An elegant proof for linear map and its adjoint has conjugate eigenvalue

This article will introduce an elegant way to prove that $T$$T$ and $T^{\ast }$$T^*$ has a conjugate eigenvalue.

$T^{\ast }$$T^*$ is adjoint of $T$$T$, defined by $⟨v,T^{\ast }w⟩=⟨Tv,w⟩$$\langle v,T^*w\rangle=\langle T v,w\rangle$

The matrix representation of $T^{\ast }$$T^*$ is the conjugate transpose.

And it has a connection to dual map $T^{\prime }$$T'$.

We know that the Riesz representation theorem gives the isomorphism $V\cong V^{\prime }$$V\cong V'$, $V^{\prime }$$V'$ is dual space

by ${\mathrm{\Phi }}_V(v)\mapsto ⟨-,v⟩$$\Phi_V(v)\mapsto\langle-,v\rangle$

Consider a linear map $T:V\to W$$T:V\to W$, its dual map $T^{\prime }:W^{\prime }\to V^{\prime }$$T':W'\to V'$ defined by $T^{\prime }(\phi )=\phi \circ T$$T'(\varphi)=\varphi\circ T$

The matrix representation is transpose $A^T$$A^T$.

And $T^{\ast }:W\to V={\mathrm{\Phi }}_V^{-1}{T}^{\prime }{\mathrm{\Phi }}_W$$T^*: W\to V=\Phi_V^{-1} T'\Phi_W$

So ${\mathrm{\Phi }}_V{T}^{\ast }=T^{\prime }{\mathrm{\Phi }}_W$$\Phi_V T^*= T'\Phi_W$ and put a $w\in W$$w\in W$ in

we get $⟨-,T^{\ast }(w)⟩=⟨T-,w⟩$$\langle-,T^*(w)\rangle=\langle T-,w\rangle$, and put $v\in V$$v\in V$ in, we get $⟨v,T^{\ast }(w)⟩=⟨Tv,w⟩$$\lang v,T^*(w)\rangle=\lang Tv,w\rangle$

And you can prove that $(T+S{)}^{\ast }=T^{\ast }+S^{\ast },(\lambda T{)}^{\ast }=\bar{\lambda }{T}^{\ast },(TS{)}^{\ast }=S^{\ast }{T}^{\ast }$$(T+S)^*=T^*+S^*, (\lambda T)^*=\overline\lambda T^*,(TS)^*=S^* T^*$

And $(T^{\ast }{)}^{\ast }=T$$(T^*)^*=T$ For $T\in \mathcal{L}(V)$$T\in\mathcal L(V)$, or $\mathrm{E}\mathrm{n}\mathrm{d}(\mathrm{V})$$\rm End(V)$

We know that for Ab, the endomorphism can be a ring

And we can view $\ast$$*$ as a isomorphism from $\mathrm{E}\mathrm{n}\mathrm{d}(\mathrm{V})\to \mathrm{E}\mathrm{n}{\mathrm{d}}^{\mathrm{o}\mathrm{p}}(\mathrm{V})$$\rm End(V)\to End^{op}(V)$

In $\mathrm{E}\mathrm{n}{\mathrm{d}}^{\mathrm{o}\mathrm{p}}(\mathrm{V})$$\rm End^{op}(V)$, the $S{\circ }_{op}T=T\circ S$$S\circ_{op} T= T\circ S$

So $(T+S{)}^{\ast }=T^{\ast }+S^{\ast },(TS{)}^{\ast }=T^{\ast }{\circ }_{op}{S}^{\ast }$$(T+S)^*=T^*+S^*,(TS)^*=T^*\circ_{op} S^*$

show us that adjoint is a ring homomorphism, and the inverse is itself because $(T^{\ast }{)}^{\ast }=T$$(T^*)^*=T$

So, adjoint is a ring isomorphism, So $T-\lambda I$$T-\lambda I$ is invertible iff $(T-\lambda I{)}^{\ast }=T^{\ast }-\bar{\lambda }I$$(T-\lambda I)^*=T^*-\overline\lambda I$ is invertible,

Thus $T$$T$ and $T^{\ast }$$T^*$ has a conjugate eigenvalue.