Mixing
What is the general method of expressing a basis in terms of another basis of a vector space?
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I want to understand basis of vector spaces more clearly. I know that the basis of a vector space is a set of linearly independent elements that span that vector space. I know that an element in a vector space can be expressed in different bases. Now most of the examples I have seen give a standard vector $ left[ begin {array}{c} b1\ b2end {array} right] $ in terms of a basis $B = [b1,b2]$ that can be expressed in the "standard basis" $S = { left[ begin {array}{c} 1\ 0end {array} right], left[ begin {array}{c} 0\ 1end {array} right] }
$.
From here follows that if I have a vector $left[ begin {array}{c} 3\ 5end {array} right]$ expressed in terms of $B$ then the change of basis matrix $C$ is the matrix with $b1,b2$ as its columns, therefore in the standard basis $left[ begin {array}{c} 3\ 5end {array} right]$ would be expressed as:
$$C left[ begin {array}{c} 3\ 5end {array} right]$$.
Now I want to know that if I have a vector say expressed in a basis $A=[a1,a2]$ how can I write that in terms of the basis $B$? In other words, how does one obtain the change of basis matrix for one basis to another other than the "standard basis"?
I have used two dimensional basis, but obviously I am interested in the general n-dimensional case.
linear-algebra linear-transformations change-of-basis hamel-basis
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
$endgroup$
add a comment |
$begingroup$
I want to understand basis of vector spaces more clearly. I know that the basis of a vector space is a set of linearly independent elements that span that vector space. I know that an element in a vector space can be expressed in different bases. Now most of the examples I have seen give a standard vector $ left[ begin {array}{c} b1\ b2end {array} right] $ in terms of a basis $B = [b1,b2]$ that can be expressed in the "standard basis" $S = { left[ begin {array}{c} 1\ 0end {array} right], left[ begin {array}{c} 0\ 1end {array} right] }
$.
From here follows that if I have a vector $left[ begin {array}{c} 3\ 5end {array} right]$ expressed in terms of $B$ then the change of basis matrix $C$ is the matrix with $b1,b2$ as its columns, therefore in the standard basis $left[ begin {array}{c} 3\ 5end {array} right]$ would be expressed as:
$$C left[ begin {array}{c} 3\ 5end {array} right]$$.
Now I want to know that if I have a vector say expressed in a basis $A=[a1,a2]$ how can I write that in terms of the basis $B$? In other words, how does one obtain the change of basis matrix for one basis to another other than the "standard basis"?
I have used two dimensional basis, but obviously I am interested in the general n-dimensional case.
linear-algebra linear-transformations change-of-basis hamel-basis
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
$endgroup$
add a comment |
$begingroup$
I want to understand basis of vector spaces more clearly. I know that the basis of a vector space is a set of linearly independent elements that span that vector space. I know that an element in a vector space can be expressed in different bases. Now most of the examples I have seen give a standard vector $ left[ begin {array}{c} b1\ b2end {array} right] $ in terms of a basis $B = [b1,b2]$ that can be expressed in the "standard basis" $S = { left[ begin {array}{c} 1\ 0end {array} right], left[ begin {array}{c} 0\ 1end {array} right] }
$.
From here follows that if I have a vector $left[ begin {array}{c} 3\ 5end {array} right]$ expressed in terms of $B$ then the change of basis matrix $C$ is the matrix with $b1,b2$ as its columns, therefore in the standard basis $left[ begin {array}{c} 3\ 5end {array} right]$ would be expressed as:
$$C left[ begin {array}{c} 3\ 5end {array} right]$$.
Now I want to know that if I have a vector say expressed in a basis $A=[a1,a2]$ how can I write that in terms of the basis $B$? In other words, how does one obtain the change of basis matrix for one basis to another other than the "standard basis"?
I have used two dimensional basis, but obviously I am interested in the general n-dimensional case.
linear-algebra linear-transformations change-of-basis hamel-basis
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
$endgroup$
I want to understand basis of vector spaces more clearly. I know that the basis of a vector space is a set of linearly independent elements that span that vector space. I know that an element in a vector space can be expressed in different bases. Now most of the examples I have seen give a standard vector $ left[ begin {array}{c} b1\ b2end {array} right] $ in terms of a basis $B = [b1,b2]$ that can be expressed in the "standard basis" $S = { left[ begin {array}{c} 1\ 0end {array} right], left[ begin {array}{c} 0\ 1end {array} right] }
$.
