Source: Friedberg, Insel, Spence – Linear Algebra, 4th Ed.
Tags: diagonalizability, eigenspace, algebraic multiplicity, geometric multiplicity, characteristic polynomial splits, direct sum, test for diagonalization
Difficulty: Intermediate | Prerequisites: Section 5.1 (Eigenvalues and Eigenvectors), null space, rank, determinants.
Section 5.1 introduced eigenvalues and eigenvectors and showed that diagonalizability is equivalent to finding a basis of eigenvectors. This section completes the picture by giving a concrete, checkable test for diagonalizability, and a systematic method for constructing the basis when the test passes. The two conditions are: (1) the characteristic polynomial must split into linear factors, and (2) for every eigenvalue, the dimension of the eigenspace must equal the algebraic multiplicity. If you are comfortable finding eigenvalues and null spaces from Section 5.1, you are ready for this section.
An operator is diagonalizable if and only if its characteristic polynomial splits and, for every eigenvalue, the eigenspace dimension equals the eigenvalue's multiplicity. When this holds, union the bases of the eigenspaces to get a basis of eigenvectors for the whole space.
Eigenspace (E_λ)
For an eigenvalue λ of T, the eigenspace is E_λ = N(T - λI) = {x ∈ V : T(x) = λx}. It is a subspace consisting of the zero vector together with all eigenvectors for λ.
Think of it as: the "λ-direction" of T, collected into a subspace. Its dimension tells you how many independent eigenvectors you can extract for that eigenvalue.
Algebraic multiplicity
The largest positive integer k such that (t - λ)^k divides the characteristic polynomial f(t). It counts how many times λ appears as a root.
In simple terms, this is how many times the factor (t - λ) repeats when you fully factor the characteristic polynomial.
Geometric multiplicity
The dimension of the eigenspace E_λ, i.e. dim(N(T - λI)). Equivalently, n - rank(A - λI).
Think of it as: the actual number of linearly independent eigenvectors you can find for λ. Always satisfies 1 ≤ geometric multiplicity ≤ algebraic multiplicity.
Splits (characteristic polynomial)
A polynomial f(t) over F splits over F if it factors completely into linear factors: f(t) = c(t - a_1)(t - a_2)···(t - a_n) with all a_i in F.
In simple terms, every root of the polynomial lives in the field you are working over. Over C this is always true (Fundamental Theorem of Algebra). Over R it can fail.
Direct sum
V = W_1 ⊕ W_2 ⊕ ··· ⊕ W_k means V is the sum of subspaces W_i and the intersection of any W_j with the sum of the others is {0}. Equivalently, every vector in V has a unique decomposition as w_1 + w_2 + ··· + w_k.
Think of it as: the subspaces partition V in a clean, non-overlapping way, like coordinate axes that span the whole space.
If v_1, v_2, ..., v_k are eigenvectors corresponding to distinct eigenvalues λ_1, λ_2, ..., λ_k, then {v_1, ..., v_k} is linearly independent.
Proved by induction: apply (T - λ_k I) to a dependence relation to eliminate the last vector, then invoke the induction hypothesis.
Corollary: If T has n distinct eigenvalues (where n = dim V), then T is diagonalizable. You simply pick one eigenvector per eigenvalue and they automatically form a basis.
The characteristic polynomial of any diagonalizable operator splits.
If T is diagonalizable with basis β of eigenvectors, then [T]_β = D is diagonal with eigenvalues on the diagonal. So det(D - tI) = (λ_1 - t)(λ_2 - t)···(λ_n - t), which clearly splits.
The converse is false: the characteristic polynomial can split without the operator being diagonalizable. The differentiation operator on P_2(R) has characteristic polynomial -t^3 (splits, only eigenvalue 0 with multiplicity 3), but dim(E_0) = 1, so it is not diagonalizable.
For any eigenvalue λ with algebraic multiplicity m:
1 ≤ dim(E_λ) ≤ m
The lower bound holds because eigenvalues always have at least one eigenvector. The upper bound is proved by extending a basis for E_λ to a basis for V and examining the block structure of the resulting matrix.
