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If $M=\{p\in P_4: p(2x)=p(x+1)\}$ find the basis for one direct complement of the subspace $M$ in vector space $P_4$.

I'm having difficulties finding the basis for $M$.

$p(x)=ax^4+bx^3+cx^2+dx+e$

$p(2x)=16ax^4+8bx^3+4cx^2+2dx+e$

$p(x+1)=a(x+1)^4+b(x+1)^3+c(x+1)^2+d(x+1)+e$

$p(2x)-p(x+1)=0 \Rightarrow 15ax^4+(7b-4a)x^3+(3c-6a-3b)x^2+(d-4a-3b-2c)x-a-b-c-d=0$

From here I get that $a=b=c=d=0$. Is that correct? $M=\{p(x)=ax^4+bx^3+cx^2+dx+e: a=b=c=d=0\}$

So I get that $M$ is a space of constant polynomials. Is then a basis for $M$ any real number? Can one direct complement be $\{x,x^2,x^3,x^4\}?$

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    Look correct to me. – 2017-02-10
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    @amd Thank you for commenting. Is the basis for $M=\{e:e\in\Bbb R\}$? – 2017-02-10
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    No, that’s an equivalent definition of $M$. For a basis, you need specific vectors, in this case, any non-zero constant polynomial. – 2017-02-10
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    @amd I see, thanks! – 2017-02-10

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