From here follows that if I have a vector $left[ begin {array}{c} 3\ 5end {array} right]$ expressed in terms of $B$ then the change of basis matrix $C$ is the matrix with $b1,b2$ as its columns, therefore in the standard basis $left[ begin {array}{c} 3\ 5end {array} right]$ would be expressed as:
$$C left[ begin {array}{c} 3\ 5end {array} right]$$.
Now I want to know that if I have a vector say expressed in a basis $A=[a1,a2]$ how can I write that in terms of the basis $B$? In other words, how does one obtain the change of basis matrix for one basis to another other than the "standard basis"?
I have used two dimensional basis, but obviously I am interested in the general n-dimensional case.
linear-algebra linear-transformations change-of-basis hamel-basis
linear-algebra linear-transformations change-of-basis hamel-basis
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
edited Jan 6 at 19:05
José Carlos Santos
157k22126227
157k22126227
asked Jan 6 at 18:42
daljit97daljit97
176111
asked Jan 6 at 18:42
daljit97daljit97
176111
asked Jan 6 at 18:42
daljit97daljit97
176111
176111
add a comment |
add a comment |
1 Answer
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Express each element of $B$ as a linear combination of elements of $A$. More precisely, if $B={b_1,ldots,b_n}$, if $A={a_1,ldots,a_n}$, and if, for each $iin{1,2,ldots,n}$,$$b_i=c_{1i}a_1+c_{2i}a_2+cdots+c_{ni}a_n,$$then$$C=begin{bmatrix}c_{11}&ldots&c_{1n}\vdots&ddots&vdots\c_{n1}&ldots&c_{nn}end{bmatrix}.$$
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Express each element of $B$ as a linear combination of elements of $A$. More precisely, if $B={b_1,ldots,b_n}$, if $A={a_1,ldots,a_n}$, and if, for each $iin{1,2,ldots,n}$,$$b_i=c_{1i}a_1+c_{2i}a_2+cdots+c_{ni}a_n,$$then$$C=begin{bmatrix}c_{11}&ldots&c_{1n}\vdots&ddots&vdots\c_{n1}&ldots&c_{nn}end{bmatrix}.$$
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
add a comment |
$begingroup$
Express each element of $B$ as a linear combination of elements of $A$. More precisely, if $B={b_1,ldots,b_n}$, if $A={a_1,ldots,a_n}$, and if, for each $iin{1,2,ldots,n}$,$$b_i=c_{1i}a_1+c_{2i}a_2+cdots+c_{ni}a_n,$$then$$C=begin{bmatrix}c_{11}&ldots&c_{1n}\vdots&ddots&vdots\c_{n1}&ldots&c_{nn}end{bmatrix}.$$
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
add a comment |
$begingroup$
Express each element of $B$ as a linear combination of elements of $A$. More precisely, if $B={b_1,ldots,b_n}$, if $A={a_1,ldots,a_n}$, and if, for each $iin{1,2,ldots,n}$,$$b_i=c_{1i}a_1+c_{2i}a_2+cdots+c_{ni}a_n,$$then$$C=begin{bmatrix}c_{11}&ldots&c_{1n}\vdots&ddots&vdots\c_{n1}&ldots&c_{nn}end{bmatrix}.$$
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Express each element of $B$ as a linear combination of elements of $A$. More precisely, if $B={b_1,ldots,b_n}$, if $A={a_1,ldots,a_n}$, and if, for each $iin{1,2,ldots,n}$,$$b_i=c_{1i}a_1+c_{2i}a_2+cdots+c_{ni}a_n,$$then$$C=begin{bmatrix}c_{11}&ldots&c_{1n}\vdots&ddots&vdots\c_{n1}&ldots&c_{nn}end{bmatrix}.$$
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 6 at 19:04
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
add a comment |
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
yes that makes sense, but the method to express each element of $B$ as a linear combination of $A$ is just calculating the expression by "intuition" and "brute force"? Or there is a more efficient/faster way?
$endgroup$
– daljit97
Jan 6 at 19:09
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
$begingroup$
In certain specific situations, there may be a shortcut, but in general you do it by hand.
$endgroup$
– José Carlos Santos
Jan 6 at 19:12
add a comment |
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