T is diagonalizable if and only if:
The characteristic polynomial of T splits.
For each eigenvalue λ, the algebraic multiplicity equals dim(E_λ).
When both conditions hold, union the bases of all eigenspaces to get a basis for V.
Practical shortcut: Condition 2 is automatic for eigenvalues of multiplicity 1. You only need to check condition 2 for repeated eigenvalues.
Compute the characteristic polynomial det(A - tI).
Factor it. If it does not split (over your field), stop: not diagonalizable.
For each eigenvalue λ with multiplicity m > 1, compute n - rank(A - λI).
If this equals m for every repeated eigenvalue, A is diagonalizable. Otherwise, it is not.
To find the basis: compute a basis for each eigenspace E_λ. Their union is the eigenvector basis.
If A is diagonalizable:
Q = matrix whose columns are the eigenvector basis
D = Q^{-1}AQ = diagonal matrix with eigenvalues on the diagonal (in the same order as the columns of Q)
Since A = QDQ^{-1}, we get A^n = QD^nQ^{-1}. For a diagonal matrix, D^n is just each diagonal entry raised to the nth power. This makes computing A^n for large n straightforward.
Example: For A = [[0, -2],[1, 3]], the eigenvalues are 1 and 2. With Q = [[-2, -1],[1, 1]], we get A^n = Q · diag(1^n, 2^n) · Q^{-1} = [[2 - 2^n, 2 - 2^{n+1}],[-1 + 2^n, -1 + 2^{n+1}]].
For x' = Ax, if A is diagonalizable with A = QDQ^{-1}, substitute y = Q^{-1}x to decouple the system into y' = Dy. Each equation y_i' = λ_i y_i has solution y_i(t) = c_i e^{λ_i t}. Then x(t) = Qy(t).
The general solution is x(t) = e^{λ_1 t} z_1 + e^{λ_2 t} z_2 + ··· where each z_i lies in E_{λ_i}.
If S_i is a linearly independent subset of E_{λ_i} for each i, then S_1 ∪ S_2 ∪ ··· ∪ S_k is linearly independent. This guarantees that assembling bases eigenspace-by-eigenspace always yields an independent set.
T is diagonalizable if and only if V = E_{λ_1} ⊕ E_{λ_2} ⊕ ··· ⊕ E_{λ_k} (the direct sum of eigenspaces). Equivalent conditions (Theorem 5.10) include: unique decomposition of every vector, and that the union of bases for each summand is a basis for V.
Test for diagonalization (condition 2): For each eigenvalue λ with multiplicity m: dim(E_λ) = n - rank(A - λI) must equal m.
Powers of a diagonalizable matrix: A^n = Q · diag(λ_1^n, λ_2^n, ..., λ_n^n) · Q^{-1}
General solution to x' = Ax (A diagonalizable): x(t) = c_1 e^{λ_1 t} v_1 + c_2 e^{λ_2 t} v_2 + ··· + c_n e^{λ_n t} v_n, where v_i are eigenvectors.
Diagonalization is the workhorse behind solving systems of linear ODEs (electrical circuits, coupled oscillators, population dynamics). The matrix exponential e^{At}, central to control theory, is trivially computed when A is diagonalizable. In data science, principal component analysis diagonalizes a covariance matrix to find the directions of greatest variance.
"If A has fewer than n distinct eigenvalues, it cannot be diagonalized." False. The identity matrix has only one eigenvalue (λ = 1) but is trivially diagonalizable. What matters is whether each eigenspace is large enough, not whether eigenvalues are distinct.
"If the characteristic polynomial splits, T is diagonalizable." False. Splitting is necessary but not sufficient. The differentiation operator on P_2(R) splits (char. poly. = -t^3) but is not diagonalizable because dim(E_0) = 1 < 3.
"Two distinct eigenvectors for the same eigenvalue are always linearly dependent." False. An eigenspace can be multi-dimensional; any two non-parallel vectors in it are independent eigenvectors for the same eigenvalue.
"Diagonalization changes the operator itself." It does not. Diagonalization is a change of basis (a coordinate relabelling). The operator T is the same; only its matrix representation changes.
