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Let $T$ be an invertible linear operator on a finite-dimensional inner product space. I just want a hint as to how I should prove that $T^{\star}$ is also invertible and $( T^{-1} )^{\star} = ( T^{\star} )^{-1}$.

$$ \circ \circ \circ ~ Answer ~ from ~ Below ~ \circ \circ \circ $$

$$ \langle(T^{-1})^*(T^*(v))\mid w\rangle\overset{1}{=} \langle T^*(v)\mid T^{-1}(w)\rangle\overset{2}{=} \langle v\mid T(T^{-1}(w))\rangle\overset{3}{=} \langle v\mid w\rangle $$

Could somebody explain steps $1$ through $3$, please?

Actually, I think @egreg is using this property:


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You need two facts about linear duals: (1) $(g \circ f)^* = f^* \circ g^*$ for any $f, g: V^* \to V^*$ and (2) $(id_V)^* = id_{V^*}$. Together, these two facts express the fact that dualizing is a contravariant functors from the category of finite dimensional vector spaces to itself. – Sammy Black Jun 3 '13 at 22:12
In other words: take the adjoint of the following $TS=ST=I$. – 1015 Jun 3 '13 at 23:13
@julien $\star$ is transpose, right? – Tyquan Pesik Jun 4 '13 at 0:33
@julien Check my answer out. What do you think? – Tyquan Pesik Jun 4 '13 at 0:39
@egreg Same to you. – Tyquan Pesik Jun 4 '13 at 0:39
up vote 2 down vote accepted

The transpose $T^*$ of an endomorphism $T$ is characterized by the property that $$ \langle v\mid T(w)\rangle=\langle T^*(v)\mid w\rangle $$ (where I denote by $\langle x\mid y\rangle$ the inner product of the two vectors $x,y\in V$). The other thing to note is that if $S_1$ and $S_2$ are endomorphisms of $V$, then $S_1=S_2$ if and only if $\langle S_1(v)\mid w\rangle=\langle S_2(v)\mid w\rangle$, for all $v,w\in V$.

By definition, for all $v,w\in V$, $$ \langle(T^{-1})^*(T^*(v))\mid w\rangle= \langle T^*(v)\mid T^{-1}(w)\rangle= \langle v\mid T(T^{-1}(w))\rangle= \langle v\mid w\rangle $$ and therefore $(T^{-1})^*\circ T^*$ is the identity.

Similarly, $$ \langle T^*((T^{-1})^*(v))\mid w\rangle= \langle (T^{-1})^*(v)\mid T(w)\rangle= \langle v\mid T^{-1}(T(w))\rangle= \langle v\mid w\rangle $$ so also $T^*\circ (T^{-1})^*$ is the identity.

Therefore $$ (T^{-1})^*=(T^*)^{-1}. $$ as we wanted to prove.

Note that no hypothesis of finite dimensionality is needed, but just the existence of the transpose endomorphism, which is true for Hilbert spaces.

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I don't quite see how yours works, but I'll consider it "the answer." – Tyquan Pesik Jun 4 '13 at 2:24

Here's a low tech hint: Pick bases! (As is the mantra of much of basic linear algebra).

Once you pick a basis, $T$ becomes matrix. Now the inverse and adjoint have very easy descriptions: namely as the inverse (!) and conjugate transpose. A little thought will now convince you that you can switch these two operations.

If you dont see it, just write down the matrix multiplications. Recall also that transpose switches the order of multiplication i.e. $(AB)^T = B^TA^T$ and conjugation distributes in the obvious way.

There are many a higher-tech solution to do this basis-free, but this is a good one to get going with.

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Let me begin by noting the following theorem:

Theorem: A linear transformation $T$ of a vector space $V$ with finite basis $\alpha_1,\dots,\alpha_n$ is nonsingular if and only if the vectors $\alpha_1T,\dots,\alpha_nT$ are linearly independent in $V$. When this is the case, $T$ has a (two-sided) linear inverse $T^{-1}$, with $TT^{-1} = T^{-1}T=I$. Now, consider the following supplemental proof (terminated by ``ὅπερ ἔδει δεῖξαι''):

Proof. First suppose $T$ is nonsingular. If there is a linear relation $\Sigma x_i(\alpha_iT)=0$ between the $\alpha_iT$, then $$ (x_1\alpha_1 + \cdots + x_n\alpha_n )T = x_1(\alpha_1 )T+\cdots + x_n(\alpha_nT)=0. $$ Since $0T=0$, and $T$ is one-one, this implies $x_1\alpha_1+\cdots + x_n\alpha_n=0$ and hence, by the independence of the $\alpha$'s, $x_1=\cdots=x_n=0$. Therefore the $\alpha_iT$ are linearly independent. Conversely, assume that the vectors $\beta_1=\alpha_1T, \dots, \beta_n=\alpha_nT$ are linearly independent, and recall that a transformation $T$ is one-one onto if and only if it has a two-sided inverse. Since $V$ is $n$-dimensional, the $n$ independent vectors $\beta_1,\dots,\beta_n$ are a basis of $V$. Now, consider the following lemma:

Lemma. If $\beta_1,\dots, \beta_m$ is any basis of the vector space $V$, and $\alpha_1,\dots, \alpha_m$ are any $m$ vectors in $W$, then there is one and only one linear transformation $T:V\rightarrow W$ with $\beta_1T=\alpha_1, \dots, \beta_mT=\alpha_m$, and this transformation is defined by $$ (x\beta_1+\cdots+x_m\beta_m)T = x_1\alpha_1 + \cdots + x_m\alpha_m.$$

By this lemma, there is a linear transformation $S$ of $V$ with $$ \beta_1S=\alpha_1,\beta_2S=\alpha_2, \cdots, \beta_nS=\alpha_n.$$ Thus for each $i=1,\dots, n$, we have that $\beta_i(ST)=\beta_i$. Since the $\beta_1,\dots, \beta_n$ are a basis, there is---in accordance with the lemma---only one linear transformation $T$ with $\beta_iR=\beta_i$ for every $i$, and this transformation is the identiy. Hence $ST=I$. Similarly, $\alpha_i(TS)=\beta_iS=\alpha_i$, and, since the $\alpha$'s are a basis, $TS=I$. Thus $S$ is the inverse of $T$, and $T$ is nonsingular; ὅπερ ἔδει δεῖξαι.

Now, a corollary of this result is that an $n\times n$ matrix $A$ is nonsingular if and only if it has a matrix inverse $A^{-1}$, such that $$ ~AA^{-1}=A^{-1}A=I,$$ so if $A$ has an inverse, so does its transpose, for upon taking the transpose of either side of the above equality, one gets $(A^{-1})^tA^t=A^t(A^{-1})^t=I$, so that we have the result $$ (A^{-1})^t=(A^t)^{-1}. $$ Hence, if $A$ is invertible, so is $A^t$; moreover, the opposite is true similarly; quod erat demonstrandum.

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Yes? No? Good? Bad? – Tyquan Pesik Jun 4 '13 at 2:44
It seems correct; but it's more complicated than necessary: just show that the matrix associated to $T^*$ is the transpose of the matrix associated to $T$ and the properties of matrices can be applied. – egreg Jun 4 '13 at 10:28

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