Eigenvalues and eigenvectors of $2 \times 2$ matrices

Recall that a matrix $A$ represents a linear transformation: it takes every vector in the plane and moves it somewhere else. We’ll use this idea to build geometric intuition for eigenvalues and eigenvectors by visualising them in 2D.

The box shows the matrix $A$, which you can edit. The arrow $\vec{v}$ is a test vector you can drag anywhere. The arrow labelled $A\vec{v}$ shows where that vector ends up after the transformation. In most directions, the transformation changes both the length and the direction of the vector.

The eigenvectors are different. The purple and green lines mark the special directions where $A$ does not rotate a vector. If $\vec{v}$ lies on one of these lines, then $A\vec{v}$ stays on the same line. The transformation only stretches, squashes, or flips it. The factor by which it is scaled is the corresponding eigenvalue.

Finally, the grid (toggle it on with the button). The blue grid is the original coordinate system. The red grid shows what happens when we apply $A$ to every point of the blue grid. Straight lines remain straight, parallel lines remain parallel, but the whole space is stretched, sheared, or compressed. This is the geometric meaning of a linear transformation: it reshapes the entire coordinate system in a uniform way.

Together, the picture ties three ideas: the grid shows how $A$ reshapes space, the test vector $\vec{v}$ shows how individual vectors are moved, and the eigenvectors reveal the privileged directions that survive the reshaping without rotation.

We know that eigenvectors and eigenvalues are defined by the equation:

$$A\vec{v} = \lambda \vec{v}.$$

This means that if $\vec{v}$ is an eigenvector, then applying the matrix $A$ to it produces a new vector that is parallel to $\vec{v}$. The scalar $\lambda$ is the eigenvalue, and it tells us how much the vector is stretched or squashed.

How can we spot this in practice? If $\vec{v}$ and $A\vec{v}$ are parallel, then the ratio of their lengths (norms) is exactly the eigenvalue:

$$\lambda = \frac{|A\vec{v}|}{|\vec{v}|}.$$

The game

• Drag the input vector $\vec{v}$ anywhere in the plane (drag from the outer edge of the arrow head).
• The diagram shows both $\vec{v}$ (blue) and $A\vec{v}$ (red).
• The top‑left box displays $|\vec{v}|$, $|A\vec{v}|$, and their ratio.
• The aim is to adjust $\vec{v}$ until the two arrows are parallel. When this happens, we’ve found an eigenvector.
• At that point, the ratio $\dfrac{|A\vec{v}|}{|\vec{v}|}$ gives us the corresponding eigenvalue.

The challenge

The matrix $A$:

$$A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix},$$

has integer eigenvalues.

Find both eigenvectors by dragging $\vec{v}$ until it lines up with $A\vec{v}$. Then read off the two eigenvalues from the ratio.

For the above matrix $A$, we can see that the eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = 3$. This is when both $\vec{v}$ and $A \vec{v}$ are pointing in the same direction (along the eigenvectors) and $A \vec{v}$ is simply a scaled version of $\vec{v}$ (scaled by exactly the corresponding eigenvalue).

More precisely, for:

• Eigenvalue $\lambda_1 = 2$, the eigenvectors are multiples of $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$.
• Eigenvalue $\lambda_2 = 3$, the eigenvectors are multiples of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

So far, we have found the eigenvalues and eigenvectors separately. In practice, the two always come together: to fully understand how a matrix acts as a transformation, we need both the eigenvalues (the scaling factors) and the eigenvectors (the invariant directions).

The procedure for a $2 \times 2$ matrix $A$ is as follows:

1. Find the eigenvalues

   • Set up the characteristic equation:
   $$\det(A - \lambda I) = 0$$
   • Expand this determinant to obtain a quadratic equation in $\lambda$.
   • Solve the quadratic to find the eigenvalues.

2. Find the eigenvectors

   • For each eigenvalue $\lambda_i$, substitute it into the eigenvector equation:
   $$(A - \lambda_i I)\vec{v}_i = \vec{0}.$$
   • Solve the resulting system of linear equations to find the general form of $\vec{v}_i$.
   • Express the eigenvector in a simple representative form (e.g. by setting a free variable to 1).

This two-step process gives the complete set of eigenvalues and eigenvectors for the matrix.

