Eigenvector of a 3×3 Matrix Notes

Published on: September 29, 2026

This Eigenvector of a 3×3 Matrix Calculator finds the eigenvectors of a \(3 \times 3\) matrix and shows the working line by line. The eigenvalues come first, since each one has its own direction. Solving \((A - \lambda I)v = 0\) for each gives a vector the matrix only stretches or shrinks, never turns. Any multiple of that vector works just as well. Following the same pattern every time keeps them checkable.

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Find the eigenvalues first.
  4. Solve for the eigenvector of each eigenvalue.

Formula:

\((A - \lambda I)\, v = 0\)
See Example 1 Hide Example 1

Problem

\(A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Find the eigenvalues first.
  4. Solve for the eigenvector of each eigenvalue.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the directions that \(\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\) only stretches or shrinks, rather than turning.

\(A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: A vector lying on such a direction is left at zero by \(A - \lambda I\), which is what this equation says.

\((A - \lambda I)\, v = 0\)

Step 3

Step 3 - Find the eigenvalues first.

In this problem: The eigenvalues come first, because each one has its own direction: \(\lambda = 1\), \(\lambda = 2\), \(\lambda = 11\).

\(\lambda = 1,\; \lambda = 2,\; \lambda = 11\)

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(3\) eigenvalues are still to be solved, so each one is taken in turn.

\(\lambda = 1,\; \lambda = 2,\; \lambda = 11\)

Step 4a

Step 4a - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 1\) into \(A - \lambda I\) gives \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 8 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}\).

\((A - 1 I) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 8 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}\)

Step 4b

Step 4b - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 2\) into \(A - \lambda I\) gives \(\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 4 & 7 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

\((A - 2 I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 4 & 7 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)

Step 4c

Step 4c - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 11\) into \(A - \lambda I\) gives \(\begin{pmatrix} -9 & 0 & 0 \\ 0 & -8 & 4 \\ 0 & 4 & -2 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ \frac{1}{2} \\ 1 \end{pmatrix}\).

\((A - 11 I) = \begin{pmatrix} -9 & 0 & 0 \\ 0 & -8 & 4 \\ 0 & 4 & -2 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ \frac{1}{2} \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = 1:\; v = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix},\quad \lambda = 2:\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad \lambda = 11:\; v = \begin{pmatrix} 0 \\ \frac{1}{2} \\ 1 \end{pmatrix}\)
See Example 2 Hide Example 2

Problem

\(A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Find the eigenvalues first.
  4. Solve for the eigenvector of each eigenvalue.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the directions that \(\begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\) only stretches or shrinks, rather than turning.

\(A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: A vector lying on such a direction is left at zero by \(A - \lambda I\), which is what this equation says.

\((A - \lambda I)\, v = 0\)

Step 3

Step 3 - Find the eigenvalues first.

In this problem: The eigenvalues come first, because each one has its own direction: \(\lambda = 1\), \(\lambda = 2\), \(\lambda = 3\).

\(\lambda = 1,\; \lambda = 2,\; \lambda = 3\)

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(3\) eigenvalues are still to be solved, so each one is taken in turn.

\(\lambda = 1,\; \lambda = 2,\; \lambda = 3\)

Step 4a

Step 4a - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 1\) into \(A - \lambda I\) gives \(\begin{pmatrix} 0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & -4 & 1 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} - \frac{1}{2} \\ 0 \\ 1 \end{pmatrix}\).

\((A - 1 I) = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & -4 & 1 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} - \frac{1}{2} \\ 0 \\ 1 \end{pmatrix}\)

Step 4b

Step 4b - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 2\) into \(A - \lambda I\) gives \(\begin{pmatrix} -1 & 2 & 0 \\ 0 & 1 & 0 \\ 2 & -4 & 0 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\).

\((A - 2 I) = \begin{pmatrix} -1 & 2 & 0 \\ 0 & 1 & 0 \\ 2 & -4 & 0 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\)

Step 4c

Step 4c - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 3\) into \(A - \lambda I\) gives \(\begin{pmatrix} -2 & 2 & 0 \\ 0 & 0 & 0 \\ 2 & -4 & -1 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{pmatrix}\).

