Mixing
how is DFT the change of basis operator?
I understand that DFT are the coefficients when we write a vector z with respect to the Fourier basis.But the following statements are giving me a vague picture about the idea but not very clear,they are
1)the DFT is the change of basis operator that converts from euclidean basis to the Fourier basis.
2)Fourier inversion formula is the change of basis formula for the Fourier basis.
So how are the change of basis operator and DFT related ?
Can somebody help me with an idea.Thanks in advance.
wavelets
asked Nov 20 '18 at 3:45
math math
184
add a comment |
I understand that DFT are the coefficients when we write a vector z with respect to the Fourier basis.But the following statements are giving me a vague picture about the idea but not very clear,they are
1)the DFT is the change of basis operator that converts from euclidean basis to the Fourier basis.
2)Fourier inversion formula is the change of basis formula for the Fourier basis.
So how are the change of basis operator and DFT related ?
Can somebody help me with an idea.Thanks in advance.
wavelets
asked Nov 20 '18 at 3:45
math math
184
Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15
add a comment |
I understand that DFT are the coefficients when we write a vector z with respect to the Fourier basis.But the following statements are giving me a vague picture about the idea but not very clear,they are
1)the DFT is the change of basis operator that converts from euclidean basis to the Fourier basis.
2)Fourier inversion formula is the change of basis formula for the Fourier basis.
So how are the change of basis operator and DFT related ?
Can somebody help me with an idea.Thanks in advance.
wavelets
asked Nov 20 '18 at 3:45
math math
184
I understand that DFT are the coefficients when we write a vector z with respect to the Fourier basis.But the following statements are giving me a vague picture about the idea but not very clear,they are
1)the DFT is the change of basis operator that converts from euclidean basis to the Fourier basis.
2)Fourier inversion formula is the change of basis formula for the Fourier basis.
So how are the change of basis operator and DFT related ?
Can somebody help me with an idea.Thanks in advance.
wavelets
wavelets
asked Nov 20 '18 at 3:45
math math
184
asked Nov 20 '18 at 3:45
math math
184
asked Nov 20 '18 at 3:45
math math
184
asked Nov 20 '18 at 3:45
math math
184
asked Nov 20 '18 at 3:45
math math
184
184
Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15
add a comment |
Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15
Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15
add a comment |
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Your statement 1 states the exact relationship: the DFT is the change of basis operator from the standard basis to the discrete Fourier basis. The discrete Fourier transform takes a vector $x$ in $mathbb C^n$ as input and returns the coordinate vector of $x$ with respect to the discrete Fourier basis as output.
– littleO
Nov 20 '18 at 3:56
yeah,got it. Likewise how can we interpret the second statement ?
– math math
Nov 20 '18 at 7:09
The inverse DFT is the change of basis operator from the discrete Fourier basis to the standard basis. This linear algebra viewpoint of the DFT is nice because once we understand the idea of a change of basis operator, and once we know what the discrete Fourier basis is (a basis of eigenvectors for the cyclic shift operator), we can immediately understand what the DFT is.
– littleO
Nov 20 '18 at 7:15