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(=>) Suppose I - xx* is singular if and only there is a y such that (I−xx*)y=0, i.e. xx* y=y. Now set λ=x* y. Then y=λx, i.e. xx* λx=λx Thus λx(x* x) = λx => x* x = 1

(<=) Suppose x*x = 1

Please help

shimura
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  • 'Singular' means that $U$ does not have an inverse. Your equation $U^2=I$ means that $U$ has an inverse, which is itself. – Empy2 Mar 23 '15 at 09:25
  • oops my bad. I was looking at the wrong problem and used that for singular. Thanks ;) – shimura Mar 23 '15 at 09:44

2 Answers2

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On top of the fact that your standing fragment of a proof is mostly incomprehensible, note that for matrices: $AB = 0$ does not imply $A = 0$ or $B = 0$.

Hint: Note that $xx^*$ has eigenvalues $0$ and $\|x\|^2 = x^*x$. A matrix is singular if and only if $0$ is one of its eigenvalues.

Alternative approach: assuming $x \neq 0$, extend $\frac{x}{\|x\|}$ into an orthonormal basis. Find the matrix of $I - xx^*$ relative to this basis.

Ben Grossmann
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  • I get that xx* has eval 0 but I want to use the fact given that xx*=1, and I don't know how to connect both of them together :( – shimura Mar 23 '15 at 09:40
  • If you know the eigenvalues of $xx^$, you can immediately deduce the eigenvalues of $I - xx^$. The fact that $x^x$ is an eigenvalue of $xx^$ is one way of "connecting both of them together". – Ben Grossmann Mar 23 '15 at 09:43
  • alrite. Thanks for the hint ;) – shimura Mar 23 '15 at 09:58
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Just start from the basics. $I-xx^*$ is singular if and only there is a $y$ such that $(I-xx^*)y=0$, i.e. such that $xx^*y=y$. Now set $\lambda=x^*y$. Then $y=\lambda x$, i.e. $xx^*\lambda x=\lambda x$, and you have almost proved one direction.

Make sure that you are aware at each step which expressions are scalars.

Carsten S
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