How do I show that if $A$ is a $m \times n$ matrix, then $E_{n} + A^{T} A$ is a nonsingular matrix? ($E_{n}$ is a identity matrix.)
I tried thinking calculations of its elements, but I could not prove this.
Asked
Active
Viewed 85 times
0
-
1The mattrix is definite symmetric therefore non-singular – Mohammed M. Zerrak Mar 27 '21 at 15:27
-
Hint: https://math.stackexchange.com/questions/1690078/positive-definite-matrix-plus-positive-semi-matrix-equals-positive-definite – user326159 Mar 27 '21 at 15:30
-
Without invoking the PSD concept, you might approach it by induction on $n$, by writing $A = (a_1, \ldots, a_n)$. – Zhanxiong Mar 27 '21 at 16:24
1 Answers
3
Let us consider arbitrary $x\neq 0$. Then $x^{\text{T}}(I_n+A^TA)x=\underbrace{||x||^2}_{>0}+\underbrace{||Ax||^2}_{\ge 0}>0$ ergo the matrix is positive definite ergo regular.
Vítězslav Štembera
- 2,307
- 1
- 10
- 19