Proving that $T_0, T_1, T_2, …$ are basis of $mathbb{R}[x]$












0














I am given that the Chebyshev polynomial $T_n(x) in mathbb{Q}[x]$ is a polynomial such that $T_0(x) = 1$, $T_1(x) = x$ and for $n ge 2$,



$T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$



Now, I am supposed to show that $T_0, T_1, T_2, ...$ are a basis of real vector space $mathbb{R}[x]$.



Basically, we want to show that $T_0, T_1, T_2, ...$ are linearly independent and span $mathbb{R}[x]$. I have no clue how to start. I should use $deg(T_n) = n$ for span, but I am not sure how to show linear independence. Could someone help? Thanks.



Also, I want to prove this without any mentioning of inner product or orthogonality and orthonormality.






linear-algebra abstract-algebra polynomials chebyshev-polynomials







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    0














    I am given that the Chebyshev polynomial $T_n(x) in mathbb{Q}[x]$ is a polynomial such that $T_0(x) = 1$, $T_1(x) = x$ and for $n ge 2$,



    $T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$



    Now, I am supposed to show that $T_0, T_1, T_2, ...$ are a basis of real vector space $mathbb{R}[x]$.



    Basically, we want to show that $T_0, T_1, T_2, ...$ are linearly independent and span $mathbb{R}[x]$. I have no clue how to start. I should use $deg(T_n) = n$ for span, but I am not sure how to show linear independence. Could someone help? Thanks.



    Also, I want to prove this without any mentioning of inner product or orthogonality and orthonormality.






    linear-algebra abstract-algebra polynomials chebyshev-polynomials







    share|cite|improve this question



























      0












      0








      0







      I am given that the Chebyshev polynomial $T_n(x) in mathbb{Q}[x]$ is a polynomial such that $T_0(x) = 1$, $T_1(x) = x$ and for $n ge 2$,



      $T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$



      Now, I am supposed to show that $T_0, T_1, T_2, ...$ are a basis of real vector space $mathbb{R}[x]$.



      Basically, we want to show that $T_0, T_1, T_2, ...$ are linearly independent and span $mathbb{R}[x]$. I have no clue how to start. I should use $deg(T_n) = n$ for span, but I am not sure how to show linear independence. Could someone help? Thanks.



      Also, I want to prove this without any mentioning of inner product or orthogonality and orthonormality.






      linear-algebra abstract-algebra polynomials chebyshev-polynomials







      share|cite|improve this question















      I am given that the Chebyshev polynomial $T_n(x) in mathbb{Q}[x]$ is a polynomial such that $T_0(x) = 1$, $T_1(x) = x$ and for $n ge 2$,



      $T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$



      Now, I am supposed to show that $T_0, T_1, T_2, ...$ are a basis of real vector space $mathbb{R}[x]$.



      Basically, we want to show that $T_0, T_1, T_2, ...$ are linearly independent and span $mathbb{R}[x]$. I have no clue how to start. I should use $deg(T_n) = n$ for span, but I am not sure how to show linear independence. Could someone help? Thanks.



      Also, I want to prove this without any mentioning of inner product or orthogonality and orthonormality.






      linear-algebra abstract-algebra polynomials chebyshev-polynomials




      linear-algebra abstract-algebra polynomials chebyshev-polynomials






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 30 '18 at 3:38







      dmsj djsl

















      asked Nov 30 '18 at 2:57









      dmsj djsldmsj djsl

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          2 Answers
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          By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,ldots,T_n$ spawn the vector subspace $mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that ${1,x,ldots,x^n}$ is a basis of $mathbb{R}_n[x]$ so by dimension you get the linear independence of ${T_0,ldots,T_n}$.






          share|cite|improve this answer





























            0














            Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.






            share|cite|improve this answer























            • Sorry, what do the roots have to do with linear independence?
              – dmsj djsl
              Nov 30 '18 at 3:33






            • 1




              For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
              – JonHales
              Nov 30 '18 at 3:44










            • My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
              – hhp2122
              Nov 30 '18 at 3:45











