Recipe

Find the eigenvalues of a matrix

Use when: you have a square matrix A and need the numbers \lambda satisfying A\mathbf v=\lambda\mathbf v for some nonzero \mathbf v

Goal: every eigenvalue of A, as roots of one polynomial equation

Taught in Eigenvalues and the characteristic equation

1. 1

   Write A back down, then subtract \lambda from each number on the main diagonal (top-left to bottom-right) only. Every other number in the matrix stays exactly as it was — don't touch them.

   Example

   Take A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. The main diagonal is 4 and 3; the off-diagonal entries are 1 and 2. Subtracting \lambda from the diagonal only: A-\lambda I=\begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix} The 1 and the 2 are untouched — that's the whole step.

   Why: this is exactly the matrix whose null space, for the right choice of \lambda, holds A's eigenvectors, since A\mathbf v=\lambda\mathbf v rearranges directly into (A-\lambda I)\mathbf v=\mathbf 0.

2. 2

   Set \det(A-\lambda I)=0, then actually multiply everything out — don't leave it in matrix form. For a 2\times2 matrix \begin{bmatrix}a&b\\c&d\end{bmatrix}, the determinant is ad-bc: multiply the two diagonal entries together, multiply the two off-diagonal entries together, and subtract the second product from the first.

   Example

   Continuing with A-\lambda I=\begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix}, multiply the diagonal entries together first: (4-\lambda)(3-\lambda) = 4\cdot3 - 4\lambda - 3\lambda + \lambda^2 = 12 - 7\lambda + \lambda^2 (that's ordinary FOIL: first \times first, outer, inner, last \times last, then combine the two middle \lambda terms.) Now multiply the off-diagonal entries: 1 \times 2 = 2. Subtract that from what you just found: \det(A-\lambda I) = (12-7\lambda+\lambda^2) - 2 = \lambda^2 - 7\lambda + 10 Set it equal to zero: \lambda^2 - 7\lambda + 10 = 0

   ⚡ Shortcut

   For ANY 2\times2 matrix, you can skip the FOIL step above entirely. The characteristic equation always comes out to \lambda^2 - \operatorname{tr}(A)\,\lambda + \det(A) = 0 where \operatorname{tr}(A) (the "trace") is just the two diagonal entries added together, and \det(A) is the ordinary determinant of A itself (no \lambda involved). For this same A: \operatorname{tr}(A) = 4+3=7 and \det(A) = (4)(3)-(1)(2)=10, so the equation is \lambda^2-7\lambda+10=0 — read straight off two numbers, matching the long way exactly.

   A 3\times3 matrix has the same kind of shortcut, one term longer: \lambda^3 - \operatorname{tr}(A)\,\lambda^2 + C\lambda - \det(A) = 0 where C is the sum of the three 2\times2 "minors" you get by crossing out one row and the matching column at a time. For A=\begin{bmatrix}2&3&2\\1&2&1\\1&1&2\end{bmatrix}: \operatorname{tr}(A)=2+2+2=6; crossing out row/column 1, 2, and 3 in turn gives minors \det\begin{bmatrix}2&1\\1&2\end{bmatrix}=3, \det\begin{bmatrix}2&2\\1&2\end{bmatrix}=2, and \det\begin{bmatrix}2&3\\1&2\end{bmatrix}=1, so C=3+2+1=6; and \det(A)=1. That gives \lambda^3-6\lambda^2+6\lambda-1=0 instantly — the same cubic you'd eventually reach by expanding the 3\times3 determinant directly, but without the row-by-row cofactor bookkeeping. Solving that cubic by hand is still real work (it's a genuine cubic, not a quadratic), which is why most people still expand the determinant directly at 3\times3 and larger — but recognizing the trace/det pattern is a fast way to double-check whatever you get.

   Why: a nonzero \mathbf v solving (A-\lambda I)\mathbf v=\mathbf 0 exists exactly when A-\lambda I is singular, and a square matrix is singular exactly when its determinant is zero.

3. 3

   If the equation factors, factor it: look for two numbers that multiply to the constant term and add to the coefficient on \lambda (with the sign flipped). If it doesn't factor by inspection, fall back to the quadratic formula.

   Example

   \lambda^2-7\lambda+10=0: look for two numbers that multiply to 10 and add to 7 — that's 5 and 2. So it factors as (\lambda-5)(\lambda-2)=0 A product is zero only when one of its factors is zero, so \lambda-5=0 or \lambda-2=0, giving \lambda=5 \quad\text{or}\quad \lambda=2

   Why: every root of this polynomial is, by construction, a value of \lambda where A-\lambda I collapses a direction, so the roots are exactly A's eigenvalues.

4. 4

   Substitute each root back into \det(A-\lambda I) by hand and confirm it actually comes out to zero — don't just trust the factoring.

   Example

   Check \lambda=5: (4-5)(3-5)-(1)(2) = (-1)(-2)-2 = 2-2=0. ✓ Check \lambda=2: (4-2)(3-2)-(1)(2) = (2)(1)-2=2-2=0. ✓ Both roots hold up, so \boxed{\lambda = 5 \text{ and } \lambda = 2 \text{ are the eigenvalues of } A}

   Why: this catches an arithmetic slip in the expansion or the factoring before it carries into the next step, finding an eigenvector for a value that was never actually a root.