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Mathematics · lesson 1 of 3

Bases, rank and linear maps

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

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Learn the ideas

Prerequisites: Systems of linear equations and vector arithmetic.

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.

Worked examples

Example 1

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

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

3

Example 2

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

Credit spanning and independence, or a nonzero determinant argument.

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.

Check your understanding

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

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

Show answer and explanation

3

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 and 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.

Credit spanning and independence, or a nonzero determinant argument.

3 Which set is a linear subspace of R²?

Show answer and explanation

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

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

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

Show answer and explanation

1

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

5 Can three vectors be linearly independent in R²?

Show answer and explanation

No

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 and 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.

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?

Show answer and explanation

4

Injectivity gives nullity zero, hence rank 4.

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

Sample answer and 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.

Credit zero membership and closure under arbitrary linear combinations.

Further reading and review references

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