# Eigenvalues and Eigenvectors

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/eigenvalues-and-eigenvectors

## Eigenvector Foundations

Most vectors change direction under a linear transformation. But special nonzero vectors are only **scaled**, meaning the transformation stretches or shrinks them without rotating.

Such a vector is an **eigenvector**, and its scale factor is the **eigenvalue** $\lambda$. The defining equation is:
$$A v = \lambda v, \qquad v \neq \mathbf{0}$$

Applying $A$ to eigenvector $v$ returns $v$ scaled by $\lambda$. Eigenvectors act as the natural axes where a transformation operates as pure scaling.

| Property | Meaning | Effect on Vector |
| --- | --- | --- |
| $\lambda > 1$ | Expansion | Stretches length |
| $0 < \lambda < 1$ | Compression | Shrinks length |
| $\lambda < 0$ | Reflection | Flips direction |

> **Common pitfall:** An eigenvector must be **nonzero** and keep its **line of direction**. The zero vector satisfies $Av = \lambda v$ for every $\lambda$, so it is excluded.

## Finding Eigenvalues & Eigenspaces

To **find** eigenvalues, rewrite $Av = \lambda v$ as $(A - \lambda I)v = \mathbf{0}$. For a nonzero solution to exist, the matrix must be **singular**.

This yields the **characteristic equation**:
$$\det(A - \lambda I) = 0$$

Solving this polynomial gives the eigenvalues. For each eigenvalue, its eigenvectors are the nonzero solutions of $(A - \lambda I)v = \mathbf{0}$, known as the **eigenspace**.

| Check | Rule | Purpose |
| --- | --- | --- |
| **Trace** | Sum of eigenvalues | Consistency check |
| **Determinant** | Product of eigenvalues | Consistency check |

> **Long-term behavior:** Under repeated application ($A^k$), the largest-magnitude eigenvalue dominates dynamics, powering Markov chains and iterative methods.

## Practice questions

4 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. The characteristic equation is det(A − λI) = 0. Why does setting a determinant to zero find exactly the eigenvalues, rather than being an arbitrary computational trick?

A. Because det(A − λI) = 0 is a polynomial identity that holds for all λ, so its roots are automatically the eigenvalues.
B. Because determinants measure area, and eigenvectors are the vectors that preserve area under A.
C. Because (A − λI) must send a nonzero vector to zero, so be singular.
D. Because the determinant of a matrix equals the sum of its eigenvalues, which is zero for the eigenvalue equation.

**Answer:** C. Because (A − λI) must send a nonzero vector to zero, so be singular.

**Why:** The rearrangement is the whole argument: Av = λv becomes (A − λI)v = 0 with v nonzero, so A − λI has a nonzero kernel and is singular, and a square matrix is singular exactly when its determinant vanishes. Note det(A − λI) is a polynomial that is *not* zero for most λ; the eigenvalues are its roots, which is why there are at most n of them.

Page: https://tryals.app/practice/mathematics-i/eigenvalues-and-eigenvectors/the-characteristic-equation-is-det-a-i-0-why-does-setting-a

### 2. A rotation of the plane by 90° has **no real eigenvectors**. Why should that be obvious before any computation?

A. Because eigenvectors exist only for symmetric matrices
B. Because a 90° rotation moves every direction off its own line
C. Because rotations are not linear transformations
D. Because the determinant of a rotation is 1, which forbids eigenvalues

**Answer:** B. Because a 90° rotation moves every direction off its own line

**Why:** Every line is moved to a perpendicular one, so no direction survives, and the characteristic polynomial λ² + 1 duly has no real roots. It is also why complex eigenvalues are not a technicality: they encode rotation, which is exactly the behaviour that has no invariant real direction.

Page: https://tryals.app/practice/mathematics-i/eigenvalues-and-eigenvectors/a-rotation-of-the-plane-by-90-has-no-real-eigenvectors-why-should

### 3. Select every TRUE statement about eigenvalues and eigenvectors.

A. If v is an eigenvector with eigenvalue lambda, then 3v is too
B. Every square matrix of real numbers has a real eigenvalue
C. An eigenvalue may be negative
D. An eigenvector must be nonzero
E. The eigenvalues of a triangular matrix are its diagonal entries

**Answer:** A. If v is an eigenvector with eigenvalue lambda, then 3v is too; C. An eigenvalue may be negative; D. An eigenvector must be nonzero; E. The eigenvalues of a triangular matrix are its diagonal entries

**Why:** Eigenvectors are nonzero and come in lines, so any nonzero multiple works; negative eigenvalues simply flip the vector. But the quarter-turn rotation [0, -1 ; 1, 0] has characteristic equation lambda squared + 1 = 0 with no real root, it preserves no real direction at all.

Page: https://tryals.app/practice/mathematics-i/eigenvalues-and-eigenvectors/select-every-true-statement-about-eigenvalues-and-eigenvectors

### 4. Let A = [3, 0 ; 0, -1], written as rows separated by a semicolon. Sort each vector by whether it is an eigenvector of A.

**Answer:**

- Eigenvector of A: (1, 0), (0, 1), (0, -5)
- Not an eigenvector of A: (1, 1), (2, 3), (0, 0)

**Why:** A multiplies the first coordinate by 3 and the second by -1, so (1,1) goes to (3,-1) and (2,3) to (6,-3), neither a multiple of the original. The zero vector satisfies A0 = lambda 0 for EVERY lambda, so it is excluded by definition, an eigenvector must be nonzero.

Page: https://tryals.app/practice/mathematics-i/eigenvalues-and-eigenvectors/let-a-3-0-0-1-written-as-rows-separated-by-a-semicolon