My linear algebra textbook gives the definition of the Adjoint Operator and then says,

You should verify the following properties:

  • Additivity: $(S + T)^* = S^* + T^*$
  • Conjugate homogeneity: $(aT)^* = \overline{a}\,T^*$
  • Adjoint of adjoint: $(T^*)^* = T$
  • Identity: $I^* = I$, where $I$ is the identity operator on $V$.

I've stared at the pages for a couple hours now. How do you verify this?

Here's my attempt at a proof for Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T

Is that correct reasoning?

BTW, this is NOT homework. Just reading for pleasure.


  • $\begingroup$ You need to write out each side of the equations using the given definition and then show that equality really does hold. What have you tried so far? $\endgroup$ – Clive Newstead Dec 20 '12 at 22:47
  • $\begingroup$ What should $T^*$ satisfy to be an adjoint? Try to check this for $S^*+T^*$. $\endgroup$ – Sigur Dec 20 '12 at 22:48
  • 1
    $\begingroup$ The adjoint is uniquely determined by a certain condition (if you haven't proven uniqueness you should prove that too). All of these properties assert that the adjoint of some operator can be described as some other operator, so what you need to verify is that that other operator satisfies the condition that uniquely determines the adjoint. $\endgroup$ – Qiaochu Yuan Dec 20 '12 at 22:50
  • $\begingroup$ Adjoint of Adjoint: (T*)* = (Tv, w) = (v, Tw)* = (Tv, w) = T $\endgroup$ – Megan Dec 20 '12 at 22:51
  • $\begingroup$ Is what I posted accurate? $\endgroup$ – Megan Dec 20 '12 at 22:53

The adjoint of a transformation is defined as the unique transformation $T^{*}$ so that $$ \langle Tx, y \rangle = \langle x , T^{*}y \rangle $$ for every $x$ and $y$.

So to prove any of your equalities above you simply need to show that the transformation you want to be the adjoint satisfies this property. Generally you can do this using the properties of an inner product.

For example, to prove $(T^{*})^{*} = T$ you need to show that for any $x,y$ $$ \langle T^{*}x , y \rangle = \langle x , Ty \rangle.$$ This follows because $\langle T^*x,y\rangle= \overline{\langle y , T^*x \rangle} = \overline{\langle Ty, x \rangle} = \langle x , Ty \rangle$.


For the $(T^*)^*=T$ problem:

$$\langle (T^*)^*v, w \rangle=\langle v, T^*w \rangle=\langle Tv, w \rangle$$

by definition of the adjoint.


I am assuming that you are working in some Hilbert space $\mathbb{H}$. The key fact is that any continuous linear functional $f:\mathbb{H} \to \mathbb{C}$ can be represented by a unique element $\phi \in \mathbb{H}$ in the sense that $f(x) = \langle \phi, x \rangle$ for all $x \in \mathbb{H}$ (and conversely any element $\phi \in \mathbb{H}$ determines a unique continuous linear functional on $\mathbb{H}$). A little work shows that $\|f\| = \|\phi\|$. The big deal above is the uniqueness.

Now suppose $T: \mathbb{H} \to \mathbb{H}$ is a continuous linear operator and $y \in \mathbb{H}$. Then $f_y(x) = \langle y, Tx \rangle$ is a continuous linear functional, and can be represented by some $\phi_{y} \in \mathbb{H}$, ie, $f_y(x) = \langle \phi_{y}, x \rangle$. It is easy to show using uniqueness that $\phi_{\lambda y} = \lambda \phi_y$ and $\phi_{y_1+y_2} = \phi_{y_1}+\phi_{y_2}$, hence the mapping $y \mapsto \phi_y$ is linear. Furthermore, since $|f_y(x)| = |\langle \phi_{y}, x \rangle| \leq \|x\| \|y\| \|T\|$, choosing $x=\frac{\phi_y}{\|\phi_y\|}$ (or zero, if $\phi_y=0$) shows that $\| \phi_y \| \leq \|T \| \|y\|$, hence the mapping $y \mapsto \phi_y$ is bounded, and hence continuous. Instead of writing $y \mapsto \phi_y$, we now use the more usual notation $T^* y = \phi_y$. We have shown that $T^*$ is linear, continuous, and completely defined by the requirement that $\langle T^*y, x \rangle = \langle y, Tx \rangle$ for all $x,y \in \mathbb{H}$.

All other properties follow from this requirement and properties of the inner product.

For example, to show additivity: $\langle (S+T)^*y, x \rangle = \langle y, (S+T)x \rangle = \langle y, Sx \rangle + \langle y, Tx \rangle = \langle S^*y, x \rangle + \langle T^*y, x \rangle = \langle (S^*+T^*)y, x \rangle$, hence $(S+T)^* = S^*+T^*$.

For the adjoint of the adjoint: $\langle (T^*)^*y, x \rangle = \langle y, T^*x \rangle = \overline{\langle T^*x,y \rangle} = \overline{\langle x,Ty \rangle} = \langle Ty,x \rangle$, hence $(T^*)^* = T$.

Now try the rest.


The adjoint of a transformation is defined as the unique transformation $T^\ast$ so that $$\langle Tx,y\rangle = \langle x,T^\ast y\rangle$$ for every $x$ and $y$. So

$$\langle T^{\ast\ast} x, y\rangle = \langle x, T^\ast y\rangle = \overline{\langle T^\ast y, x\rangle} = \overline{\langle y, Tx\rangle} = \langle Tx,y\rangle.$$

And we'll get that $\langle T^{\ast\ast}x, y\rangle = \langle Tx, y\rangle$ implies that $T^{\ast\ast} = T$.

  • $\begingroup$ This answer is very poorly argumented. $\endgroup$ – user228113 May 20 '15 at 18:03

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