⚠️ The two-condition test (splits + multiplicity = eigenspace dimension) is the core exam concept. Memorise it and be ready to apply it step by step.
⚠️ When testing condition 2, you need to compute rank(A - λI) for each repeated eigenvalue. Practise row reduction under pressure.
⚠️ Computing A^n = QD^nQ^{-1} is a standard exam problem. Be able to form Q, compute Q^{-1}, and multiply.
⚠️ Systems of differential equations x' = Ax solved via diagonalization appear frequently in applied exam problems.
True or false: A diagonalizable matrix with only one eigenvalue must be a scalar matrix (λI). A: True. If the only eigenvalue is λ, then D = λI, so A = Q(λI)Q^{-1} = λI.
True or false: If E_{λ_1} ∩ E_{λ_2} = {0} for all pairs of eigenvalues, that alone proves diagonalizability. A: False. This intersection property always holds (by Theorem 5.5). Diagonalizability requires additionally that the eigenspaces are large enough to span V.
Fill in the blank: T is diagonalizable if and only if V = E_{λ_1} ⊕ E_{λ_2} ⊕ ··· ⊕ E_{λ_k}, i.e. V is the ______ of the eigenspaces. A: Direct sum.
Q: Test whether A = [[3, 1, 0],[0, 3, 0],[0, 0, 4]] is diagonalizable.
A: Characteristic polynomial: -(t-4)(t-3)^2. It splits. Eigenvalue λ = 4 has multiplicity 1 (automatically fine). Eigenvalue λ = 3 has multiplicity 2. Check: A - 3I = [[0,1,0],[0,0,0],[0,0,1]], which has rank 2. So dim(E_3) = 3 - 2 = 1 ≠ 2. Therefore A is not diagonalizable.
Q: Let A = [[1,1],[1,1]]. Is A diagonalizable? If so, find Q and D.
A: Characteristic polynomial: t(t-2). Eigenvalues 0 and 2, both multiplicity 1. Two distinct eigenvalues in a 2×2 matrix, so A is diagonalizable. E_0: null space of A, spanned by (1,-1). E_2: null space of A-2I = [[-1,1],[1,-1]], spanned by (1,1). So Q = [[1,1],[-1,1]], D = [[0,0],[0,2]].
Q: Given A = QDQ^{-1} with D = diag(1, 2), find A^{10}.
A: A^{10} = Q · diag(1^{10}, 2^{10}) · Q^{-1} = Q · diag(1, 1024) · Q^{-1}.
Q: Why is the operator T(f) = f' on P_2(R) not diagonalizable, even though its characteristic polynomial splits?
A: The characteristic polynomial is -t^3, so the only eigenvalue is 0 with multiplicity 3. But E_0 = N(T) = {constant polynomials}, which is 1-dimensional. Since dim(E_0) = 1 ≠ 3, condition 2 fails.
Q: For the system x'_1 = 3x_1 + x_2 + x_3, x'_2 = 2x_1 + 4x_2 + 2x_3, x'_3 = -x_1 - x_2 + x_3, write the general solution using diagonalization.
A: The coefficient matrix has eigenvalues 2 (multiplicity 2) and 4 (multiplicity 1). The general solution is x(t) = e^{2t}z_1 + e^{4t}z_2, where z_1 is an arbitrary vector in E_2 and z_2 is an arbitrary vector in E_4.
Diagonalizability feeds directly into Section 5.3, where the convergence of A^m depends on eigenvalues lying inside the unit disk. The case where diagonalization fails leads to the Jordan canonical form (Chapter 7), which handles non-diagonalizable operators. Direct sums (introduced here optionally) become essential in Chapter 7. The trace and determinant, which equal the sum and product of eigenvalues respectively, connect back to Chapters 2 and 4.
diagonalizability test, eigenspace dimension, algebraic multiplicity, geometric multiplicity, characteristic polynomial splits, direct sum of eigenspaces, change of basis matrix Q, diagonal matrix D, Q inverse A Q, matrix powers, systems of ODEs, decoupled system