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Matrix Representions [on hold]

Matrix representation of linear transformationThe real Matrix of change of basis. Not really. Only in $mathbb{R}^n$.Change Bases of Linear TransformationNeed help with a better understanding of change of basis matrix and corresponding theoremsWhat am I doing wrong? - Change of basis matrixFinding bases such that the matrix representation is a block matrix where one submatrix is the identity matrixFind the Transformation matrix from M(nxn) space to ROrthogonal Basis Transformation MatrixGive a transformation matrix with respect two bases, find the basesUse a change of basis matrix to compute the matrix T.

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Let $ V={ fintext{func} (mathbb{R}, mathbb{C}) : f(t) =alpha cos (t) +beta sin(t), alpha, beta in mathbb{C} } $.

(a) show that cos$(t) $, sin$(t) $, and exp(-it), exp(it) both form a basis for $ V$.

(b) Find the change of basis matrix.

(c) Find the matrix representation of $ D:Vto V $ with respect to both bases and check that the change of basis matrix gives the correct relationship between these two matrices. Where $D$ is the derivative function.

I am trying to get insight into this problem from linear algebra. It appears to have some intersection with Fourier analysis, which I don't know about. Any help is appreciated.

linear-algebra fourier-analysis

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edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

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• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, Travis, Leucippus, Eevee Trainer, Alex Provost

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0

$begingroup$

Let $ V={ fintext{func} (mathbb{R}, mathbb{C}) : f(t) =alpha cos (t) +beta sin(t), alpha, beta in mathbb{C} } $.

(a) show that cos$(t) $, sin$(t) $, and exp(-it), exp(it) both form a basis for $ V$.

(b) Find the change of basis matrix.

(c) Find the matrix representation of $ D:Vto V $ with respect to both bases and check that the change of basis matrix gives the correct relationship between these two matrices. Where $D$ is the derivative function.

I am trying to get insight into this problem from linear algebra. It appears to have some intersection with Fourier analysis, which I don't know about. Any help is appreciated.

linear-algebra fourier-analysis

share|cite|improve this question

edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

307313

$endgroup$

put on hold as off-topic by David Hill, Travis, Leucippus, Eevee Trainer, Alex Provost yesterday

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, Travis, Leucippus, Eevee Trainer, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please ask only one question per post. Having multiple questions in the same post is discouraged and such posts may be put on hold, see meta.
  $endgroup$
  – Alex Provost
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let $ V={ fintext{func} (mathbb{R}, mathbb{C}) : f(t) =alpha cos (t) +beta sin(t), alpha, beta in mathbb{C} } $.

(a) show that cos$(t) $, sin$(t) $, and exp(-it), exp(it) both form a basis for $ V$.

(b) Find the change of basis matrix.

(c) Find the matrix representation of $ D:Vto V $ with respect to both bases and check that the change of basis matrix gives the correct relationship between these two matrices. Where $D$ is the derivative function.

I am trying to get insight into this problem from linear algebra. It appears to have some intersection with Fourier analysis, which I don't know about. Any help is appreciated.

linear-algebra fourier-analysis

share|cite|improve this question

edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

307313

$endgroup$

share|cite|improve this question

edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

307313

share|cite|improve this question

edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

307313

edited yesterday

user85503

5031415

edited yesterday

user85503

5031415

asked yesterday

Jerome TurnerJerome Turner

307313

asked yesterday

Jerome TurnerJerome Turner

307313

1 Answer
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Since the vectors are given as "$alpha cos(x)+ beta sin(x)$" all that you need to do is show that cos(x) and sin(x) are independent. That is, that A cos(x)+ B sin(x)= 0 (for all x) only A= B= 0. And that's pretty close to "trivial". Take x= 0 and $x= pi/2$ and see what happens.

To see that $e^{ix}$ and $e{-ix}$ also form a basis use the fact that $cos(x)= frac{e^{ix}+ e^{ix}}{2}$ and $sin(x)= frac{e^{ix}- e^{-ix}}{2i}$.

For the "change of basis matrix" you want a matrix that multiplied by a column matrix with coefficients from the first basis gives the column matrix with coefficients from the second basis. The first "basis vector", cos(x)= 1(cos(x))+ 0(sin(x)), would, as I said above, be written as $frac{e^{ix}+ e^{-ix}}{2}= frac{1}{2}e^{ix}+ frac{1}{2}e^{-ix}$. So we want our matrix to map $begin{pmatrix}1 \ 0 end{pmatrix}$ to $begin{pmatrix}frac{1}{2} \ frac{1}{2}end{pmatrix}$. Similarly with sin(x) to $frac{e^{ix}- e^{-ix}}{2i}$ the matrix must map $begin{pmatrix}0 \ 1 end{pmatrix}$ to $begin{pmatrix} frac{1}{2i} \ -frac{1}{2i} end{pmatrix}$. Such a matrix is $begin{pmatrix}frac{1}{2} & frac{1}{2i} \ frac{1}{2} & -frac{1}{2i} end{pmatrix}$.

