Problem on dual basis












1















Let V be the real vector space of all polynomials, in a single
variable and with real coefficients, of degree at most $3$.



Let $V^*$ be its dual space.



Let $t_1 = 1, t_2 = 2, t_3 = 3, t_4 = 4.$



Which of the following sets of functionals ${f_i |1 leq i leq 4}$
form a basis for $V^*$?



a). For $1 leq i leq 4,$ and for all $p in V , f_i(p) = p(t_i)$.



b). For all $p in V , f_i(p) = p(t_i)$ for $i = 1, 2$, $f_3(p) =
p'(t_1)$ and $f_4(p) = p'(t_2)$.



c). For all $p in V , f_i(p) =
p(t_i)$ for $1 leq i leq 3$ and $f_4(p)=int_{a}^{b} p'(t)dt$





For option a) i'm doing just take $f=af_1+bf_2+cf_3+df_4=0$. There exists a $pin V$ with $p(1)=1,p(2)=p(3)=p(4)=0$ Since $f_i(v_j) = delta_{ij}$. Applying $f$ to $p$ we find $a=0$, Similarly $b=c=d=0$ hence basis. but don't know how doing for b and c ????Thank you for help






linear-algebra







share|cite|improve this question






















  • For (a) you can probably use Lagrange polynomial.
    – Rohan Rajagopal
    Nov 26 '18 at 20:25


















1















Let V be the real vector space of all polynomials, in a single
variable and with real coefficients, of degree at most $3$.



Let $V^*$ be its dual space.



Let $t_1 = 1, t_2 = 2, t_3 = 3, t_4 = 4.$



Which of the following sets of functionals ${f_i |1 leq i leq 4}$
form a basis for $V^*$?



a). For $1 leq i leq 4,$ and for all $p in V , f_i(p) = p(t_i)$.



b). For all $p in V , f_i(p) = p(t_i)$ for $i = 1, 2$, $f_3(p) =
p'(t_1)$ and $f_4(p) = p'(t_2)$.



c). For all $p in V , f_i(p) =
p(t_i)$ for $1 leq i leq 3$ and $f_4(p)=int_{a}^{b} p'(t)dt$





For option a) i'm doing just take $f=af_1+bf_2+cf_3+df_4=0$. There exists a $pin V$ with $p(1)=1,p(2)=p(3)=p(4)=0$ Since $f_i(v_j) = delta_{ij}$. Applying $f$ to $p$ we find $a=0$, Similarly $b=c=d=0$ hence basis. but don't know how doing for b and c ????Thank you for help






linear-algebra







share|cite|improve this question






















  • For (a) you can probably use Lagrange polynomial.
    – Rohan Rajagopal
    Nov 26 '18 at 20:25
















1












1








1








Let V be the real vector space of all polynomials, in a single
variable and with real coefficients, of degree at most $3$.



Let $V^*$ be its dual space.



Let $t_1 = 1, t_2 = 2, t_3 = 3, t_4 = 4.$



Which of the following sets of functionals ${f_i |1 leq i leq 4}$
form a basis for $V^*$?



a). For $1 leq i leq 4,$ and for all $p in V , f_i(p) = p(t_i)$.



b). For all $p in V , f_i(p) = p(t_i)$ for $i = 1, 2$, $f_3(p) =
p'(t_1)$ and $f_4(p) = p'(t_2)$.



c). For all $p in V , f_i(p) =
p(t_i)$ for $1 leq i leq 3$ and $f_4(p)=int_{a}^{b} p'(t)dt$





For option a) i'm doing just take $f=af_1+bf_2+cf_3+df_4=0$. There exists a $pin V$ with $p(1)=1,p(2)=p(3)=p(4)=0$ Since $f_i(v_j) = delta_{ij}$. Applying $f$ to $p$ we find $a=0$, Similarly $b=c=d=0$ hence basis. but don't know how doing for b and c ????Thank you for help






linear-algebra







share|cite|improve this question














Let V be the real vector space of all polynomials, in a single
variable and with real coefficients, of degree at most $3$.



Let $V^*$ be its dual space.



Let $t_1 = 1, t_2 = 2, t_3 = 3, t_4 = 4.$



Which of the following sets of functionals ${f_i |1 leq i leq 4}$
form a basis for $V^*$?



a). For $1 leq i leq 4,$ and for all $p in V , f_i(p) = p(t_i)$.



b). For all $p in V , f_i(p) = p(t_i)$ for $i = 1, 2$, $f_3(p) =
p'(t_1)$ and $f_4(p) = p'(t_2)$.



c). For all $p in V , f_i(p) =
p(t_i)$ for $1 leq i leq 3$ and $f_4(p)=int_{a}^{b} p'(t)dt$





For option a) i'm doing just take $f=af_1+bf_2+cf_3+df_4=0$. There exists a $pin V$ with $p(1)=1,p(2)=p(3)=p(4)=0$ Since $f_i(v_j) = delta_{ij}$. Applying $f$ to $p$ we find $a=0$, Similarly $b=c=d=0$ hence basis. but don't know how doing for b and c ????Thank you for help






linear-algebra




linear-algebra






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share|cite|improve this question











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share|cite|improve this question










asked Nov 26 '18 at 19:25









RAM_3R

527214




527214












  • For (a) you can probably use Lagrange polynomial.
    – Rohan Rajagopal
    Nov 26 '18 at 20:25




















  • For (a) you can probably use Lagrange polynomial.
    – Rohan Rajagopal
    Nov 26 '18 at 20:25


















For (a) you can probably use Lagrange polynomial.
– Rohan Rajagopal
Nov 26 '18 at 20:25






For (a) you can probably use Lagrange polynomial.
– Rohan Rajagopal
Nov 26 '18 at 20:25












1 Answer
1






active

oldest

votes


















1














One approach is as follows: just as the linear map
$$
p mapsto pmatrix{p(1)\p(2)\p(3)\p(4)}
$$
has a trivial kernel, show that the map
$$
p mapsto pmatrix{p(1)\p(2)\p'(1)\p'(2)}
$$
has a trivial kernel.