\((A - 3 I) = \begin{pmatrix} -2 & 2 & 0 \\ 0 & 0 & 0 \\ 2 & -4 & -1 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = 1:\; v = \begin{pmatrix} - \frac{1}{2} \\ 0 \\ 1 \end{pmatrix},\quad \lambda = 2:\; v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},\quad \lambda = 3:\; v = \begin{pmatrix} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{pmatrix}\)
See Example 3 Hide Example 3

Problem

\(A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Find the eigenvalues first.
  4. Solve for the eigenvector of each eigenvalue.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the directions that \(\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\) only stretches or shrinks, rather than turning.

\(A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: A vector lying on such a direction is left at zero by \(A - \lambda I\), which is what this equation says.

\((A - \lambda I)\, v = 0\)

Step 3

Step 3 - Find the eigenvalues first.

In this problem: The eigenvalues come first, because each one has its own direction: \(\lambda = -1\), \(\lambda = 3\), \(\lambda = 3\).

\(\lambda = -1,\; \lambda = 3,\; \lambda = 3\)

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(3\) eigenvalues are still to be solved, so each one is taken in turn.

\(\lambda = -1,\; \lambda = 3,\; \lambda = 3\)

Step 4a

Step 4a - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = -1\) into \(A - \lambda I\) gives \(\begin{pmatrix} 4 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\).

\((A - -1 I) = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\)

Step 4b

Step 4b - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 3\) into \(A - \lambda I\) gives \(\begin{pmatrix} 0 & 0 & 0 \\ 0 & -2 & 2 \\ 0 & 2 & -2 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

\((A - 3 I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -2 & 2 \\ 0 & 2 & -2 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)

Step 4c

Step 4c - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 3\) into \(A - \lambda I\) gives \(\begin{pmatrix} 0 & 0 & 0 \\ 0 & -2 & 2 \\ 0 & 2 & -2 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).

\((A - 3 I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -2 & 2 \\ 0 & 2 & -2 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = -1:\; v = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix},\quad \lambda = 3:\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad \lambda = 3:\; v = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\)
See Example 4 Hide Example 4

Problem

\(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Find the eigenvalues first.
  4. Solve for the eigenvector of each eigenvalue.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the directions that \(\begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\) only stretches or shrinks, rather than turning.

\(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: A vector lying on such a direction is left at zero by \(A - \lambda I\), which is what this equation says.

\((A - \lambda I)\, v = 0\)

Step 3

Step 3 - Find the eigenvalues first.

In this problem: The eigenvalues come first, because each one has its own direction: \(\lambda = 2\), \(\lambda = 4\), \(\lambda = 5\).

\(\lambda = 2,\; \lambda = 4,\; \lambda = 5\)

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(3\) eigenvalues are still to be solved, so each one is taken in turn.

\(\lambda = 2,\; \lambda = 4,\; \lambda = 5\)

Step 4a

Step 4a - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 2\) into \(A - \lambda I\) gives \(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

\((A - 2 I) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)

Step 4b

Step 4b - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 4\) into \(A - \lambda I\) gives \(\begin{pmatrix} -2 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} \frac{1}{2} \\ 1 \\ 0 \end{pmatrix}\).

\((A - 4 I) = \begin{pmatrix} -2 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} \frac{1}{2} \\ 1 \\ 0 \end{pmatrix}\)

Step 4c

Step 4c - Solve for the eigenvector of each eigenvalue.

In this problem: Putting \(\lambda = 5\) into \(A - \lambda I\) gives \(\begin{pmatrix} -3 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\).

\((A - 5 I) = \begin{pmatrix} -3 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = 2:\; v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad \lambda = 4:\; v = \begin{pmatrix} \frac{1}{2} \\ 1 \\ 0 \end{pmatrix},\quad \lambda = 5:\; v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\)