Question

Suppose A is an n×n invertible matrix. Show that if v₁ ,…,vₙ​ is a basis of Rⁿ , then the set of vectors Av₁​,…,Avₙ is also a basis of Rⁿ

Answer 1

If A is an n×n invertible matrix and v₁ ,…,vₙ are the basis vectors of Rⁿ, then the set of vectors Av₁​,…,Avₙ is also a basis of Rⁿ. This can be shown by demonstrating that the set of vectors Av₁​,…,Avₙ spans Rⁿ and is linearly independent.

Step-by-step explanation:

Suppose A is an n×n invertible matrix and v₁ ,…,vₙ are the basis vectors of Rⁿ. To show that the set of vectors Av₁​,…,Avₙ is also a basis of Rⁿ, we need to show two things:

1. The set of vectors Av₁​,…,Avₙ spans Rⁿ.
2. The set of vectors Av₁​,…,Avₙ is linearly independent.

To show that the set of vectors Av₁​,…,Avₙ spans Rⁿ:

Let's take an arbitrary vector in Rⁿ, say u. Since v₁ ,…,vₙ are the basis vectors of Rⁿ, we can write u as a linear combination of v₁ ,…,vₙ:

u = c₁v₁ + c₂v₂ + ... + cₙvₙ, for some constants c₁, c₂, ..., cₙ.

Multiplying both sides of the equation by A, we get:

Au = Ac₁v₁ + Ac₂v₂ + ... + Acₙvₙ.

Since A is invertible, it has an inverse, A⁻¹. We can multiply both sides of the equation by A⁻¹ to get:

A⁻¹(Au) = c₁(A⁻¹v₁) + c₂(A⁻¹v₂) + ... + cₙ(A⁻¹vₙ).

The left side of the equation simplifies to:

u = c₁(A⁻¹v₁) + c₂(A⁻¹v₂) + ... + cₙ(A⁻¹vₙ).

This shows that any arbitrary vector u in Rⁿ can be expressed as a linear combination of Av₁​,…,Avₙ. Therefore, the set of vectors Av₁​,…,Avₙ spans Rⁿ.

To show that the set of vectors Av₁​,…,Avₙ is linearly independent:

Suppose we have a linear combination of Av₁​,…,Avₙ that equals the zero vector, 0:

c₁Av₁ + c₂Av₂ + ... + cₙAvₙ = 0.

Multiplying both sides of the equation by A⁻¹, we get:

A⁻¹(c₁Av₁ + c₂Av₂ + ... + cₙAvₙ) = A⁻¹(0).

This simplifies to:

c₁(A⁻¹Av₁) + c₂(A⁻¹Av₂) + ... + cₙ(A⁻¹Avₙ) = 0.

Since A is invertible, A⁻¹A = I, the identity matrix. Therefore, the equation becomes:

c₁v₁ + c₂v₂ + ... + cₙvₙ = 0.

Since v₁ ,…,vₙ are linearly independent (as they form a basis), the only solution to the equation is c₁ = c₂ = ... = cₙ = 0.

This shows that the only linear combination of Av₁​,…,Avₙ that equals the zero vector is the trivial combination (where all the constants are zero). Therefore, the set of vectors Av₁​,…,Avₙ is linearly independent.

Since the set of vectors Av₁​,…,Avₙ spans Rⁿ and is linearly independent, it is a basis of Rⁿ.