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Module 3/Eigenvalues & Diagonalization

Multiplicities, Trace, Determinant & Similarity

Two eigenvalues can be the same number yet behave differently. This lesson separates algebraic multiplicity (how often λ\lambda is a root of the characteristic polynomial) from geometric multiplicity (how many independent eigenvectors it has), uses the eigenvalues to read off the trace and determinant, pins down the bound 1≤gemu⁡(λ)≤almu⁡(λ)≤n1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda)\le n, and shows which spectral data survive transposing or passing to a similar matrix.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

For A=(210020002)A=\begin{pmatrix}2&1&0\\0&2&0\\0&0&2\end{pmatrix} the only eigenvalue is λ=2\lambda=2. Compute its geometric multiplicity gemu⁡(2)=3−rank⁡(A−2I)\operatorname{gemu}(2)=3-\operatorname{rank}(A-2I).

A 3×33\times3 matrix AA has characteristic polynomial −(λ−2)2(λ−5)-(\lambda-2)^2(\lambda-5), and its eigenspace E2=ker⁡(A−2I)E_2=\ker(A-2I) is one-dimensional. Which statement is correct?

What you’ll be able to do

  • State Definitions 7.3 and 7.4 and compute the algebraic multiplicity almu⁡(λ)\operatorname{almu}(\lambda) from the characteristic polynomial and the geometric multiplicity gemu⁡(λ)=n−rank⁡(A−λI)\operatorname{gemu}(\lambda)=n-\operatorname{rank}(A-\lambda I).
  • Use Corollary 7.2 to obtain det⁡A=∏iλi\det A=\prod_i\lambda_i and tr⁡A=∑iλi\operatorname{tr}A=\sum_i\lambda_i from the eigenvalues counted with algebraic multiplicity.
  • Apply the bound 1≤gemu⁡(λ)≤almu⁡(λ)≤n1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda)\le n (Theorem 7.6) and identify defective matrices (some eigenvalue with gemu⁡<almu⁡\operatorname{gemu}<\operatorname{almu}).
  • Explain why AA and ATA^{T}, and any two similar matrices, share the same characteristic polynomial, eigenvalues, and multiplicities (Theorems 7.4 and 7.5).
  • Use Theorem 7.3 to bound the number of real eigenvalues of an n×nn\times n matrix.

In your course

· MATH2015 · Linear Algebra & Probability
§7.1 Eigenvalues and eigenvectors
  • Definition 7.3Algebraic multiplicity
    λ0\lambda_0 has algebraic multiplicity kk if cA(λ)=(λ0−λ)kg(λ)c_A(\lambda)=(\lambda_0-\lambda)^k g(\lambda) with g(λ0)≠0g(\lambda_0)\neq 0; written almu⁡(λ0)=k\operatorname{almu}(\lambda_0)=k.
  • Definition 7.4Geometric multiplicity
    gemu⁡(λ)=dim⁡Eλ=nullity⁡(A−λIn)=n−rank⁡(A−λIn)\operatorname{gemu}(\lambda)=\dim E_\lambda=\operatorname{nullity}(A-\lambda I_n)=n-\operatorname{rank}(A-\lambda I_n).
  • Theorem 7.3Number of eigenvalues
    An n×nn\times n matrix has at most nn real eigenvalues counted with algebraic multiplicity; if nn is odd, at least one real eigenvalue.
  • Corollary 7.2Eigenvalues, determinant, and trace
    det⁡A=λ1⋯λn\det A=\lambda_1\cdots\lambda_n and tr⁡A=λ1+⋯+λn\operatorname{tr}A=\lambda_1+\cdots+\lambda_n, with eigenvalues listed by algebraic multiplicity.
  • Theorem 7.4Eigenvalues of the transpose
    AA and ATA^{T} have the same characteristic polynomial, eigenvalues, and multiplicities.
  • Theorem 7.5Eigenvalues of similar matrices
    If B=S−1ASB=S^{-1}AS with SS invertible, then AA and BB have the same characteristic polynomial, eigenvalues, and multiplicities.
  • Theorem 7.6Algebraic versus geometric multiplicity
    1≤gemu⁡(λ)≤almu⁡(λ)≤n1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda)\le n.
  • Proposition 7.1Complex conjugate eigenvalues (optional)
    For a real matrix, a complex eigenvalue λ=μ+iβ\lambda=\mu+i\beta with eigenvector v=x+iyv=x+iy is accompanied by λ‾=μ−iβ\overline{\lambda}=\mu-i\beta with eigenvector v‾=x−iy\overline{v}=x-iy.
Based on §7.1 Definitions 7.3–7.4, Theorems 7.3–7.6, Corollary 7.2, and Proposition 7.1, with Examples 7.3 (diagonalizable) and 7.4 (defective). The term 'defective' (some eigenvalue with gemu⁡(λ)<almu⁡(λ)\operatorname{gemu}(\lambda)<\operatorname{almu}(\lambda)) is introduced just before Corollary 7.3. Because eigenvectors are only defined up to a nonzero scalar, the auto-graded items deliberately target multiplicities and the trace/determinant invariants rather than asking for specific eigenvectors.
1

