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From Friedberg's Linear Algebra

Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.

Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $\lambda$. Then for any $x \in V$,

$0 = \langle 0,x \rangle = \langle(T - \lambda\ I)(v),x \rangle = \langle v,(T-\lambda\ I)^*(x)\rangle = \langle v, (T^*-\bar{\lambda} I)(x)\rangle$,

and hence $v$ is orthogonal to the range of $T^* -\bar{\lambda}I$. So $T^* -\bar{\lambda}I$ is not onto and hence is not one-to-one. Thus $T^* -\bar{\lambda}I$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T*$ with correspoinding eigenvalue $\bar{\lambda}.$

I'm unable to see why $T^* -\bar{\lambda}I$ is not one-to-one, and why $T^* -\bar{\lambda}I$ has a nonzero null space.

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    $\begingroup$ If the range of $T* -\bar{\lambda}I$ was all of $V$, then how can $v$ be orthogonal to all of it? Unless $v$ is zero, but it cannot be zero because it is an eigenvector of something, and zero vectors are not eigenvectors of anything. $\endgroup$ – Marcus Aurelius Jul 22 '18 at 16:49
  • $\begingroup$ I thought that would mean that it's not onto, but I still can't see why it's not one-to-one. $\endgroup$ – K.M Jul 22 '18 at 17:14
  • $\begingroup$ This is just the dimension formula. Dimension of kernel + dimension of image = dimension of V (or more generally, the domain). $\endgroup$ – Marcus Aurelius Jul 22 '18 at 17:15
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Not onto: Assume it's onto, then exists $x'\in V$ s.t. $0=\langle v,(T^*-\bar\lambda I)(x')\rangle=\langle v,v\rangle,$ so $v=0,$ which contradicts $v\not=0.$

Not one-one: Since $T^*:V\to V, I_V:V\to V,$ clearly $(T^*-\bar\lambda I):V\to V,$ so $$\dim(V)=Nullity(T^*-\bar\lambda I)+Rank(T^*-\bar\lambda I),$$ but $Rank(T^*-\bar\lambda I)\lt\dim(V)$ since we just observed, so $Nullity(T^*-\bar\lambda I)\gt0$.

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Lauds to our colleague Vizag for his elegant demonstration that

$\lambda \; \text{an eigenvalue of} \; T \Longrightarrow \bar \lambda \; \text{an eigenvalue of} \; T^\dagger; \tag 1$

however, his work on this subject leaves unaddressed the title question, that is,

$\text{"Why is} \; T^\dagger - \bar \lambda I \; \text{not one-to-one?"} \tag 2$

I wish to take up this specific topic here, and provide a sort of "classic" answer; specifically, I wish to demonstrate the essential and well-known result,

"A linear map $S:V \to V$ from a finite dimensional vector space to itself is one-to-one if and only if it is onto."

Note: in what follows we allow $V$ to be a vector space over any base field $\Bbb F$.

Proof: The argument is based upon elementary notions of basis and linear independence.

We first assume $S:V \to V$ is onto. Then we let $w_1, w_2, \dots w_n$ be a basis for $V$ over the field $\Bbb F$, and we see, since $S$ is surjective, that there must be a set of vectors $v_i$, $1 \le i \le n$, with

$Sv_i = w_i, \; 1 \le i \le n; \tag 3$

I claim the set $\{ v_i \mid 1 \le i \le n \}$ is linearly independent over $\Bbb F$; for if not, there would exist $\alpha_i \in \Bbb F$, not all zero, with

$\displaystyle \sum_1^n \alpha_i v_i = 0; \tag 4$

then

$\displaystyle \sum_1^n \alpha_i w_i = \sum_1^n \alpha_i Sv_i = S \left (\sum_1^n \alpha_i v_i \right ) = S(0) = 0; \tag 5$

but this contradicts the linear independence of the $w_i$ unless

$\alpha_i = 0, \; 1 \le i \le n; \tag 6$

but condition (6) is precluded by our assumption that not all the $\alpha_i = 0$; therefore the $v_i$ are linearly independent over $\Bbb F$ and hence form a basis for $V$; then any $x \in V$ may be written

$x = \displaystyle \sum_1^n x_i v_i, \; x_i \in \Bbb F; \tag 7$

now suppose $S$ were not injective. Then we could find $x_1, x_2 \in V$ with

$Sx_1 = Sx_2; \tag 8$

if, in accord with (7) we set

$x_1 = \displaystyle \sum_1^n \alpha_i v_i, \tag 9$

$x_2 = \displaystyle \sum_1^n \beta_i v_i, \tag{10}$

then from (8)-(10),

$\displaystyle \sum_1^n \alpha_i w_i = \sum_1^n \alpha_i Sv_i = S \left (\sum_1^n \alpha_i v_i \right ) = S \left (\sum_1^n \beta_i v_i \right) = \sum_1^n \beta_i Sv_i = \sum_1^n \beta_i w_i, \tag{11}$

whence

$\displaystyle \sum_1^n (\alpha_i - \beta_i) w_i = 0; \tag{12}$

now the linear independence of the $w_i$ forces

$\alpha_i = \beta_i, \; 1 \le i \le n, \tag{13}$

whence again via (9)-(10)

$x_1 = x_2, \tag{14}$

and we see that $S$ is injective.

Going the other way, we now suppose $S$ is injective; and let the set $\{v_i \mid 1 \le i \le n \}$ form a basis for $V$. I claim that the vectors $Sv_1, Sv_2, \ldots, Sv_n$ also form a basis; for if not, they must be linearly dependent and we may find $\alpha_i \in \Bbb F$ such that

$\displaystyle S \left ( \sum_1^n \alpha_i v_i \right ) = \sum_1^n \alpha_i Sv_i = 0; \tag{15}$

now with $S$ injective this forces

$\displaystyle \sum_1^n \alpha_i v_i = 0, \tag{16}$

impossible by the assumed linear independence of the $v_i$; thus the $Sv_i$ do form a basis and hence any $y \in V$ may be written

$y = \displaystyle \sum_1^n \beta_i Sv_i = S \left ( \sum_1^n \beta_i v_i \right ); \tag{17}$

thus every $y \in V$ lies in the image of $S$ which at last seen to be onto. End: Proof.

If we apply this result to $T^\dagger - \bar \lambda I$ as in the body of the question, we see that, having shown that $T^\dagger - \bar \lambda I$ is not onto, we may conclude it is also not injective by the preceding basic demonstration; but not injective implies the null space is not $\{ 0 \}$, since if $x_1 \ne x_2$ but $Sx_1 = Sx_2$, we have $x_1 - x_2 \ne 0$ but

$S(x_1 - x_2) = Sx_1 - Sx_2 = 0, \tag{18}$

whence $0 \ne x_1 - x_2 \in \ker S \ne \{ 0 \}$.

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