# Bases, rank and linear maps

> A basis both spans a vector space and is linearly independent.

Course: [Linear algebra: spaces, eigenvectors and fitting](https://studybrix.com/en/courses/linear-algebra/index.md)
Level: university
Prerequisites: Systems of linear equations and vector arithmetic.

## Notes

### Basis

A basis both spans a vector space and is linearly independent.

### Rank

The rank is the dimension of the image, or equivalently the number of pivot columns.

### Nullity

Rank plus nullity equals the dimension of the domain.

### Subspace test

A candidate subspace must contain zero and be closed under addition and scalar multiplication.

## Practice questions

### 1. A linear map from R^7 to R^5 has rank 4. Find its nullity.

Answer: 3

Explanation: Rank-nullity uses the 7-dimensional domain: 7−4=3.

### 2. Show that vectors (1,0) and (1,1) form a basis of R².

Sample answer / marking guidance: A linear combination a(1,0)+b(1,1)=(a+b,b) can represent every (x,y) using b=y,a=x−y. The zero combination forces a=b=0.

Explanation: Credit spanning and independence, or a nonzero determinant argument.

### 3. Which set is a linear subspace of R²?

A. The unit circle
B. All (x,y) with x+y=0
C. All points with x>0
D. All (x,y) with x+y=1

Answer: All (x,y) with x+y=0

Explanation: The homogeneous solution set contains zero and is closed under linear combinations.

### 4. What is the rank of [[1,2],[2,4]]?

Answer: 1

Explanation: The second row is twice the first and the matrix is not zero.

### 5. Can three vectors be linearly independent in R²?

A. Yes, if all entries are nonzero
B. Yes, if all lengths differ
C. Only when perpendicular in pairs
D. No

Answer: No

Explanation: No independent set has more vectors than the dimension.

### 6. Explain why the pivot columns of a row-reduced matrix identify basis columns in the original matrix rather than replacing them.

Sample answer / marking guidance: Row operations preserve linear relations among columns, so pivot positions select independent original columns spanning the original column space. The reduced columns may live in a different column space.

Explanation: Credit the distinction between selecting original columns and using transformed columns.

### 7. A linear map from R^4 to R^4 is injective. What is its rank?

Answer: 4

Explanation: Injectivity gives nullity zero, hence rank 4.

### 8. Prove that the kernel of a linear map is a subspace.

Sample answer / marking guidance: T(0)=0. If T(u)=T(v)=0 then T(au+bv)=aT(u)+bT(v)=0 for any scalars a,b.

Explanation: Credit zero membership and closure under arbitrary linear combinations.

## 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.