Eigenvector of a 2×2 Matrix Notes

Published on: September 29, 2026

This Eigenvector of a 2×2 Matrix Calculator finds the eigenvectors of a \(2 \times 2\) 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 makes eigenvectors easy to check and repeat.

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

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

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 \(\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\) only stretches or shrinks, rather than turning.

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

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: \(\lambda = 1\), \(\lambda = 3\).

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

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(2\) eigenvalues are still to be solved, so each one is taken in turn.

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

Step 4a

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

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

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

Step 4b

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

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

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

Final Answer

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

Problem

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

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 \(\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\) only stretches or shrinks, rather than turning.

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

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: \(\lambda = 2\), \(\lambda = 5\).

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

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(2\) eigenvalues are still to be solved, so each one is taken in turn.

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

Step 4a

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

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

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

Step 4b

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

In this problem: Putting \(\lambda = 5\) into \(A - \lambda I\) gives \(\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

\((A - 5 I) = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = 2:\; v = \begin{pmatrix} - \frac{1}{2} \\ 1 \end{pmatrix},\quad \lambda = 5:\; v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)
See Example 3 Hide Example 3

Problem

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

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 \(\begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\) only stretches or shrinks, rather than turning.

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

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: \(\lambda = 1\), \(\lambda = 6\).

\(\lambda = 1,\; \lambda = 6\)

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(2\) eigenvalues are still to be solved, so each one is taken in turn.

\(\lambda = 1,\; \lambda = 6\)

Step 4a

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

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

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

Step 4b

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

In this problem: Putting \(\lambda = 6\) into \(A - \lambda I\) gives \(\begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix}\), and the vectors it sends to zero point along \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\).

\((A - 6 I) = \begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix} \;\Rightarrow\; v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\)

Final Answer

\(\lambda = 1:\; v = \begin{pmatrix} - \frac{1}{2} \\ 1 \end{pmatrix},\quad \lambda = 6:\; v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\)
See Example 4 Hide Example 4

Problem

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

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 \(\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\) only stretches or shrinks, rather than turning.

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

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: \(\lambda = 2\), \(\lambda = 3\).

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

Step 4

Step 4 - Solve for the eigenvector of each eigenvalue.

In this problem: \(2\) eigenvalues are still to be solved, so each one is taken in turn.

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

Step 4a

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

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

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

Step 4b

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

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

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

Final Answer

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