Two multiplicities for one eigenvalue

Every eigenvalue carries two counts. The algebraic multiplicity almu⁡(λ0)\operatorname{almu}(\lambda_0) is the multiplicity of λ0\lambda_0 as a root of the characteristic polynomial cA(λ)c_A(\lambda): we can factor cA(λ)=(λ0−λ)k g(λ)c_A(\lambda)=(\lambda_0-\lambda)^k\,g(\lambda) with g(λ0)≠0g(\lambda_0)\neq 0, and then almu⁡(λ0)=k\operatorname{almu}(\lambda_0)=k (Definition 7.3). The geometric multiplicity gemu⁡(λ)=dim⁡Eλ\operatorname{gemu}(\lambda)=\dim E_\lambda counts independent eigenvectors; since Eλ=ker⁡(A−λIn)E_\lambda=\ker(A-\lambda I_n), rank–nullity gives gemu⁡(λ)=nullity⁡(A−λIn)=n−rank⁡(A−λIn)\operatorname{gemu}(\lambda)=\operatorname{nullity}(A-\lambda I_n)=n-\operatorname{rank}(A-\lambda I_n) (Definition 7.4). To find almu⁡\operatorname{almu} you factor a polynomial; to find gemu⁡\operatorname{gemu} you row-reduce a matrix. The two numbers need not agree.

2

Trace and determinant from the spectrum

Corollary 7.2 turns the eigenvalues into two familiar numbers. If the eigenvalues of an n×nn\times n matrix AA, listed with algebraic multiplicity, are λ1,…,λn\lambda_1,\dots,\lambda_n, then det⁡A=λ1λ2⋯λn\det A=\lambda_1\lambda_2\cdots\lambda_n and tr⁡A=λ1+λ2+⋯+λn\operatorname{tr}A=\lambda_1+\lambda_2+\cdots+\lambda_n. The phrase 'listed with algebraic multiplicity' matters: a double eigenvalue is counted twice. Two quick consequences follow: AA is invertible exactly when 00 is not an eigenvalue (because det⁡A=0\det A=0 iff some λi=0\lambda_i=0), and the trace gives a free sanity check — the sum of the eigenvalues must equal the sum of the diagonal entries.

3

Invariants under transpose and similarity

The characteristic polynomial is unchanged by transposing or by a change of basis. Theorem 7.4: cAT(λ)=cA(λ)c_{A^{T}}(\lambda)=c_A(\lambda) because det⁡(A−λI)=det⁡((A−λI)T)\det(A-\lambda I)=\det\big((A-\lambda I)^{T}\big), so AA and ATA^{T} have the same eigenvalues with the same algebraic and geometric multiplicities. Theorem 7.5: if B=S−1ASB=S^{-1}AS for an invertible SS, then cB(λ)=cA(λ)c_B(\lambda)=c_A(\lambda), so similar matrices share eigenvalues and both multiplicities. In both cases the eigenvectors generally differ (Remarks 7.2 and 7.3). These invariants are exactly what make diagonalization meaningful: a diagonal matrix DD similar to AA displays the spectrum of AA on its diagonal.

4

Defective matrices

Theorem 7.6 relates the two multiplicities: for every eigenvalue λ\lambda of an n×nn\times n matrix, 1≤gemu⁡(λ)≤almu⁡(λ)≤n1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda)\le n. Geometric multiplicity is at least 11 (an eigenvalue has at least one eigenvector) and never exceeds algebraic multiplicity. A matrix with some eigenvalue for which gemu⁡(λ)<almu⁡(λ)\operatorname{gemu}(\lambda)<\operatorname{almu}(\lambda) is called defective, and a defective matrix is not diagonalizable — it cannot supply enough independent eigenvectors to form an eigenbasis. Because a simple eigenvalue (almu⁡=1\operatorname{almu}=1) always has gemu⁡=1\operatorname{gemu}=1, only repeated eigenvalues can make a matrix defective.

Theorem 7.3 — Number of (real) eigenvalues

An n×nn\times n matrix has at most nn real eigenvalues, counted with their algebraic multiplicities. Moreover, if nn is odd, then the matrix has at least one real eigenvalue.

Intuition. The characteristic polynomial has degree nn, so by the Fundamental Theorem of Algebra it has exactly nn roots in C\mathbb{C} counted with multiplicity — at most nn of them real. An odd-degree real polynomial runs to +∞+\infty at one end and −∞-\infty at the other, so it must cross zero at least once, forcing a real eigenvalue.
Corollary 7.2 — Determinant and trace from eigenvalues

If the eigenvalues of an n×nn\times n matrix AA, listed with algebraic multiplicity, are λ1,…,λn\lambda_1,\dots,\lambda_n, then det⁡A=λ1λ2⋯λn\det A=\lambda_1\lambda_2\cdots\lambda_n and tr⁡A=λ1+λ2+⋯+λn\operatorname{tr}A=\lambda_1+\lambda_2+\cdots+\lambda_n.

