Classifying a Map from Its Polynomial Formula
A formula for a linear map between polynomial spaces can hide its essential features in the coefficients. Differentiation lowers degree, evaluation produces a scalar, and multiplication by a polynomial can raise degree. To classify a particular map, we need to keep track of which coefficients are free to vary and which conditions force an input into the kernel.
Throughout, scalars are real. For \(n\geq 0\), let \(P_n(\mathbb{R})\) denote the vector space of real polynomials of degree at most \(n\), together with the zero polynomial. Each \(p\in P_n(\mathbb{R})\) has a unique expression \(p(x)=a_0+a_1x+\cdots+a_nx^n\). Thus \((1,x,\ldots,x^n)\) is a basis and \(\dim P_n(\mathbb{R})=n+1\). The uniqueness of the coefficients also means that two polynomials are equal exactly when their corresponding coefficients are equal.
For a polynomial-space map, write a general input with unknown coefficients. Then calculate the output as a polynomial and compare coefficients. This single calculation often verifies linearity and reveals both the kernel and image. A map is linear when \(T(ap+bq)=aT(p)+bT(q)\) for all inputs \(p,q\) and real scalars \(a,b\); its formula must respect both addition and scalar multiplication.
Proof. Let \(p(x)=a_0+a_1x+\cdots+a_mx^m\). By linearity, \(T(p)=a_0T(1)+a_1T(x)+\cdots+a_mT(x^m)\), so every output belongs to the stated span. Conversely, every vector in that span has the form \(a_0T(1)+\cdots+a_mT(x^m)\). By linearity this is \(T(a_0+a_1x+\cdots+a_mx^m)\), and the polynomial in parentheses belongs to \(P_m(\mathbb{R})\). Therefore every vector in the span is an output. \(\square\)
This result gives a finite way to describe the image: determine the outputs of the monomials and find their span. It does not require the map to be onto. The kernel is found differently: solve \(T(p)=0\), using an arbitrary input polynomial and setting every output coefficient equal to zero.
A Map That Is Onto but Not One-to-One
Worked Example: A Derivative and an Evaluation Term
Classify \(T:P_3(\mathbb{R})\to P_2(\mathbb{R})\) defined by \(T(p)=p'(x)+p(0)x\).
Write \(p(x)=a+bx+cx^2+dx^3\). Then \(p'(x)=b+2cx+3dx^2\) and \(p(0)=a\), so
This output has degree at most \(2\), so it belongs to the stated codomain. The coefficient formula is linear in \(a,b,c,d\): replacing those coefficients by the coefficients of \(rp+s q\) replaces the output coefficients by the same linear combinations. Equivalently, differentiation, evaluation at \(0\), and multiplication by the fixed polynomial \(x\) each respect addition and scalar multiplication. Hence \(T\) is linear.
For the kernel, set each output coefficient to zero. The equations are \(b=0\), \(a+2c=0\), and \(3d=0\), giving \(b=d=0\) and \(a=-2c\). Thus \(p(x)=c(x^2-2)\), so
For the image, every output is in \(P_2(\mathbb{R})\). Conversely, given any \(q(x)=u+vx+wx^2\), choose \(p(x)=v+ux+\frac{w}{3}x^3\). Its coefficients are \(a=v\), \(b=u\), \(c=0\), \(d=w/3\), so \(T(p)=u+vx+wx^2=q\). Therefore \(\operatorname{im}T=P_2(\mathbb{R})\), and \(\operatorname{rank}T=3\).
Check: For a general kernel element \(p(x)=c(x^2-2)\), we have \(p'(x)=2cx\) and \(p(0)x=-2cx\), so \(T(p)=0\). For the preimage used above, \(p'(x)=u+wx^2\) and \(p(0)x=vx\), which indeed gives \(q(x)\). The map is not injective because its kernel contains \(x^2-2\ne0\), but it is surjective because its image is the whole codomain.
Extracting the Image and Kernel from Coefficients
The image and kernel answer different questions. The image consists of the outputs the map can produce; the kernel consists of inputs that produce zero. A restriction on output coefficients can make the image smaller than the codomain, while a nonzero polynomial can still lie in the kernel. Neither property should be inferred from the other without checking the formula.
Proof. Suppose \(T\) is injective and \(T(\mathbf{v})=\mathbf{0}\). Since \(T(\mathbf{0})=\mathbf{0}\), injectivity applied to \(T(\mathbf{v})=T(\mathbf{0})\) gives \(\mathbf{v}=\mathbf{0}\). Thus the kernel is trivial. Conversely, suppose \(\ker T=\{\mathbf{0}\}\), and suppose \(T(\mathbf{u})=T(\mathbf{v})\). Linearity gives \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}\), so \(\mathbf{u}-\mathbf{v}\in\ker T\). The kernel assumption implies \(\mathbf{u}-\mathbf{v}=\mathbf{0}\), hence \(\mathbf{u}=\mathbf{v}\). Therefore \(T\) is injective. \(\square\)
For surjectivity, the definition is the direct test: show that every polynomial in the codomain can be written as \(T(p)\) for some allowed input \(p\). To show that a map is not surjective, it is enough to identify a codomain polynomial that cannot occur as an output.
Worked Example: An Image Missing the Quadratic Term
Classify \(S:P_2(\mathbb{R})\to P_2(\mathbb{R})\) defined by \(S(p)=p'(x)+p(1)\), where \(p(1)\) is regarded as a constant polynomial.
