LINEAR-ALGEBRA

Cracking the Characteristic Equation

IKey Formulas

Formula Name Description
Ax=λxA\mathbf{x} = \lambda \mathbf{x} Eigenvalue equation λ\lambda = eigenvalue, x\mathbf{x} = eigenvector
det⁡(A−λI)=0\det(A - \lambda I) = 0 Characteristic equation Polynomial whose roots are eigenvalues
(A−λiI)x=0(A - \lambda_i I)\mathbf{x} = \mathbf{0} Eigenvector equation Null space gives eigenvectors
∑λi=tr(A),∏λi=det⁡(A)\sum \lambda_i = \text{tr}(A), \quad \prod \lambda_i = \det(A) Sanity checks Trace & determinant relations

IIVisualization 1 — Eigenvectors Don't Rotate

Only along an eigenvector does AxA\mathbf{x} stay on the same line — a pure stretch, no turn.

💡 The two stretch factors λ=2\lambda = 2 and λ=4\lambda = 4 are exactly the roots of the characteristic equation λ2−6λ+8=0\lambda^2 - 6\lambda + 8 = 0 — the equation this lesson solves.

IIIVisualization 2 — Adding cIcI Shifts the Eigenvalues

Add cIcI to a matrix and every eigenvalue slides by cc — yet no eigenvector moves.

💡 The shift is exact: (P+cI)x=Px+cx=(λ+c)x(P + cI)\mathbf{x} = P\mathbf{x} + c\mathbf{x} = (\lambda + c)\mathbf{x}, so cIcI hands its own eigenvalue cc to every direction at once.

IVVisualization 3 — The Characteristic Polynomial

The eigenvalues of AA are the roots of p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I) — where its curve crosses zero.

💡 For any 2×22\times 2 matrix the characteristic equation is λ2−(tr⁡A) λ+det⁡A=0\lambda^2 - (\operatorname{tr}A)\,\lambda + \det A = 0 — read its coefficients straight off the trace and determinant.

VQuiz Questions

Question 1 · Compute the Characteristic Polynomial

For the matrix A=[5225]A = \begin{bmatrix} 5 & 2 \\ 2 & 5 \end{bmatrix}, what is the characteristic polynomial det⁡(A−λI)\det(A - \lambda I)?

✅ Correct! You properly subtracted the off-diagonal product.

❌ Not quite. You forgot to subtract 2×2=42 \times 2 = 4 from the diagonal product.

❌ Sign error. The coefficient of λ\lambda should be −(a+d)-(a+d), so −10-10 not +10+10.

❌ Not quite. Remember the trace is 5+5=105+5=10, not 77.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

Form A−λI=[5−λ225−λ]A - \lambda I = \begin{bmatrix} 5-\lambda & 2 \\ 2 & 5-\lambda \end{bmatrix}.

Compute the determinant: det⁡(A−λI)=(5−λ)(5−λ)−(2)(2)=25−10λ+λ2−4=λ2−10λ+21\det(A - \lambda I) = (5-\lambda)(5-\lambda) - (2)(2) = 25 - 10\lambda + \lambda^2 - 4 = \lambda^2 - 10\lambda + 21

Sanity check: trace =5+5=10= 5 + 5 = 10 (coefficient of −λ-\lambda), determinant =25−4=21= 25 - 4 = 21 (constant term).

The eigenvalues are roots: λ=7\lambda = 7 and λ=3\lambda = 3.

Question 2 · Finding the Eigenvector

For A=[4213]A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}, one eigenvalue is λ=5\lambda = 5. Which vector is a corresponding eigenvector?

✅ Correct! A[2,1]T=[10,5]T=5⋅[2,1]TA[2,1]^T = [10, 5]^T = 5 \cdot [2,1]^T.

❌ Components swapped. Check: A[1,2]T=[8,7]TA[1,2]^T = [8, 7]^T, which is NOT a scalar multiple of [1,2][1,2].

