Recipe

Find a matrix's eigenvalues and determine whether it's diagonalizable

Use when: you have a square matrix A and need its eigenvalues, plus a settled yes or no call on whether A=PDP^{-1} for some invertible P

Goal: every eigenvalue of A, plus a verdict on whether A has enough independent eigenvectors to diagonalize

Taught in Eigenvalues, eigenvectors, and diagonalization, revisited

1. 1

   Subtract \lambda from each diagonal entry only, then set the determinant of the result equal to zero.

   Example

   For A=\begin{pmatrix}5&2\\2&5\end{pmatrix}: A-\lambda I=\begin{pmatrix}5-\lambda&2\\2&5-\lambda\end{pmatrix}, so \det(A-\lambda I)=(5-\lambda)^2-4=0.

   Why: an eigenvalue is exactly a \lambda that makes A-\lambda I singular, sending some nonzero v to \mathbf 0.

2. 2

   Expand and solve for \lambda. Every root found this way is an eigenvalue of A.

   Example

   (5-\lambda)^2-4=0 \Rightarrow (5-\lambda)^2=4 \Rightarrow 5-\lambda=\pm2, giving \lambda=3 or \lambda=7, two distinct roots.

   • every root turns out distinct — each eigenvalue automatically gets a full, independent eigenvector; skip to Step 4.
   • a root repeats — continue to Step 3 before drawing any conclusion about diagonalizability.

   ⚡ Shortcut

   For any 2\times2 matrix, skip straight to \lambda=\frac{\operatorname{tr}A\pm\sqrt{(\operatorname{tr}A)^2-4\det A}}{2}. Here \operatorname{tr}A=5+5=10 and \det A=25-4=21, so \lambda=\frac{10\pm\sqrt{100-84}}{2}=\frac{10\pm4}{2}=7 or 3, matching the long way exactly.

   Why: the roots are the only candidates for eigenvalues; the shortcut only reaches the same equation faster for a 2\times2 matrix.

3. 3

   Find how many independent vectors solve the homogeneous system for that repeated \lambda. If that count matches the eigenvalue's multiplicity, it contributes enough eigenvectors; if it falls short, diagonalization fails outright.

   Example

   For D=\begin{pmatrix}4&0\\0&4\end{pmatrix} with repeated \lambda=4: (D-4I)v=\mathbf 0 is \mathbf 0v=\mathbf 0, so every vector works, a full 2-dimensional eigenspace. For E=\begin{pmatrix}4&1\\0&4\end{pmatrix}, also repeated \lambda=4: (E-4I)v=\mathbf 0 forces the second coordinate to 0, leaving only a 1-dimensional eigenspace, one short.

   • the eigenspace's dimension matches the eigenvalue's multiplicity — this eigenvalue is fine; check whether every other eigenvalue also has enough.
   • the eigenspace's dimension falls short of the multiplicity — A is not diagonalizable, and no other eigenvalue can fix that.

   Why: P needs a full set of n independent eigenvector columns to be invertible, and a repeated eigenvalue only threatens that when its own eigenspace comes up short.

4. 4

   Declare A diagonalizable once every eigenvalue supplied its full share of independent eigenvectors. If A is also symmetric, those eigenvectors are automatically orthogonal across different eigenvalues.

   Example

   For A=\begin{pmatrix}5&2\\2&5\end{pmatrix} (symmetric), the eigenvectors are v_1=(1,1) for \lambda=7 and v_2=(1,-1) for \lambda=3: v_1\cdot v_2=(1)(1)+(1)(-1)=0, orthogonal with no extra effort.

   • A is symmetric — eigenvectors from different eigenvalues are guaranteed orthogonal, with no extra work.
   • A is diagonalizable but not symmetric — independence still holds, but perpendicularity is not guaranteed; check it directly if it's needed.

   Why: orthogonality is a bonus that symmetry specifically guarantees, not a property every diagonalizable matrix's eigenvectors share.