How Do You Diagonalize a 2×2 Matrix?

You diagonalize a $2\times2$ matrix $A$ by finding an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$. The columns of $P$ are eigenvectors of $A$, and the corresponding diagonal entries of $D$ are its eigenvalues.

The Method

This is diagonalization, covered in IB Mathematics: Applications and Interpretation AHL 1.15. IB examination questions restrict diagonalization to $2\times2$ matrices with distinct real eigenvalues.

Diagonalization changes the basis so that the transformation acts by simple scaling along two eigenvector directions. This is why it makes repeated matrix multiplication much more efficient.

1. Find the eigenvalues by solving the characteristic equation:

$$\det(A-\lambda I)=0.$$

2. For each eigenvalue $\lambda$, find a corresponding eigenvector $\mathbf{v}$ by solving:

$$(A-\lambda I)\mathbf{v}=\mathbf{0}.$$

3. Place the eigenvectors as columns of $P$. Place their corresponding eigenvalues in the same order along the diagonal of $D$.

Consider

$$A=\begin{pmatrix}4&1\\2&3\end{pmatrix}.$$

The characteristic equation is

$$\det\begin{pmatrix}4-\lambda&1\\2&3-\lambda\end{pmatrix}=0,$$

so

$$(4-\lambda)(3-\lambda)-2=0.$$

Expanding gives $\lambda^2-7\lambda+10=0$, so the eigenvalues are $\lambda=5$ and $\lambda=2$. Because these eigenvalues are distinct, their eigenvectors are linearly independent, so they can form the columns of an invertible matrix .

Solving the corresponding eigenvector equations gives

$$\lambda=5:\quad \mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}, \qquad \lambda=2:\quad \mathbf{v}_2=\begin{pmatrix}1\\-2\end{pmatrix}.$$

Therefore,

$$P=\begin{pmatrix}1&1\\1&-2\end{pmatrix}, \qquad D=\begin{pmatrix}5&0\\0&2\end{pmatrix}.$$

Since $\det(P)=-3\neq0$, $P$ is invertible, and therefore $A=PDP^{-1}$. For this example, . Multiplying reproduces , which verifies both the calculations and the column ordering.

Diagonalization makes powers easier because

$$A^n=PD^nP^{-1}.$$

In particular, $D^n=\operatorname{diag}(5^n,2^n)$, while direct multiplication of $A$ repeatedly would be longer and more error-prone.

A common misconception is that the eigenvalues can be reordered without changing $P$. They may be reordered, but each eigenvalue in $D$ must remain aligned with its eigenvector column in $P$.

Exam Technique

On an IB Math AI HL paper, show the characteristic equation and eigenvector equations even when using a GDC. If asked to verify the diagonalization, calculate $PDP^{-1}$ and confirm that it equals $A$. A GDC may check the result, but calculator output should not replace the mathematical setup.