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Ryan's question at Yahoo! Answers (Eigenvalues of A^*A)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Ryan,

Suppose $\lambda$ is an eigenvalue of $A^*A\in \mathbb{C}^{n\times n}$ then, there exists $x=(x_1,\ldots,x_n)^T\in\mathbb{C}^n$ such that $A^*Ax=\lambda x$. Multypling both sides by $x^*=(\overline{x_1},\ldots,\overline{x_n})$ we get $$x^*A^*Ax=(Ax)^*(Ax)=\lambda x^*x$$ But $y=Ax=(y_1,\ldots,y_m)^T$ and $x\neq 0$ (i.e. $||x||\neq0$), so $$\lambda=\frac{y^*y}{x^*x}=\frac{||y||^2}{||x||^2}\ge 0$$ On the other hand, $$\det (A^*A+I)=0\Leftrightarrow \det (A^*A-(-1)I)=0\Leftrightarrow -1\mbox{ is eigenvalue of }A^*A$$ This is a contradiction, so $\det(A^*A+I)\neq0$, as a consequence $A^*A+I$ is invertible.

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