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            2 Answers
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            active

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            2 Answers
            2






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            active

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            active

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            1














            By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,ldots,T_n$ spawn the vector subspace $mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that ${1,x,ldots,x^n}$ is a basis of $mathbb{R}_n[x]$ so by dimension you get the linear independence of ${T_0,ldots,T_n}$.






            share|cite|improve this answer


























              1














              By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,ldots,T_n$ spawn the vector subspace $mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that ${1,x,ldots,x^n}$ is a basis of $mathbb{R}_n[x]$ so by dimension you get the linear independence of ${T_0,ldots,T_n}$.






              share|cite|improve this answer
























                1












                1








                1






                By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,ldots,T_n$ spawn the vector subspace $mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that ${1,x,ldots,x^n}$ is a basis of $mathbb{R}_n[x]$ so by dimension you get the linear independence of ${T_0,ldots,T_n}$.






                share|cite|improve this answer












                By induction, the degree of $T_n$ is $n$ for every non-negative integer number $n$. So, you can prove inductively that $T_0,ldots,T_n$ spawn the vector subspace $mathbb{R}_n[x]$ of polynomials of degree less or equal to $n$. We know that ${1,x,ldots,x^n}$ is a basis of $mathbb{R}_n[x]$ so by dimension you get the linear independence of ${T_0,ldots,T_n}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 3:44









                Dante GrevinoDante Grevino

                94319




                94319























                    0














                    Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.






                    share|cite|improve this answer























                    • Sorry, what do the roots have to do with linear independence?
                      – dmsj djsl
                      Nov 30 '18 at 3:33






                    • 1




                      For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                      – JonHales
                      Nov 30 '18 at 3:44










                    • My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                      – hhp2122
                      Nov 30 '18 at 3:45
















                    0














                    Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.






                    share|cite|improve this answer























                    • Sorry, what do the roots have to do with linear independence?
                      – dmsj djsl
                      Nov 30 '18 at 3:33






                    • 1




                      For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                      – JonHales
                      Nov 30 '18 at 3:44










                    • My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                      – hhp2122
                      Nov 30 '18 at 3:45














                    0












                    0








                    0






                    Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.






                    share|cite|improve this answer














                    Try induction on the degree part. Polynomials of different degree will be linearly independent for obvious reasons. Then spanning will follow from the degree argument.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 30 '18 at 3:46

























                    answered Nov 30 '18 at 3:12









                    hhp2122hhp2122

                    163




                    163












                    • Sorry, what do the roots have to do with linear independence?
                      – dmsj djsl
                      Nov 30 '18 at 3:33






                    • 1




                      For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                      – JonHales
                      Nov 30 '18 at 3:44










                    • My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                      – hhp2122
                      Nov 30 '18 at 3:45


















                    • Sorry, what do the roots have to do with linear independence?
                      – dmsj djsl
                      Nov 30 '18 at 3:33






                    • 1




                      For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                      – JonHales
                      Nov 30 '18 at 3:44










                    • My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                      – hhp2122
                      Nov 30 '18 at 3:45
















                    Sorry, what do the roots have to do with linear independence?
                    – dmsj djsl
                    Nov 30 '18 at 3:33




                    Sorry, what do the roots have to do with linear independence?
                    – dmsj djsl
                    Nov 30 '18 at 3:33




                    1




                    1




                    For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                    – JonHales
                    Nov 30 '18 at 3:44




                    For linear independence it probably would be easier to say that if they weren't linearly independent then you could write one of the polynomials in terms of the others, say $c_1T_{i_1} + c_2 T_{i_2} + cdots + c_nT_{i_n} = T_j$ but if you compare the degrees of the left and right hand side you get a contradiction.
                    – JonHales
                    Nov 30 '18 at 3:44












                    My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                    – hhp2122
                    Nov 30 '18 at 3:45




                    My possible mistake, actually. I had the lagrange interpolation polynomials in mind. You just need to get the degree part, which is induction. Polynomials of different degree will be linearly independent for obvious reasons.
                    – hhp2122
                    Nov 30 '18 at 3:45


















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