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answered yesterday

user247327user247327

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1

$begingroup$

Since the vectors are given as "$alpha cos(x)+ beta sin(x)$" all that you need to do is show that cos(x) and sin(x) are independent. That is, that A cos(x)+ B sin(x)= 0 (for all x) only A= B= 0. And that's pretty close to "trivial". Take x= 0 and $x= pi/2$ and see what happens.

To see that $e^{ix}$ and $e{-ix}$ also form a basis use the fact that $cos(x)= frac{e^{ix}+ e^{ix}}{2}$ and $sin(x)= frac{e^{ix}- e^{-ix}}{2i}$.

For the "change of basis matrix" you want a matrix that multiplied by a column matrix with coefficients from the first basis gives the column matrix with coefficients from the second basis. The first "basis vector", cos(x)= 1(cos(x))+ 0(sin(x)), would, as I said above, be written as $frac{e^{ix}+ e^{-ix}}{2}= frac{1}{2}e^{ix}+ frac{1}{2}e^{-ix}$. So we want our matrix to map $begin{pmatrix}1 \ 0 end{pmatrix}$ to $begin{pmatrix}frac{1}{2} \ frac{1}{2}end{pmatrix}$. Similarly with sin(x) to $frac{e^{ix}- e^{-ix}}{2i}$ the matrix must map $begin{pmatrix}0 \ 1 end{pmatrix}$ to $begin{pmatrix} frac{1}{2i} \ -frac{1}{2i} end{pmatrix}$. Such a matrix is $begin{pmatrix}frac{1}{2} & frac{1}{2i} \ frac{1}{2} & -frac{1}{2i} end{pmatrix}$.

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answered yesterday

user247327user247327

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add a comment | 

1

$begingroup$

Since the vectors are given as "$alpha cos(x)+ beta sin(x)$" all that you need to do is show that cos(x) and sin(x) are independent. That is, that A cos(x)+ B sin(x)= 0 (for all x) only A= B= 0. And that's pretty close to "trivial". Take x= 0 and $x= pi/2$ and see what happens.

To see that $e^{ix}$ and $e{-ix}$ also form a basis use the fact that $cos(x)= frac{e^{ix}+ e^{ix}}{2}$ and $sin(x)= frac{e^{ix}- e^{-ix}}{2i}$.

For the "change of basis matrix" you want a matrix that multiplied by a column matrix with coefficients from the first basis gives the column matrix with coefficients from the second basis. The first "basis vector", cos(x)= 1(cos(x))+ 0(sin(x)), would, as I said above, be written as $frac{e^{ix}+ e^{-ix}}{2}= frac{1}{2}e^{ix}+ frac{1}{2}e^{-ix}$. So we want our matrix to map $begin{pmatrix}1 \ 0 end{pmatrix}$ to $begin{pmatrix}frac{1}{2} \ frac{1}{2}end{pmatrix}$. Similarly with sin(x) to $frac{e^{ix}- e^{-ix}}{2i}$ the matrix must map $begin{pmatrix}0 \ 1 end{pmatrix}$ to $begin{pmatrix} frac{1}{2i} \ -frac{1}{2i} end{pmatrix}$. Such a matrix is $begin{pmatrix}frac{1}{2} & frac{1}{2i} \ frac{1}{2} & -frac{1}{2i} end{pmatrix}$.

share|cite|improve this answer

answered yesterday

user247327user247327

11.4k1516

$endgroup$

add a comment | 

$begingroup$

Since the vectors are given as "$alpha cos(x)+ beta sin(x)$" all that you need to do is show that cos(x) and sin(x) are independent. That is, that A cos(x)+ B sin(x)= 0 (for all x) only A= B= 0. And that's pretty close to "trivial". Take x= 0 and $x= pi/2$ and see what happens.

To see that $e^{ix}$ and $e{-ix}$ also form a basis use the fact that $cos(x)= frac{e^{ix}+ e^{ix}}{2}$ and $sin(x)= frac{e^{ix}- e^{-ix}}{2i}$.

For the "change of basis matrix" you want a matrix that multiplied by a column matrix with coefficients from the first basis gives the column matrix with coefficients from the second basis. The first "basis vector", cos(x)= 1(cos(x))+ 0(sin(x)), would, as I said above, be written as $frac{e^{ix}+ e^{-ix}}{2}= frac{1}{2}e^{ix}+ frac{1}{2}e^{-ix}$. So we want our matrix to map $begin{pmatrix}1 \ 0 end{pmatrix}$ to $begin{pmatrix}frac{1}{2} \ frac{1}{2}end{pmatrix}$. Similarly with sin(x) to $frac{e^{ix}- e^{-ix}}{2i}$ the matrix must map $begin{pmatrix}0 \ 1 end{pmatrix}$ to $begin{pmatrix} frac{1}{2i} \ -frac{1}{2i} end{pmatrix}$. Such a matrix is $begin{pmatrix}frac{1}{2} & frac{1}{2i} \ frac{1}{2} & -frac{1}{2i} end{pmatrix}$.

share|cite|improve this answer

answered yesterday

user247327user247327

11.4k1516

$endgroup$

share|cite|improve this answer

answered yesterday

user247327user247327

11.4k1516

answered yesterday

user247327user247327

11.4k1516

answered yesterday

user247327user247327

11.4k1516