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Let $A$ be a square matrix with complex entries , then is it true that $I+A^*A$ is non-singular ? where $A^*$ denotes the conjugate transpose of $A$ http://en.wikipedia.org/wiki/Conjugate_transpose

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  • $\begingroup$ What does $A^*$ represent? $\endgroup$
    – Learnmore
    Oct 30 '14 at 4:35
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    $\begingroup$ @learnmore Adjoint, aka conjugate transpose. $\endgroup$
    – user147263
    Oct 30 '14 at 4:59
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Wealll, here's a quick way to see it without introducing the concepts Hermitian, eigen-stuff, or diagonalization. Look at

$\langle x, (I + A^\ast A)x \rangle = \langle x, x \rangle + \langle x, A^\ast A x \rangle = \langle x, x \rangle + \langle Ax, Ax \rangle \ge \langle x, x \rangle, \tag{1}$

since $\langle Ax, Ax \rangle \ge 0$. This shows that for any $x \ne 0$, we have $(I + A^\ast A)x \ne 0$, whence $I + A^\ast A$ must, by definition, be nonsingular. QED.

Hope this helps. Cheerio,

and as ever,

Fiat Lux!!!

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  • $\begingroup$ Very nice! I did not know. $\endgroup$ Oct 30 '14 at 6:01
  • $\begingroup$ @Bombyxmori: and I thank you, my friend! $\endgroup$ Oct 30 '14 at 6:03
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Yes. $A^{*}A$ is Hermitian, so it is diagonalizable. Further, all its eigenvalues are non-negative, since $x^{*}A^{*}Ax=\|Ax\|^2\geq0$. Thus, the eigenvalues of $I+A^{*}A$ are all greater than or equal to one. Therefore, non-singular.

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  • $\begingroup$ How to show $A^*A$ is hermitian and hence diagonable $\endgroup$
    – Learnmore
    Oct 30 '14 at 5:01
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    $\begingroup$ @learnmore, $(A^{*}A)^{*}=A^{*}A$ is the definition of Hermitian and follows from $(AB)^{*}=B^{*}A^{*}$ and $(A^{*})^{*}=A$. It is diagonalizable by the spectral theorem $\endgroup$ Oct 30 '14 at 5:34
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Since $A^{*}A$ is Hermitian, under an appropriate change of coordinate we have $A^{*}A\rightarrow D$, where $D$ is a diagonal matrix. The same change of coordinates should leave $I$ invariant. So as a linear operator $I+A^{*}A$ is really the same as $I+D$. We know that $D$ is semi-positive-definite because $$ \langle x, A^{*}Ax\rangle=\langle Ax, Ax\rangle\ge 0 $$ Therefore all of the entries of $D$ are greater or equal to zero. Now it should be clear that $I+A^{*}A$ must be non-singular. In fact a formal inverse can be given by

$$ (I+A^{*}A)^{-1}=I+A^{*}A-(A^{*}A)^{2}\cdots $$

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  • $\begingroup$ My dear Bombyx, you might want to take a look at your formal series; the signs should alternate, unless I err. Regards. $\endgroup$ Oct 30 '14 at 8:01
  • $\begingroup$ @RobertLewis: Sorry! $\endgroup$ Oct 30 '14 at 13:44

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