# $I+A^*A$ is non-singular whenever $A$ is a square matrix with complex entries? [closed]

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

• What does $A^*$ represent? Oct 30 '14 at 4:35
• @learnmore Adjoint, aka conjugate transpose.
– user147263
Oct 30 '14 at 4:59

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!!!

• Very nice! I did not know. Oct 30 '14 at 6:01
• @Bombyxmori: and I thank you, my friend! Oct 30 '14 at 6:03

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.

• How to show $A^*A$ is hermitian and hence diagonable Oct 30 '14 at 5:01
• @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 Oct 30 '14 at 5:34

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

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

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