Final answer:
In (a), it is not necessarily true that AAᵀ=Iₙ. In (b), it is necessarily true that AAᵀ=Iₙ. In (c), AᵀA equals Iₙ when the columns of A are orthonormal.
Step-by-step explanation:
(a) Consider an n×m matrix A such that Aᵀ A= Iₙ. Is it necessarily true that AAᵀ =Iₙ? Explain or give a counter-example.
No, it is not necessarily true that AAᵀ = Iₙ. A counter-example can be shown by considering a 2x3 matrix A such that A = [1 0 0; 0 1 0]. In this case, Aᵀ A = I₂ but AAᵀ ≠ I₂, because the dimensions do not match.
(b) Consider an n×n matrix A such that Aᵀ A=Iₙ. Is it necessarily true that AAᵀ = Iₙ? Explain or give a counter-example.
Yes, it is necessarily true that AAᵀ = Iₙ. This can be proven by considering the singular value decomposition (SVD) of matrix A. From the SVD, it can be shown that AAᵀ = UΣ²Uᵀ = Iₙ, where U is a unitary matrix and Σ is a diagonal matrix containing the singular values of A.
(c) Consider an n×n matrix A which has orthonormal columns. Compute Aᵀ A, what is it equal to? Is it the same as AAᵀ?
Aᵀ A is equal to the identity matrix Iₙ when the columns of A are orthonormal. This is because the transpose of a matrix with orthonormal columns is the inverse of the matrix. However, AAᵀ is not necessarily equal to Iₙ.