Find the basis of a kernel and the dimension of the image

  • #1
72
0

Homework Statement


Let ##n>1\in\, \mathbb{N}##. A map ##A:\mathbb{R}_{n}[x]\to\mathbb{R}_{n}[x]## is given with the rule ##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)##
a)Proof that this map is linear
b)Find some basis of the kernel
b)Find the dimension of the image

Homework Equations


##\mathbb{R}_{n}[x]##
is defined as the set of all polynomial with real coeficient that have the power less or equal to n
##kerA=\{x;Ax=0\}##
##imA=\{Ax,x\in\mathbb{R}_{n}[x]\}##
##p(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^2+a_{1}x^1+a_{0}x^0##

The Attempt at a Solution


a)Prove that this map is linear
##p,q\in A \\
(A(p+q))(x)=(x^n+1)(p+q)(1)+(p+q)^{'''}(x)=((x^n+1)(p)(1)+(p)^{'''}(x)+(x^n+1)(q)(1)+(q)^{'''}(x)=Ap(x)+Aq(x)
\\
\text{ }
\\
A(\theta p)(x)=\theta((x^n+1)(p)(1)+(p)^{'''}(x))=\theta(Ap)(x)
##
c)
I started doing this by writing out the possible polynomial
##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)\\
p(1)=a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\ldots+a_{2}+a_{1}+a_{0}\\
p^{'''}(x)=6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
##
now I placed all of this together
##
(Ap)(x)=(x^n+1)\displaystyle\sum_{i=0}^{n}a_{i}+6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
##
Then I paired all of the same coeficients together and got
##
a_{n}(x^{n}+1+6\binom{n}{3}*x^{n-3}) \\
a_{n-1}(x^{n}+1+6\binom{n-1}{3}*x^{n-4}) \\
a_{n-2}(x^{n}+1+6\binom{n-2}{3}*x^{n-5}) \\
\vdots\\
a_{3}(x^n+7) \\
a_{2}(x^n+1) \\
a_{1}(x^n+1) \\
a_{0}(x^n+1) \\
##
Here I noticed that the bottom 3 functions are linearly dependent, which means that If I want to find the basis or dimension I should take ##a_{0}\, and\, a_{1}## out. Then I also noticed that all of the above (an to a3) are also linearly dependent on a2 so I subtracted them and got
##
a_{n}(x^{n-3}) \\
a_{n-1}(x^{n-4}) \\
a_{n-2}(x^{n-5}) \\
\vdots\\
a_{3}(6) \\
a_{2}(x^n+1) \\
##
as my basis for the image of A therefore the ##dimension(imA)=n-1##
This is as far as I have gotten. I've tried solving b) by setting the whole polynomial I got equal to 0 however I have no idea how to continue from there.
thank you
 
Last edited:

Answers and Replies

  • #2
member 587159

Homework Statement


Let ##n>1\in\, \mathbb{N}##. A map ##A:\mathbb{R}_{n}[x]\to\mathbb{R}_{n}[x]## is given with the rule ##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)##
a)Proof that this map is linear
b)Find some basis of the kernel
b)Find the dimension of the image

Homework Equations


##\mathbb{R}_{n}[x]##
is defined as the set of all polynomial with real coeficient that have the power less or equal to n
##kerA=\{x;Ax=0\}##
##imA=\{Ax,x\in\mathbb{R}_{n}[x]\}##
##p(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^2+a_{1}x^1+a_{0}x^0##

The Attempt at a Solution


a)Prove that this map is linear
##p,q\in A \\
(A(p+q))(x)=(x^n+1)(p+q)(1)+(p+q)^{'''}(x)=((x^n+1)(p)(1)+(p)^{'''}(x)+(x^n+1)(q)(1)+(q)^{'''}(x)=Ap(x)+Aq(x)
\\
\text{ }
\\
A(\theta p)(x)=\theta((x^n+1)(p)(1)+(p)^{'''}(x))=\theta(Ap)(x)
##
c)
I started doing this by writing out the possible polynomial
##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)\\
p(1)=a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\ldots+a_{2}+a_{1}+a_{0}\\
p^{'''}(x)=6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
##
now I placed all of this together
##
(Ap)(x)=(x^n+1)\displaystyle\sum_{i=0}^{n}a_{i}+6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
##
Then I paired all of the same coeficients together and got
##
a_{n}(x^{n}+1+6\binom{n}{3}*x^{n-3}) \\
a_{n-1}(x^{n}+1+6\binom{n-1}{3}*x^{n-4}) \\
a_{n-2}(x^{n}+1+6\binom{n-2}{3}*x^{n-5}) \\
\vdots\\
a_{3}(x^n+7) \\
a_{2}(x^n+1) \\
a_{1}(x^n+1) \\
a_{0}(x^n+1) \\
##
Here I noticed that the bottom 3 functions are linearly dependent, which means that If I want to find the basis or dimension I should take ##a_{0}\, and\, a_{1}## out. Then I also noticed that all of the above (an to a3) are also linearly dependent on a2 so I subtracted them and got
##
a_{n}(x^{n-3}) \\
a_{n-1}(x^{n-4}) \\
a_{n-2}(x^{n-5}) \\
\vdots\\
a_{3}(6) \\
a_{2}(x^n+1) \\
##
as my basis for the image of A therefore the ##dimension(imA)=n-1##
This is as far as I have gotten. I've tried solving b) by setting the whole polynomial I got equal to 0 however I have no idea how to continue from there.
thank you

Find n + 1 polynomials that are in the kernel and are linear independent and you are done.

Also, if you would have made the b question first, you could have deduced the dimension of the kernel, and using the second dimension theorem ##dim(Ker(A)) + dim(Im(A)) = dim(\mathbb{R}_n[X])## you would immediately know the dimension of the image of the map without finding a basis.
 

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