Let \(p(x)=a+bx+cx^2\). Then \(p'(x)=b+2cx\) and \(p(1)=a+b+c\). Hence
The output has degree at most \(1\), so it is in \(P_2(\mathbb{R})\). The displayed coefficients depend linearly on \(a,b,c\), which verifies that \(S\) is linear.
For \(S(p)=0\), the coefficient of \(x\) gives \(2c=0\), so \(c=0\). The constant coefficient then gives \(a+2b=0\), or \(a=-2b\). Therefore
The formula shows that every output lies in \(\operatorname{span}\{1,x\}\). To see that every polynomial in this span occurs, take an arbitrary \(u+vx\) and choose \(p(x)=u-\frac{v}{2}+\frac{v}{2}x^2\). Here \(a=u-v/2\), \(b=0\), and \(c=v/2\), so \(a+2b+c=u\) and \(2c=v\). Thus \(S(p)=u+vx\), proving \(\operatorname{im}S=\operatorname{span}\{1,x\}\) and \(\operatorname{rank}S=2\).
Check: \(S(x-2)=1+(1-2)=0\), as required for a kernel element. For the chosen preimage, \(p'(x)=vx\) and \(p(1)=u\), so \(S(p)=u+vx\). The map is not injective because \(x-2\ne0\) lies in its kernel. It is not surjective because \(x^2\in P_2(\mathbb{R})\), but every output has zero coefficient of \(x^2\).
In these examples the dimensions also provide a consistency check. The Rank-Nullity Theorem states that for a linear map with finite-dimensional domain, \(\dim V=\operatorname{nullity}T+\operatorname{rank}T\). Here the theorem applies because each polynomial domain is finite-dimensional. For \(S\), the domain has dimension \(3\), while the computed nullity and rank are \(1\) and \(2\), which sum to \(3\). This check supports the calculations, but it does not replace the direct work needed to identify the kernel and image.
Multiplication by a Polynomial: Injective but Not Surjective
Worked Example: Multiplication by \(x\)
Classify \(U:P_2(\mathbb{R})\to P_3(\mathbb{R})\) defined by \(U(p)=xp\).
For \(p,q\in P_2(\mathbb{R})\) and real \(r,s\), \(U(rp+sq)=x(rp+sq)=r(xp)+s(xq)=rU(p)+sU(q)\). Thus \(U\) is linear. Also, multiplying a polynomial of degree at most \(2\) by \(x\) gives a polynomial of degree at most \(3\), so the formula has values in \(P_3(\mathbb{R})\).
Write \(p(x)=a+bx+cx^2\). Then \(U(p)=ax+bx^2+cx^3\). This is zero exactly when \(a=b=c=0\), by uniqueness of polynomial coefficients. Therefore \(\ker U=\{0\}\), the nullity is \(0\), and the kernel test proves that \(U\) is injective.
The image consists exactly of polynomials \(ax+bx^2+cx^3\), so \(\operatorname{im}U=\operatorname{span}\{x,x^2,x^3\}\). These three monomials are independent by uniqueness of coefficients; hence the rank is \(3\). The map is not surjective: the constant polynomial \(1\) belongs to \(P_3(\mathbb{R})\), but no output of \(U\) has a nonzero constant coefficient.
Check: The input \(p(x)=a+bx+cx^2\) gives \(U(p)=ax+bx^2+cx^3\), so every polynomial in the stated image has a preimage of the form \(a+bx+cx^2\). The only way this output can be zero is \(a=b=c=0\), confirming the kernel calculation. The rank and nullity satisfy \(3+0=\dim P_2(\mathbb{R})=3\).
Common Pitfall: Confusing the Codomain with the Image
A map into \(P_2(\mathbb{R})\) does not necessarily produce every polynomial in \(P_2(\mathbb{R})\). For the map \(S\) above, one might incorrectly argue that the image is \(P_2(\mathbb{R})\) just because the codomain is \(P_2(\mathbb{R})\). The formula corrects that conclusion: \(S(a+bx+cx^2)=(a+2b+c)+2cx\), which has no \(x^2\) term. In particular, the incorrect claim would imply that \(x^2=S(p)\) for some \(p\), but comparing \(x^2\) with the output formula would require its \(x^2\) coefficient \(1\) to equal \(0\), a contradiction. The correct image is \(\operatorname{span}\{1,x\}\), a proper subspace of the codomain.
Another useful check is to keep domain and codomain dimensions distinct. In the multiplication example, the domain has dimension \(3\) and the codomain has dimension \(4\). Injectivity is possible even though surjectivity fails. In the first example, the map is onto despite having a nontrivial kernel. The dimensions help describe what is possible, but the coefficient calculations establish what happens for the particular formula.
Use independent coefficients, such as \(p(x)=a_0+\cdots+a_mx^m\).
Collect like powers of \(x\), verify the output is in the codomain, and check that the coefficient formulas are linear.
Set every output coefficient to zero and express the resulting polynomials as a span.
Identify all possible output coefficients, then test injectivity using the kernel and surjectivity using the codomain.
Check Your Understanding
For each map, use a general polynomial input and compare coefficients. State the kernel, image, rank, nullity, and injectivity or surjectivity status.
- For \(A:P_2(\mathbb{R})\to P_1(\mathbb{R})\), let \(A(p)=p'(x)\).
- For \(B:P_2(\mathbb{R})\to P_2(\mathbb{R})\), let \(B(p)=p(0)+p'(0)x\).
- For \(C:P_1(\mathbb{R})\to P_2(\mathbb{R})\), let \(C(p)=xp(x)\).
- Explain why knowing the codomain alone does not determine the image of a map.
- State the kernel test for injectivity and justify both directions.
