x)$ is the scalar product between $e_i$ and $x$. Why is the following true? $$x^i = (e_i, x)$$
My lecturer mentioned something about projections, but I don't understand the relation.
linear-algebra vector-spaces vectors linear-transformations
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
$endgroup$
So if $x in X$ is an element in a vector space $X$, then $forall x in X$:
$$x = e_ix^i$$
where $e_i$ is a basis for $X$. However, I encountered that this is equivalent to:
$$x = e_i (e_i, x)$$
where $(e_i, x)$ is the scalar product between $e_i$ and $x$. Why is the following true? $$x^i = (e_i, x)$$
My lecturer mentioned something about projections, but I don't understand the relation.
linear-algebra vector-spaces vectors linear-transformations
linear-algebra vector-spaces vectors linear-transformations
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
asked Dec 5 '18 at 0:21
daljit97daljit97
178111
178111
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Imagine the basis $e_i$ is orthonormal, that is $(e_i, e_j) = delta_{ij}$ ($1$ if $i = j$ or $0$ otherwise). You can expand any vector $x$ in this bases
$$
x = x^1 e_1 + x^2 e_2 + cdots = x^ie_i
$$
Now multiple both sides by some vector of the basis, say the first one
$$
(e^1, x) = x^1underbrace{(e_1, e_1)}_{=1} + x^2underbrace{(e_2, e_1)}_{=0} + x^3underbrace{(e_3, e_1)}_{=0} + cdots = x^1
$$
You can repeat the same experiment for all the other components, and quickly realize that
$$
(e^i, x) = x^i
$$
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine the basis $e_i$ is orthonormal, that is $(e_i, e_j) = delta_{ij}$ ($1$ if $i = j$ or $0$ otherwise). You can expand any vector $x$ in this bases
$$
x = x^1 e_1 + x^2 e_2 + cdots = x^ie_i
$$
Now multiple both sides by some vector of the basis, say the first one
$$
(e^1, x) = x^1underbrace{(e_1, e_1)}_{=1} + x^2underbrace{(e_2, e_1)}_{=0} + x^3underbrace{(e_3, e_1)}_{=0} + cdots = x^1
$$
You can repeat the same experiment for all the other components, and quickly realize that
$$
(e^i, x) = x^i
$$
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
add a comment |
$begingroup$
Imagine the basis $e_i$ is orthonormal, that is $(e_i, e_j) = delta_{ij}$ ($1$ if $i = j$ or $0$ otherwise). You can expand any vector $x$ in this bases
$$
x = x^1 e_1 + x^2 e_2 + cdots = x^ie_i
$$
Now multiple both sides by some vector of the basis, say the first one
$$
(e^1, x) = x^1underbrace{(e_1, e_1)}_{=1} + x^2underbrace{(e_2, e_1)}_{=0} + x^3underbrace{(e_3, e_1)}_{=0} + cdots = x^1
$$
You can repeat the same experiment for all the other components, and quickly realize that
$$
(e^i, x) = x^i
$$
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
$endgroup$
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
add a comment |
$begingroup$
Imagine the basis $e_i$ is orthonormal, that is $(e_i, e_j) = delta_{ij}$ ($1$ if $i = j$ or $0$ otherwise). You can expand any vector $x$ in this bases
$$
x = x^1 e_1 + x^2 e_2 + cdots = x^ie_i
$$
Now multiple both sides by some vector of the basis, say the first one
$$
(e^1, x) = x^1underbrace{(e_1, e_1)}_{=1} + x^2underbrace{(e_2, e_1)}_{=0} + x^3underbrace{(e_3, e_1)}_{=0} + cdots = x^1
$$
You can repeat the same experiment for all the other components, and quickly realize that
$$
(e^i, x) = x^i
$$
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
$endgroup$
Imagine the basis $e_i$ is orthonormal, that is $(e_i, e_j) = delta_{ij}$ ($1$ if $i = j$ or $0$ otherwise). You can expand any vector $x$ in this bases
$$
x = x^1 e_1 + x^2 e_2 + cdots = x^ie_i
$$
Now multiple both sides by some vector of the basis, say the first one
$$
(e^1, x) = x^1underbrace{(e_1, e_1)}_{=1} + x^2underbrace{(e_2, e_1)}_{=0} + x^3underbrace{(e_3, e_1)}_{=0} + cdots = x^1
$$
You can repeat the same experiment for all the other components, and quickly realize that
$$
(e^i, x) = x^i
$$
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
answered Dec 5 '18 at 0:27
caveraccaverac
14.5k31130
14.5k31130
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
add a comment |
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
$begingroup$
So does the basis $e_i$ need to be orthonormal?
$endgroup$
– daljit97
Dec 5 '18 at 1:03
1
1
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
$begingroup$
@daljit97 Yes, but remember that you can always use the Gram-Schmidt algorithm, so it is not such a big thing to assume
$endgroup$
– caverac
Dec 5 '18 at 1:07
add a comment |
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