Question

Let D = diag [d₁, d₂, d₃], where none of d₁, d₂, d₃ is 0; prove that D^-1 = diag [d₁^-1, d₂^-1, d₃^-1].

Answer 1

Given: D = diag [d₁, d₂, d₃]

It is also given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0

A diagonal matrix D = diag(d₁, d₂, …d_n) is invertible iff all diagonal entries are non – zero, 

i.e. di ≠ 0 for 1 ≤ i ≤ n 

If D is invertible then D^-1 = diag(d1^-1, …dn^-1) 

By the Inverting Diagonal Matrices Theorem, which states that 

Here, it is given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0 

∴ D is invertible 

⇒ D^-1 = diag [d₁^-1, d₂^-1, d₃^-1] 

Hence Proved.