For c, note that we can rewrite $f_4(p) = p(b) - p(a)$. So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.






share|cite|improve this answer





















  • why we have to just find trivial kernel???
    – RAM_3R
    Nov 26 '18 at 19:46










  • For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
    – Omnomnomnom
    Nov 26 '18 at 19:51












  • ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
    – RAM_3R
    Nov 26 '18 at 20:01










  • @RAM_3R we also need to know what $a$ is
    – Omnomnomnom
    Nov 26 '18 at 20:03










  • suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
    – RAM_3R
    Nov 26 '18 at 20:07













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1














One approach is as follows: just as the linear map
$$
p mapsto pmatrix{p(1)\p(2)\p(3)\p(4)}
$$
has a trivial kernel, show that the map
$$
p mapsto pmatrix{p(1)\p(2)\p'(1)\p'(2)}
$$
has a trivial kernel.



For c, note that we can rewrite $f_4(p) = p(b) - p(a)$. So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.






share|cite|improve this answer





















  • why we have to just find trivial kernel???
    – RAM_3R
    Nov 26 '18 at 19:46










  • For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
    – Omnomnomnom
    Nov 26 '18 at 19:51












  • ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
    – RAM_3R
    Nov 26 '18 at 20:01










  • @RAM_3R we also need to know what $a$ is
    – Omnomnomnom
    Nov 26 '18 at 20:03










  • suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
    – RAM_3R
    Nov 26 '18 at 20:07


















1














One approach is as follows: just as the linear map
$$
p mapsto pmatrix{p(1)\p(2)\p(3)\p(4)}
$$
has a trivial kernel, show that the map
$$
p mapsto pmatrix{p(1)\p(2)\p'(1)\p'(2)}
$$
has a trivial kernel.



For c, note that we can rewrite $f_4(p) = p(b) - p(a)$. So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.






share|cite|improve this answer





















  • why we have to just find trivial kernel???
    – RAM_3R
    Nov 26 '18 at 19:46










  • For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
    – Omnomnomnom
    Nov 26 '18 at 19:51












  • ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
    – RAM_3R
    Nov 26 '18 at 20:01










  • @RAM_3R we also need to know what $a$ is
    – Omnomnomnom
    Nov 26 '18 at 20:03










  • suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
    – RAM_3R
    Nov 26 '18 at 20:07
















1












1








1






One approach is as follows: just as the linear map
$$
p mapsto pmatrix{p(1)\p(2)\p(3)\p(4)}
$$
has a trivial kernel, show that the map
$$
p mapsto pmatrix{p(1)\p(2)\p'(1)\p'(2)}
$$
has a trivial kernel.



For c, note that we can rewrite $f_4(p) = p(b) - p(a)$. So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.






share|cite|improve this answer












One approach is as follows: just as the linear map
$$
p mapsto pmatrix{p(1)\p(2)\p(3)\p(4)}
$$
has a trivial kernel, show that the map
$$
p mapsto pmatrix{p(1)\p(2)\p'(1)\p'(2)}
$$
has a trivial kernel.



For c, note that we can rewrite $f_4(p) = p(b) - p(a)$. So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 19:36









Omnomnomnom

126k788176




126k788176












  • why we have to just find trivial kernel???
    – RAM_3R
    Nov 26 '18 at 19:46










  • For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
    – Omnomnomnom
    Nov 26 '18 at 19:51












  • ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
    – RAM_3R
    Nov 26 '18 at 20:01










  • @RAM_3R we also need to know what $a$ is
    – Omnomnomnom
    Nov 26 '18 at 20:03










  • suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
    – RAM_3R
    Nov 26 '18 at 20:07




















  • why we have to just find trivial kernel???
    – RAM_3R
    Nov 26 '18 at 19:46










  • For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
    – Omnomnomnom
    Nov 26 '18 at 19:51












  • ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
    – RAM_3R
    Nov 26 '18 at 20:01










  • @RAM_3R we also need to know what $a$ is
    – Omnomnomnom
    Nov 26 '18 at 20:03










  • suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
    – RAM_3R
    Nov 26 '18 at 20:07


















why we have to just find trivial kernel???
– RAM_3R
Nov 26 '18 at 19:46




why we have to just find trivial kernel???
– RAM_3R
Nov 26 '18 at 19:46












For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
– Omnomnomnom
Nov 26 '18 at 19:51






For part a: showing that the map $T:p mapsto (p(1),p(2),p(3),p(4))^T$ has a trivial kernel means showing that $p(1) = cdots = p(4) = 0$, then $p$ must be zero. If this kernel is non-zero, then we may choose non-zero vector $v = (a,b,c,d)^T in operatorname{im}(T)^perp$, which would satisfy $v^T T(p) = 0$. That is to say, the $f_i$ would be linearly dependent.
– Omnomnomnom
Nov 26 '18 at 19:51














ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
– RAM_3R
Nov 26 '18 at 20:01




ok thank you understand Now for c if b=4 then can i say it's linearly independent ??
– RAM_3R
Nov 26 '18 at 20:01












@RAM_3R we also need to know what $a$ is
– Omnomnomnom
Nov 26 '18 at 20:03




@RAM_3R we also need to know what $a$ is
– Omnomnomnom
Nov 26 '18 at 20:03












suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
– RAM_3R
Nov 26 '18 at 20:07






suppose a=1 ....then we can write $f_4(b)=p(4)-p(1)$
– RAM_3R
Nov 26 '18 at 20:07




















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