Chapter 35 — Quiz
Twelve conceptual questions. Try each before opening the answer. The goal is to confirm you can separate the transformation from its matrix, and to recognize kernel/image, rank–nullity, and isomorphism in disguise.
Q1. Which two properties define a linear transformation $T:V\to W$, and what are their everyday names from calculus when $T=\frac{d}{dx}$?
Answer
Additivity, $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$, and homogeneity, $T(c\mathbf{v})=cT(\mathbf{v})$. For differentiation these are the *sum rule* ($(p+q)'=p'+q'$) and the *constant-multiple rule* ($(cp)'=cp'$). Together they say $\frac{d}{dx}$ is linear.Q2. Is the map $T(p)(x)=p(x)+1$ on polynomials linear? Why or why not?
Answer
No. It fails the fastest test: $T(\mathbf{0})=0+1=1\neq\mathbf{0}$, and a linear map must send the zero polynomial to the zero polynomial. (It also fails additivity: $T(p+q)=p+q+1$ but $T(p)+T(q)=p+q+2$.) It is an *affine* map, not a linear one.Q3. A single linear operator can have many different matrices. What determines which matrix you get, and how are any two of them related?
Answer
The choice of basis (or bases) determines the matrix — its $j$-th column is the image of the $j$-th basis vector, in the codomain's basis. Any two matrices of the same operator are *similar*: $[T]_{\tilde B}=P^{-1}[T]_B P$, where $P$ is the change-of-basis matrix. The operator is the invariant; the matrix is a basis-dependent photograph.Q4. The kernel and the image of $T:V\to W$ each generalize a "four fundamental subspaces" object from Chapter 13. Which is which, and in which space does each live?
Answer
The kernel is the **null space** freed from coordinates — a subspace of the *domain* $V$ (the vectors sent to $\mathbf{0}$). The image is the **column space** freed from coordinates — a subspace of the *codomain* $W$ (the vectors actually reached). They live in different spaces when $V\neq W$.Q5. What is the kernel of the differentiation operator $D$ on polynomials, and how does it explain the "$+C$" of indefinite integration?
Answer
$\ker D$ is the constant polynomials, $\operatorname{span}\{1\}$ — the only polynomials with zero derivative. Because $D$ has a nonzero kernel, it is not injective: many polynomials share a derivative, differing by a constant. Integrating a derivative recovers the polynomial only up to that constant, which is exactly the "$+C$." The kernel of $D$ *is* the integration constant.Q6. State the abstract Rank–Nullity theorem and verify it for $D:\mathbb{P}_3\to\mathbb{P}_3$.
Answer
$\dim(\ker T)+\dim(\operatorname{im}T)=\dim V$. For $D$ on $\mathbb{P}_3$: $\dim\ker D=1$ (the constants) and $\dim\operatorname{im}D=3$ (the image is $\mathbb{P}_2$), and $1+3=4=\dim\mathbb{P}_3$. Dimension is conserved — what differentiation kills plus what it keeps equals the whole domain.Q7. Why is the differentiation matrix on $\mathbb{P}_3$ nilpotent, and what is the smallest power that vanishes?
Answer
Because differentiating a polynomial of degree $\le 3$ four times leaves nothing: $D^4=0$ on $\mathbb{P}_3$. Each application of $D$ lowers the degree by one, so after four applications even a cubic is gone. (On $\mathbb{P}_n$, the smallest vanishing power is $D^{n+1}$.) A nilpotent operator has only the eigenvalue $0$.Q8. What is an isomorphism, and what does it mean to say "every $n$-dimensional vector space is isomorphic to $\mathbb{R}^n$"?
Answer
An isomorphism is a linear map that is bijective (injective *and* surjective); when one exists, the two spaces are "the same up to relabeling." Every $n$-dimensional real space $V$ is isomorphic to $\mathbb{R}^n$ via the *coordinate map* $\mathbf{v}\mapsto[\mathbf{v}]_B$ for a chosen basis $B$. This is precisely *why coordinates work*: we may compute with coordinate vectors and matrices while thinking about abstract maps, because the two are faithfully the same.Q9. Two matrices represent the same operator in different bases. Name three quantities they must share and one that they need not.
Answer
They must share their **determinant, trace, rank, characteristic polynomial, and eigenvalues** (any three of these) — these are invariants of the operator, blind to the coordinate system. They need *not* share individual matrix entries, nor properties like "is upper-triangular" or "is symmetric," which are artifacts of the chosen basis.Q10. The differentiation operator $D$ on $\mathbb{P}_3$ has matrix $\left[\begin{smallmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{smallmatrix}\right]$ in the monomial basis but a clean shift (all $1$s on the superdiagonal) in the factorial basis. What general principle does this illustrate?
Answer
That the matrix of an operator depends on the basis, and a well-chosen basis can make the matrix dramatically simpler — even though the operator is unchanged. The two matrices are similar. It is the same principle behind diagonalization (Chapter 25) and PCA (Chapter 32): fix the operator, hunt for the basis that makes its matrix simplest.Q11. Is the evaluation map $\operatorname{ev}_0:\mathbb{P}_3\to\mathbb{R}$, $p\mapsto p(0)$, injective? Surjective? Give its rank and nullity.
Answer
Not injective (its kernel is all polynomials with $p(0)=0$, a 3-dimensional space — many polynomials share the value $0$). It *is* surjective onto $\mathbb{R}$ (every real number is some polynomial's value at $0$). Rank $=1$, nullity $=3$, and $1+3=4=\dim\mathbb{P}_3$, confirming rank–nullity.Q12. For a linear operator $T:V\to V$ on a finite-dimensional space, "injective" and "surjective" are equivalent. Why — and why does this fail for differentiation on the space of all polynomials?