User:J D Bowen/Math725 HW10

1) We aim to show that if $$M \ $$ is an $$m\times n \ $$ matrix with linearly independent columns, then $$M^HM \ $$ is invertible.

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4) Suppose $$U:V\to V \ $$ satisfies $$U^H = U^{-1} \ $$, and that $$\lambda \ $$ is an eigenvalue.  Then $$\lambda ^{-1} \ $$ is the corresponding eigenvalue for $$U^{1} \ $$, and for some vector $$v, \ Uv=\lambda v \ $$.

Observe

$$\lambda \langle v,v, \rangle =\langle \lambda v, v \rangle = \langle Uv, v \rangle = \langle v, U^{-1} v \rangle = \langle v, \lambda^{-1} v \rangle = \overline{\lambda^{-1}}\langle v,v \rangle \ $$

5) Observe that $$\frac{T+T^H}{2}+\frac{T-T^H}{2} = \frac{T+T^H+T-T^H}{2} = \frac{2T}{2} = T \ $$. Now, for any two vectors $$v,w \ $$, we have

$$\langle \frac{T+T^H}{2} v, w \rangle = \frac{1}{2}\langle Tv + T^H v, w \rangle = \frac{1}{2} \left({ \langle Tv,w \rangle +\langle T^H v, w \rangle }\right) = \frac{1}{2} \left({ \langle v,T^H w \rangle + \langle v, (T^H)^H w \rangle }\right) = \frac{1}{2} \langle v, (T^H+T)w \rangle \ $$

$$ = \langle v, \overline{\frac{1}{2}} (T+T^H)w \rangle = \langle v, \frac{T+T^H}{2}w\rangle \ $$,

and so $$\frac{T+T^H}{2} \ $$ is Hermitian.

Further observe

$$\langle \frac{T-T^H}{2} v, w \rangle = \frac{1}{2}\langle Tv - T^H v, w \rangle = \frac{1}{2} \left({ \langle Tv,w \rangle -\langle T^H v, w \rangle }\right) = \frac{1}{2} \left({ \langle v,T^H w \rangle - \langle v, (T^H)^H w \rangle }\right) = \frac{1}{2} \langle v, (T^H-T)w \rangle \ $$

$$ = \langle v, -\overline{\frac{1}{2}} (T^H-T)w \rangle = \langle v, \frac{-(T^H-T)}{2}w\rangle \ $$

and so $$\left({ \frac{T-T^H}{2} }\right)^H = - \frac{T-T^H}{2} \ $$, meaning this is anti-Hermitian.