❌ Not quite. A[1,1]T=[6,4]TA[1,1]^T = [6, 4]^T, not a scalar multiple of [1,1][1,1].

❌ Sign error. Row reduction gives x1=2x2x_1 = 2x_2, so components have the same sign.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

Form A−5I=[4−5213−5]=[−121−2]A - 5I = \begin{bmatrix} 4-5 & 2 \\ 1 & 3-5 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix}.

Row reduce: R2←R2+R1R_2 \leftarrow R_2 + R_1 gives [−1200]\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}.

From row 1: −x1+2x2=0-x_1 + 2x_2 = 0, so x1=2x2x_1 = 2x_2. Setting x2=1x_2 = 1 gives x=[2,1]\mathbf{x} = [2, 1].

Verify: A[2,1]T=[4(2)+2(1),1(2)+3(1)]T=[10,5]T=5⋅[2,1]TA[2,1]^T = [4(2)+2(1), 1(2)+3(1)]^T = [10, 5]^T = 5 \cdot [2, 1]^T

Question 3 · The Shift Property

Suppose the matrix BB has eigenvalues {−2,1,5}\{-2, 1, 5\} with eigenvectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3. What are the eigenvalues and eigenvectors of B+4IB + 4I?

✅ Correct! Adding cIcI shifts each eigenvalue by cc while eigenvectors stay fixed.

❌ Not quite. Eigenvalues DO change (they shift), but eigenvectors stay the same.

❌ That's for multiplication. Eigenvalues of cAcA are cλic\lambda_i. But we're adding 4I4I, which shifts rather than scales.

❌ Eigenvalues are right, but eigenvectors don't rotate. Adding cIcI leaves eigenvectors completely unchanged.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The shift property: if Bv=λvB\mathbf{v} = \lambda \mathbf{v}, then (B+cI)v=Bv+cv=λv+cv=(λ+c)v(B + cI)\mathbf{v} = B\mathbf{v} + c\mathbf{v} = \lambda\mathbf{v} + c\mathbf{v} = (\lambda + c)\mathbf{v}

So adding cIcI:

  • Shifts each eigenvalue by cc: {−2,1,5}→{2,5,9}\{-2, 1, 5\} \to \{2, 5, 9\}
  • Preserves all eigenvectors unchanged

This works because the identity matrix II has every vector as an eigenvector, so it automatically shares eigenvectors with any matrix.

Question 4 · True or False — Eigenvalues of a Sum

Claim: For any two square matrices AA and BB of the same size, the eigenvalues of A+BA + B equal the sum of eigenvalues of AA and BB.

Is this claim true or false?

✅ Correct! The rule requires a shared eigenvector — that's why cIcI is the special case where it always works.

❌ This is a common trap. The algebra only works if x\mathbf{x} is an eigenvector of both AA and BB, which is usually NOT the case.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The claim is FALSE in general.

The flawed argument goes: if Ax=λxA\mathbf{x} = \lambda \mathbf{x} and Bx=αxB\mathbf{x} = \alpha \mathbf{x}, then (A+B)x=(λ+α)x(A+B)\mathbf{x} = (\lambda + \alpha)\mathbf{x}.

The hidden assumption is that the same vector x\mathbf{x} is an eigenvector of both AA and BB. Different matrices usually have eigenvectors pointing in completely different directions!

Why cIcI escapes this trap: the identity matrix II has every nonzero vector as an eigenvector, so it automatically shares eigenvectors with any matrix AA. That's the special case where additivity works.

Counterexample: Let A=[1000]A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} (eigenvalues 0, 1) and B=[0110]B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} (eigenvalues −1,1-1, 1). Then A+B=[1110]A+B = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} has eigenvalues 1±52\frac{1 \pm \sqrt{5}}{2} — definitely NOT any sum of {0,1}\{0,1\} and {−1,1}\{-1,1\}.

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