Quiz — The Characteristic Polynomial and How to Find Eigenvalues
Twelve quick checks on the chapter's core ideas. Answer before expanding. Each answer ends with a one-line explanation.
Q1. Why is $\lambda$ an eigenvalue of $A$ exactly when $\det(A - \lambda I) = 0$?
Answer
Because an eigenvalue needs a *nonzero* $\mathbf{v}$ with $(A - \lambda I)\mathbf{v} = \mathbf{0}$, i.e. a nontrivial null space, i.e. $A - \lambda I$ is singular — and a square matrix is singular precisely when its determinant is zero (Chapter 11). *The determinant of $A - \lambda I$ is the singularity detector.*Q2. What is wrong with writing $\det(A - \lambda I) = \det(A) - \lambda^n$?
Answer
The determinant is **not** additive: $\det(X + Y) \neq \det(X) + \det(Y)$ in general. You must form the matrix $A - \lambda I$ (subtract $\lambda$ from each diagonal entry) and then take *one* determinant. *There is no way to split the determinant across the subtraction.*Q3. For a $2 \times 2$ matrix, what are the coefficients of the characteristic polynomial in terms of the matrix?
Answer
$p_A(\lambda) = \lambda^2 - \operatorname{tr}(A)\,\lambda + \det(A)$. The linear coefficient is minus the trace; the constant term is the determinant. *Trace and determinant alone determine a $2 \times 2$'s eigenvalues.*Q4. A $3 \times 3$ matrix has eigenvalues $1, 2, 3$. What are $\operatorname{tr}(A)$ and $\det(A)$?
Answer
$\operatorname{tr}(A) = 1 + 2 + 3 = 6$ (sum of eigenvalues) and $\det(A) = 1 \cdot 2 \cdot 3 = 6$ (product of eigenvalues). *These identities hold for every square matrix, by comparing coefficients of the characteristic polynomial.*Q5. What is the eigenspace $E_\lambda$, and why is it a subspace?
Answer
$E_\lambda = N(A - \lambda I)$, the set of all eigenvectors for $\lambda$ together with $\mathbf{0}$. It is a subspace because it is a *null space* (Chapter 13): closed under addition and scalar multiplication and containing the zero vector. *Its dimension is the geometric multiplicity.*Q6. Distinguish algebraic multiplicity from geometric multiplicity.
Answer
Algebraic multiplicity $m_a(\lambda)$ is how many times $\lambda$ is a *root* of $p_A$; geometric multiplicity $m_g(\lambda) = \dim E_\lambda$ is how many *independent eigenvectors* $\lambda$ has. They satisfy $1 \le m_g \le m_a$. *Roots versus dimensions — two different counts.*Q7. What is a defective matrix?
Answer
A square matrix for which some eigenvalue has $m_g < m_a$ — strictly fewer eigenvectors than its root-multiplicity promises. Equivalently, it lacks a full basis of eigenvectors and cannot be diagonalized (Chapter 25). *The shear $\begin{bmatrix} 2 & 1 \\ 0 & 2\end{bmatrix}$ is the canonical example.*Q8. A matrix has a double eigenvalue. Is it necessarily defective?
Answer
No. You must compute the eigenspace dimension. $2I$ has a double eigenvalue $2$ with a two-dimensional eigenspace ($m_g = 2 = m_a$), so it is **not** defective; the shear has the same double eigenvalue but $m_g = 1 < 2$, so it **is**. *Always row-reduce $A - \lambda I$ and count free variables for a repeated root.*Q9. What are the eigenvalues of an upper-triangular matrix?
Answer
Exactly its diagonal entries. Because $A - \lambda I$ stays triangular and the determinant of a triangular matrix is the product of its diagonal, $p_A(\lambda) = \prod_i (a_{ii} - \lambda)$, whose roots are the $a_{ii}$. *No row reduction needed for the eigenvalues themselves.*Q10. A real matrix's characteristic polynomial has roots $2 \pm 3i$. Does the matrix have real eigenvectors? What does this signal geometrically?
Answer
No real eigenvectors — the eigenvalues are complex, so the eigenvectors are complex too. Geometrically the transformation has no real invariant line; it *rotates* (and scales) the plane, with the complex eigenvalues encoding the angle and scaling (Chapter 26). *Complex eigenvalues = a rotation in disguise.*Q11. Why is $\lambda = 1$ always an eigenvalue of a column-stochastic (Markov) matrix?
Answer
Subtracting $1$ from each diagonal entry makes every column of $M - I$ sum to $0$ (since each column of $M$ summed to $1$). A matrix with all columns summing to zero has dependent rows — $(1,\dots,1)$ times it is the zero row — so it is singular, $\det(M - I) = 0$, and $1$ is an eigenvalue. *Its eigenvector is the stationary distribution (PageRank, Chapter 29).*Q12. Why don't real software libraries find eigenvalues by computing the characteristic polynomial and rooting it?
Answer
Two reasons: (1) by Abel–Ruffini there is no radical formula for roots of degree $\ge 5$, so it fails in principle for large matrices; (2) forming the polynomial and rooting it is *numerically unstable* — tiny coefficient errors move the roots enormously. Libraries iterate instead (power iteration, the QR algorithm; Chapter 38). *The polynomial is for understanding; iteration is for computing.*Q13. (Bonus) If $\lambda$ is an eigenvalue of invertible $A$ with eigenvector $\mathbf{v}$, what is an eigenvalue of $A^{-1}$ and its eigenvector?