We proved that if the number of non-zero singular values of a square matrix A equals the number of its columns, then A is invertible.
Let A be invertible and all the singular values of A are non-zero.
Then there exists two orthogonal matrices
U ∈ R(n×n)
V ∈ R(m×m)
Σ ∈ R(n×m)
It states that:
Σ(1,1) ≥ Σ(2,2) ≥ ... ≥ 0 such that A=UΣV^{⊤}, and U and V are invertible.
Since the diagonal of Σ is non-zero, Σ is invertible.
So A=UΣV^{⊤} is invertible.
Also if now we assume A is invertible.
Then we can show that:
σ1(A)σ1(A^{−1}) ≥ 1
where σ1(A) and σ1(A^{−1}) represents the largest singular value of A and A^{−1}.
To learn more about invertible matrix link is here
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