Intuition. Compare the characteristic polynomial cA(λ)=(−1)nλn+(−1)n−1tr⁡(A)λn−1+⋯+det⁡Ac_A(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}\operatorname{tr}(A)\lambda^{n-1}+\cdots+\det A (Theorem 7.2) with its factored form ∏i(λi−λ)\prod_i(\lambda_i-\lambda). Setting λ=0\lambda=0 reads off the constant term det⁡A=∏iλi\det A=\prod_i\lambda_i, and matching the λn−1\lambda^{n-1} coefficients gives tr⁡A=∑iλi\operatorname{tr}A=\sum_i\lambda_i.
Theorem 7.6 — Geometric vs algebraic multiplicity

For every eigenvalue λ\lambda of an n×nn\times n matrix, 1≤gemu⁡(λ)≤almu⁡(λ)≤n1\le\operatorname{gemu}(\lambda)\le\operatorname{almu}(\lambda)\le n.

Intuition. Take a basis of the mm-dimensional eigenspace Eλ0E_{\lambda_0} and extend it to a basis of Rn\mathbb{R}^n. In that basis AA becomes block upper-triangular with the scalar block λ0Im\lambda_0 I_m in the corner, so the factor (λ0−λ)m(\lambda_0-\lambda)^m divides cA(λ)c_A(\lambda); hence almu⁡(λ0)≥m=gemu⁡(λ0)\operatorname{almu}(\lambda_0)\ge m=\operatorname{gemu}(\lambda_0). The outer bounds just say 'at least one eigenvector' and 'degree nn'.
Theorems 7.4 & 7.5 — Transpose and similarity invariance

A matrix AA and its transpose ATA^{T} have the same characteristic polynomial, hence the same eigenvalues with the same algebraic and geometric multiplicities (Theorem 7.4). Likewise, if B=S−1ASB=S^{-1}AS for some invertible SS, then AA and BB have the same characteristic polynomial and the same eigenvalues and multiplicities (Theorem 7.5).

Intuition. Both proofs ride on determinant identities: det⁡(A−λI)=det⁡((A−λI)T)\det(A-\lambda I)=\det\big((A-\lambda I)^{T}\big) handles the transpose, while det⁡(S(A−λI)S−1)=det⁡(A−λI)\det\big(S(A-\lambda I)S^{-1}\big)=\det(A-\lambda I) handles similarity. Ranks are preserved in both cases, so the geometric multiplicities match too. What is not preserved is the actual eigenvectors.

Worked examples

Example 1

Let A=(0−1−1121112)A=\begin{pmatrix}0&-1&-1\\1&2&1\\1&1&2\end{pmatrix} (Example 7.3), with characteristic polynomial cA(λ)=−(λ−1)2(λ−2)c_A(\lambda)=-(\lambda-1)^2(\lambda-2). Find det⁡A\det A and tr⁡A\operatorname{tr}A from the eigenvalues, and check both directly.

  1. 1

    Read the eigenvalues off the factored polynomial: λ=1\lambda=1 with almu⁡(1)=2\operatorname{almu}(1)=2, and λ=2\lambda=2 with almu⁡(2)=1\operatorname{almu}(2)=1. Listed with algebraic multiplicity, the eigenvalues are 1,1,21,1,2.

  2. 2

    Determinant via Corollary 7.2: det⁡A=λ1λ2λ3=1⋅1⋅2=2\det A=\lambda_1\lambda_2\lambda_3=1\cdot 1\cdot 2=2.

  3. 3

    Trace via Corollary 7.2: tr⁡A=1+1+2=4\operatorname{tr}A=1+1+2=4.

  4. 4

    Check the trace against the diagonal entries: 0+2+2=40+2+2=4. Match.

  5. 5

    Check the determinant by cofactor expansion along the first row: det⁡A=0⋅(4−1)−(−1)⋅(2−1)+(−1)⋅(1−2)=0+1+1=2\det A=0\cdot(4-1)-(-1)\cdot(2-1)+(-1)\cdot(1-2)=0+1+1=2. Match.

Answer. det⁡A=2\det A=2 and tr⁡A=4\operatorname{tr}A=4, both confirmed by direct computation.
Example 2

Let A=(1211−11201)A=\begin{pmatrix}1&2&1\\1&-1&1\\2&0&1\end{pmatrix} (Example 7.4), with cA(λ)=−(λ+1)2(λ−3)c_A(\lambda)=-(\lambda+1)^2(\lambda-3). Find the algebraic and geometric multiplicity of each eigenvalue, and decide whether AA is defective.

Example 3

Let A=(3103)A=\begin{pmatrix}3&1\\0&3\end{pmatrix}. Find its eigenvalue with both multiplicities, verify tr⁡A\operatorname{tr}A and det⁡A\det A against the eigenvalues, and confirm that ATA^{T} has the same spectral data.