Diagonalize a 2×2 matrix in 60 seconds

The method in 4 steps

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

Step 1 — Eigenvalues. Solutions of $\lambda^2 - t\lambda + \det = 0$ where:

• $t = \text{tr}(A) = a + d$ (trace)
• $\det = ad - bc$ (determinant)

Step 2 — Diagonalizability test. If $\Delta = t^2 - 4\det > 0$ (two distinct eigenvalues), then $A$ is diagonalizable over $\mathbb{R}$.

Step 3 — Eigenvectors. For each $\lambda$, solve $(A - \lambda I) v = 0$.

Step 4 — Diagonal form. $A = P D P^{-1}$ with $D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ and $P$ having the eigenvectors as columns.

Complete example: $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$

• Trace: $t = 5$. Det: $\det = 6$.
• Characteristic polynomial: $\lambda^2 - 5\lambda + 6 = 0$ → $\lambda_1 = 2, \lambda_2 = 3$.
• Eigenvector for $\lambda_1 = 2$: solve $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} v = 0$ → $v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
• Eigenvector for $\lambda_2 = 3$: solve $\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix} v = 0$ → $v_2 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
• $D = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$, $P = \begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$.

Case $\Delta = 0$: watch out

Single double eigenvalue. Matrix is diagonalizable iff the eigenspace has dimension 2. Otherwise, only triangularizable.

Case $\Delta < 0$

No real eigenvalues → not diagonalizable over $\mathbb{R}$, but is over $\mathbb{C}$.

?Your turn

Eigenvalues of A = ((4, 1), (2, 3)) are:

A.4 and 3B.5 and 2C.2 and 3D.4 and 2

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