Compute the eigenvalues and eigenvectors of the matrix:

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

Solution

Step 1: Find the eigenvalues

Set up the characteristic equation $\det(A - \lambda I) = 0$, by first calculating $A - \lambda I$:

$$A - \lambda I = \begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix}.$$

Take the determinant and set to $0$:

$$(4 - \lambda)(3 - \lambda) - (2)(1) = 0.$$

Expanding gives the quadratic equation:

$$\lambda^2 - 7\lambda + 10 = 0.$$

Solve the quadratic:

$$\lambda = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2}.$$

So the eigenvalues are:

$$\lambda_1 = 5, \quad \lambda_2 = 2.$$

Step 2: Find the eigenvectors

For $\lambda_1 = 5$:

Solve $(A - 5I)\vec{v} = \vec{0}$:

$$\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Which gives the equations:

$$-v_1 + v_2 = 0, \quad 2v_1 - 2v_2 = 0.$$

Both reduce to $v_1 = v_2$.

So the general eigenvector is:

$$\vec{v}_1 = \begin{bmatrix} t \\ t \end{bmatrix} = t \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad t \neq 0.$$

A representative choice is to set $t=1$:

$$\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

For $\lambda_2 = 2$:

Solve $(A - 2I)\vec{v} = \vec{0}$:

$$\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Which gives the equation (both are the same):

$$2v_1 + v_2 = 0.$$

And so $v_2 = -2v_1$.

Let $v_1 = t$ and so the general eigenvector is:

$$\vec{v}_2 = \begin{bmatrix} t \\ -2t \end{bmatrix} = t \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \quad t \neq 0.$$

A representative choice is to set $t=1$:

$$\vec{v}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}.$$

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Putting it together we have:

• Eigenvalues: $\lambda_1 = 5$, $\lambda_2 = 2$
• Corresponding eigenvectors:
  • For $\lambda_1 = 5$: $\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
  • For $\lambda_2 = 2$: $\vec{v}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$.

Note that any non-zero scalar multiple of each of the above eigenvectors is also an eigenvector for the eigenvalue.

Once we’ve calculated the eigenvalues and eigenvectors of a matrix, it’s straightforward to verify whether a number $\lambda$ and a non-zero vector $\vec{v}$ actually form an eigenvalue-eigenvector pair for a matrix $A$. Simply check if the equation

$$A\vec{v} = \lambda\vec{v}$$

is satisfied.

In practice, this means:

1. Multiply the matrix $A$ by the vector $\vec{v}$.
2. Multiply the vector $\vec{v}$ by the scalar $\lambda$.
3. Compare the results. If they are equal, then $\lambda$ and $\vec{v}$ are indeed an eigenvalue and eigenvector for $A$.

Since an $n \times n$ matrix has $n$ eigenvalues (counting multiplicities), we need to verify this relationship for each eigenvalue-eigenvector pair we find.

Let

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

Below are five candidate eigenvalue-eigenvector pairs for $A$:

1. $\lambda = 5$, $\vec{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$
2. $\lambda = 2$, $\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$
3. $\lambda = 4$, $\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$
4. $\lambda = 5$, $\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
5. $\lambda = 2$, $\vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Which pairs are valid eigenvalue-eigenvector pairs for $A$?

Solution

To verify each pair, we check whether $A\vec{v} = \lambda\vec{v}$.

Pair 1: $\lambda = 5$, $\vec{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

Compute $A\vec{v}$:

$$\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$$

Compute $5\vec{v}$:

$$5 \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$$

Valid: Yes, $A\vec{v} = 5\vec{v}$.

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Pair 2: $\lambda = 2$, $\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

Compute $A\vec{v}$:

$$\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix}$$

Compute $2\vec{v}$:

$$2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix}$$

Valid: Yes, $A\vec{v} = 2\vec{v}$.

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Pair 3: $\lambda = 4$, $\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

Compute $A\vec{v}$:

$$\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$$

Compute $4\vec{v}$:

$$4 \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$$

Valid: No, $A\vec{v} \neq 4\vec{v}$.

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Pair 4: $\lambda = 5$, $\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Compute $A\vec{v}$:

$$\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \end{bmatrix}$$

Compute $5\vec{v}$:

$$5 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix}$$

Valid: No, $A\vec{v} \neq 5\vec{v}$.

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Pair 5: $\lambda = 2$, $\vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Compute $A\vec{v}$:

$$\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$$

Compute $2\vec{v}$:

$$2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$$

Valid: Yes, $A\vec{v} = 2\vec{v}$.

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So, the valid eigenvalue-eigenvector pairs for $A$ are:

• Pair 1: $\lambda = 5$, $\vec{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$
• Pair 2: $\lambda = 2$, $\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$
• Pair 5: $\lambda = 2$, $\vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$