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How to prove a general matrix invertible given by as below?

How to prove that $A^TA+I$ is always invertible for $\forall A \in \mathbb{R}^{n\times n}$?

Dominic Michaelis
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wanyancan
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1 Answers1

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Following Daniel Fischer's comment

$$\forall x\ne0,\quad\langle (A^TA+I)x,x\rangle=\langle x,x\rangle+\langle A^TAx,x\rangle=||x||^2+||Ax||^2>0$$ so the given matrix is positive definite so it's invertible.

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    Maybe you should write why a positive definite matrix is invertible – Dominic Michaelis Aug 31 '13 at 12:37
  • So how to find the inverse? – Supriyo Aug 31 '13 at 13:06
  • @Samprity : Sami's answer shows that $A^T A + I$ has a trivial nullspace (consisting of just the zero vector). A square matrix is invertible if and only if its nullspace is trivial. You don't need to know anything about positive definite matrices to understand the answer. The question didn't ask how to find the inverse. – Stefan Smith Aug 31 '13 at 13:39