Compute eigenvalues and eigenvectors of a $3 \times 3$ matrix

Primary Tool

Eigenvalue & Eigenvector Calculator

The dedicated eigenvalue tool keeps the algebra in one place. That lets the page spend its time on interpretation: where the eigenvalues come from, what the eigenspaces look like, and why the matrix is diagonalizable.

Problem

Compute the eigenvalues and eigenvectors of $A = 110110003$ , and decide whether the matrix is diagonalizable.

This is a good first 3 by 3 example because the matrix is not trivial, but it also does not fight you at every line. The structure is visible enough that the main ideas stay in view.

Why this matrix stays readable

The matrix almost breaks into two pieces. The lower-right entry already gives one eigenvalue, $3$ , with eigenvector $001$ .

The remaining work sits inside the top-left block $[1111]$ , whose eigenvalues are $2$ and $0$ . So the example still feels like linear algebra, not just symbolic bookkeeping.

Step-by-step walkthrough

Treat the calculator output as a checkpoint, not the point of the exercise. The real job is to pair each eigenvalue with its eigenspace and then count whether you have enough directions.

1Write the characteristic polynomial: $det (A - λ I) = (3 - λ) det [1 - λ 1 1 1 - λ] = (3 - λ) λ (λ - 2)$ .
2Read off the eigenvalues: $λ = 3, 2, 0$ .
3For $λ = 2$ , solve $(A - 2 I) v = 0$ and get an eigenvector such as $110$ .
4For $λ = 0$ , solve $A v = 0$ and get an eigenvector such as $1 - 1 0$ .
5For $λ = 3$ , solve $(A - 3 I) v = 0$ and get $001$ .
6These three eigenvectors are linearly independent, so they form a basis of eigenvectors. That is exactly what diagonalizability requires.

Common Pitfall

Finding the eigenvalues is only part of the job. Diagonalization depends on having enough independent eigenvectors to build a basis, not just on listing the roots of the characteristic polynomial.

Try a Variation

Replace the bottom-right entry $3$ with $2$ . The eigenvalue $2$ then repeats. Does the matrix still have enough independent eigenvectors to be diagonalizable?

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Worked ExampleLinear AlgebraMath Workspace (CAS)

Diagonalize a $2 \times 2$ matrix and compute $A^{n}$

Diagonalization is one of the first places students see why eigenvalues matter computationally. Once $A = P D P^{- 1}$ , powers of $A$ become powers of a diagonal matrix.

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ProofLinear AlgebraProof Builder

Proof that eigenvectors for distinct eigenvalues are linearly independent

Distinct eigenvalues do more than label eigenspaces. They force eigenvectors to be independent, which is exactly why diagonalization becomes possible so often in practice.

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ExplanationLinear AlgebraMath Workspace (CAS)

What eigenvectors and eigenvalues mean geometrically

An eigenvector is a direction that survives a linear transformation without rotating away from its own line. The eigenvalue tells you the scale and possible sign flip along that direction.

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