How to Diagonalize a 3×3 Matrix

To diagonalize a 3×3 matrix, you need to find its eigenvalues and eigenvectors, then check whether you can choose three linearly independent eigenvectors. If you can, those vectors form the columns of a matrix that changes the matrix into diagonal form.

The method works over a specified number system, usually the real or complex numbers. A matrix with real entries may have complex eigenvalues, so it may be diagonalizable over the complex numbers even when it is not diagonalizable over the reals.

What diagonalization means

A matrix $A$ is diagonalizable if it can be written as

$$A=PDP^{-1},$$

where $P$ is invertible and $D$ is diagonal. The columns of $P$ are eigenvectors of $A$, in the same order as their corresponding eigenvalues appear on the diagonal of $D$.

Equivalently, $AP=PD$. This form is often convenient for checking your result. If $A$ is diagonalizable, its powers are also easy to compute: $A^n=PD^nP^{-1}$.

Step-by-step method

1. Find the characteristic polynomial. Calculate $\det(A-\lambda I)$ and solve $\det(A-\lambda I)=0$. The solutions are the eigenvalues. Using $\det(\lambda I-A)$ instead is also standard; it gives the same eigenvalues.
2. Find each eigenspace. For each eigenvalue $\lambda$, solve $(A-\lambda I)v=0$. The nonzero solutions are its eigenvectors.
3. Choose three independent eigenvectors. For each eigenvalue, find a basis for its eigenspace. Altogether, you need three linearly independent vectors in $\mathbb{R}^3$ (or $\mathbb{C}^3$).
4. Form $P$ and $D$. Put your chosen eigenvectors in the columns of $P$. Put the matching eigenvalues on the diagonal of $D$, in the same order.
5. Check the result. Confirm that $P$ is invertible and that $AP=PD$. If both hold, then $A=PDP^{-1}$.

A 3×3 matrix with three distinct eigenvalues is always diagonalizable over a field containing those eigenvalues. If some eigenvalues repeat, diagonalizability depends on the number of independent eigenvectors, not just the characteristic polynomial.

Worked example

Consider

$$A=\begin{pmatrix} 4&1&0\\ 0&2&0\\ 0&0&3 \end{pmatrix}.$$

Because $A$ is triangular, its eigenvalues are its diagonal entries: $4$, $2$, and $3$. They are distinct, so the matrix is diagonalizable. Find an eigenvector for each one by solving $(A-\lambda I)v=0$.

For $\lambda=4$, the equations give $y=z=0$, so choose $v_1=(1,0,0)^T$. For $\lambda=2$, the first row gives $2x+y=0$ and the third gives $z=0$; choose $v_2=(-1,2,0)^T$. For $\lambda=3$, the equations give $x=y=0$, so choose $v_3=(0,0,1)^T$.

Use these vectors as the columns of $P$, keeping their eigenvalues in the corresponding order in $D$:

$$P=\begin{pmatrix} 1&-1&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}, \qquad D=\begin{pmatrix} 4&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix}.$$

The determinant of $P$ is $2$, so $P$ is invertible. Its inverse is

$$P^{-1}=\begin{pmatrix} 1&\tfrac12&0\\ 0&\tfrac12&0\\ 0&0&1 \end{pmatrix}.$$

Each column of $P$ is an eigenvector, so multiplying $A$ by that column scales it by its eigenvalue. Thus $AP=PD$, and the diagonalization is $A=PDP^{-1}$.

When a 3×3 matrix is not diagonalizable

A matrix is not diagonalizable if its eigenspaces do not together provide three linearly independent eigenvectors. For example, a repeated eigenvalue does not automatically prevent diagonalization. The identity matrix has the single eigenvalue $1$ with a three-dimensional eigenspace, so it is already diagonal. But a repeated eigenvalue whose eigenspace has too few independent vectors can make diagonalization impossible.

For each eigenvalue, its algebraic multiplicity is how many times it occurs as a root of the characteristic polynomial. Its geometric multiplicity is the dimension of its eigenspace. A 3×3 matrix is diagonalizable precisely when the dimensions of its eigenspaces add up to 3. For each eigenvalue, geometric multiplicity cannot exceed algebraic multiplicity.

Common mistakes

• Using eigenvalues as the columns of $P$. The columns must be eigenvectors; the eigenvalues go on the diagonal of $D$.
• Mismatching the order. If a column of $P$ is an eigenvector for $\lambda$, the corresponding diagonal entry of $D$ must be $\lambda$.
• Assuming repeated eigenvalues rule out diagonalization. Check the eigenspace dimensions instead.
• Assuming three eigenvalues counted with repetition guarantee diagonalization. You need three independent eigenvectors, not merely three roots of the characteristic polynomial.
• Trying to invert a singular $P$. If your three selected vectors are dependent, they cannot form a diagonalizing matrix.
• Forgetting the number system. Check whether the required eigenvalues and eigenvectors are real or whether complex numbers are needed.