Problem. Let \(\mathbb{F}\) be a field and let \(V=\mathbb{F}[X]_{\le3}\).
We shall define the linear map \(T\colon V\to V\) by \[(T(p))(X)=p(X)-p(1-X).\] Find bases for \(\ker(T)\) and for \(\operatorname{Im}(T)\). Prove your answers.
Proof. The standard basis of \(V\) is \(\left(1,x,x^{2},x^{3}\right)\).
Let us compute \(T\) on the standard basis.
\[T(1)=1-1=0\]
\[T(x)=x-(1-x)=2x-1\]
\[T(x^{2})=x^{2}-(1-x)^{2}=x^{2}-(1-2x+x^{2})=2x-1\]
\[T(x^{3})=x^{3}-(1-x)^{3}=x^{3}-(1-x)^{2}(1-x)=x^{3}-(1-2x+x^{2})(1-x)=\]
\[x^{3}-(1-3x+3x^{2}-x^{3})=2x^{3}-3x^{2}+3x-1\]
Let us recall that if \(S\) spans \(V\) and the domain of a linear transformation \(T\) is \(V\) then \(T(S)\) spans \(\operatorname{im}(T)\).
This means that \[\text{Span}\left(0,2x-1,2x-1,2x^{3}-3x^{2}+3x-1\right)=\operatorname{Im}(T)\]
Let us recall that for a span of a sequence of vectors, if one vector is a linear combination of the others, we can remove it and not change the span.
In any field, \[0\cdot(2x-1)=0\quad\wedge1\cdot(2x-1)=2x-1\]
\[\therefore\quad\text{Span}\left(2x-1,2x^{3}-3x^{2}+3x-1\right)=\text{Im}\left(T\right)\]
Let \(a,b\in\mathbb{F}\)
\[a(2x-1)+b(2x^{3}-3x^{2}+3x-1)=0\]
\[2bx^{3}-3bx^{2}+(3b+2a)x+(-a-b)\cdot1=0\]
Since \(\left(1,x,x^{2},x^{3}\right)\) is a basis of \(V\) there is exactly one way to reach any element of \(V\) as a linear combination of it.
This gives us
\[2b=0\wedge-3b=0\quad\therefore\quad b=3b-2b=0\]
\[-a-b\underset{\text{plug in b=0}}{=}-a\underset{-a-b=0}{=}0\therefore a=0\] Recall the claim that a sequence is linearly independent if the only way to reach zero as a linear combination of its elements is by setting all of the scalars to zero, then \(\left(2x-1,2x^{3}-3x^{2}+3x-1\right)\) is linearly independent.
This means \(\left(2x-1,2x^{3}-3x^{2}+3x-1\right)\) is a base for \(\operatorname{Im}(T)\) since it is linearly independent and spans it.
Now the kernel. It’s all of the values that are mapped to \(0_{V}\). i.e. since we know the basis for \(V\), every
\[p=\alpha\cdot1+\beta\cdot x+\gamma\cdot x^{2}+\delta\cdot x^{3}\quad s.t.\quad T(p)=0_{V}\]
We already have a basis for \(\text{\text{Im} }T\), which means there is one way to represent \(T(p)\) as a linear combination of the vectors in the basis:
\[T(p)=T\left(\alpha\cdot1+\beta\cdot x+\gamma\cdot x^{2}+\delta\cdot x^{3}\right)=0\]
This is a linear transformation so we get
\[T(\alpha\cdot1)+T(\beta x)+T(\gamma x^{2})+T(\delta x^{3})=0\] \[\alpha\cdot T(1)+\beta\cdot T(x)+\gamma\cdot T(x^{2})+\delta\cdot T(x^{3})=0\]
\[\alpha\cdot0+\beta\cdot\left(2x-1\right)+\gamma\cdot\left(2x-1\right)+\delta\cdot\left(2x^{3}-3x^{2}+3x-1\right)=0\]
\[(\beta+\gamma)\left(2x-1\right)+\delta\left(2x^{3}-3x^{2}+3x-1\right)=0\]
And since we know that the only way to reach zero as a linear combination of \((2x-1,2x^{3}-3x^{2}+3x-1)\) is by taking all of the scalars as zero, we get
\[\beta=-\gamma,\quad\delta=0\] Therefore
\[\ker(T)=\left\{ \alpha+\left(-\gamma\right)\cdot x+\gamma\cdot x^{2}\mid\alpha,\gamma\in\mathbb{F}\right\} =\left\{ \alpha\cdot1+\gamma\cdot(x^{2}-x)\mid\alpha,\gamma\in\mathbb{F}\right\} =\text{Span}\left(1,x^{2}-x\right)\]
\[a\cdot1+b\cdot(x^{2}-x)=0\iff bx^{2}-bx+a=0\]
Again, there is a unique way to represent 0 using the base \((1,x,x^{2},x^{3})\) and this gives us \(b=0,a=0\) therefore this sequence spans the kernel and is linearly independent, i.e. a base for the kernel. ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.