# Eigenvectors and diagonalization

> Av=λv requires a nonzero vector v.

Course: [Linear algebra: spaces, eigenvectors and fitting](https://studybrix.com/en/courses/linear-algebra/index.md)
Level: university
Prerequisites: Matrix multiplication and characteristic polynomials.

## Notes

### Eigenpair

Av=λv requires a nonzero vector v.

### Diagonalization

A=PDP⁻¹ exists when there is a basis of eigenvectors over the chosen field.

### Repeated eigenvalues

Algebraic multiplicity does not guarantee enough independent eigenvectors.

### Symmetric matrices

A real symmetric matrix has an orthonormal eigenbasis over the reals.

## Practice questions

### 1. For A=[[3,1],[1,3]], find the eigenvalue of (1,1).

Answer: 4

Explanation: A(1,1)=(4,4)=4(1,1).

### 2. For A=[[3,1],[1,3]], find the eigenvalue corresponding to (1,−1).

Answer: 2

Explanation: A(1,−1)=(2,−2)=2(1,−1).

### 3. Why is the zero vector excluded from eigenvectors?

A. Its entries are undefined
B. It satisfies Av=λv for every λ and gives no eigendirection
C. It cannot be multiplied by a matrix
D. It is never in a vector space

Answer: It satisfies Av=λv for every λ and gives no eigendirection

Explanation: The definition needs a nonzero vector to encode a meaningful eigenvalue relation.

### 4. Is [[1,1],[0,1]] diagonalizable over the reals?

A. No; its eigenspace has dimension 1
B. No matrix with eigenvalue 1 is diagonalizable
C. Yes; repeated eigenvalues always suffice
D. Yes; every triangular matrix is diagonalizable

Answer: No; its eigenspace has dimension 1

Explanation: The only eigenvalue is 1, and eigenvectors have second component zero.

### 5. A diagonal matrix has diagonal entries 2,3,5. Find the trace of its square.

Answer: 38

Explanation: Squaring gives diagonal entries 4,9,25; their sum is 38.

### 6. Show that eigenvectors for two distinct eigenvalues are linearly independent.

Sample answer / marking guidance: If v₂=cv₁ with c≠0, applying A gives λ₂cv₁=cλ₁v₁, forcing λ₂=λ₁, a contradiction.

Explanation: Credit use of nonzero vectors and the distinct-eigenvalue contradiction.

### 7. A real symmetric matrix always admits what decomposition?

A. Orthogonal diagonalization
B. No diagonal representation
C. Only a complex basis with no real eigenvalues
D. A unique eigenvector in total

Answer: Orthogonal diagonalization

Explanation: The real spectral theorem provides an orthonormal basis of eigenvectors.

### 8. Explain why diagonalization helps calculate A^n.

Sample answer / marking guidance: From A=PDP⁻¹, adjacent P⁻¹P factors cancel so A^n=PD^nP⁻¹. Powers of a diagonal matrix are computed entrywise.

Explanation: Credit the cancellation argument and the correct order of factors.

## Further reading

- https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/
- https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.