2
$\begingroup$

Consider $\Bbb C^n$. Let $A^*$ be the conjugate transpose of matrix $A$, I see in many materials $A^*$ is also called the adjoint of $A$. I am now confused by that does $〈Ax,y⟩=〈x,A^*y⟩$ hold just for the standard inner product (whose proof is simple), or any inner product? Thanks!

$\endgroup$
3
$\begingroup$

It does not hold for every inner product. The analogous situation in the real plane illustrates this. Consider $$ H = \pmatrix{2 & 0 \\ 0 & 1} $$ and the inner product $x \cdot y = x^t H y$. Let $$ A = \pmatrix{0 & 1 \\ 1 & 0} $$ $x = \pmatrix{1\\0}$ and $y = \pmatrix{0\\1}$.

Then $Ax \cdot y = 1$, but $x \cdot A^t y = 2$.

Since the real plane is a subspace of the complex plane, this suffices as a counterexample in $\Bbb C^2$ as well (at least if you extend the definition of the dot product by writing $x \cdot y = x^t H \bar{y}$).

$\endgroup$
  • $\begingroup$ With the right commutativity condition on $H$ and $A$, then $\langle Ax,y\rangle=\langle x,A^*y\rangle$. $\endgroup$ – alex.jordan Oct 5 '17 at 4:00
  • $\begingroup$ your definition doesn't hold in $\mathbb C^2$ for conjugate symmetry, e.g., consider $x=[i, 1]^T, y=[1, i]^T$, then $ <x, y>=3i= <y, x>$, but the later one should be $-3i$. $\endgroup$ – vita nova Oct 5 '17 at 6:59
  • $\begingroup$ alex.jordan: yes. I was hoping that OP might discover that by working through the definitions. @novavita: it's true. You have to properly generalize my definition of "dot" by writing $x \cdot y = x^t H \bar{y}$. I should probably have mentioned this in my last sentence. $\endgroup$ – John Hughes Oct 5 '17 at 11:07
1
$\begingroup$

Usually, the answer is no, and depends on the structure of the inner product. More specifically, it requires that the orthogonal basis under the inner product is a unitary matrix.

Let's clarify some notations first. Denote by $T$ the operator transforming $x$ to $Ax$. Then $T$ is a linear operator from $\mathbb C^n$ to itself, i.e., $T\in \mathcal L (\mathbb C^n)$, we prefer to use $T^*$ as the "adjoint" operator defined by the equation: $$ <Tx, y>=<x, T^* y> $$ for any $x, y \in \mathbb C^n$.

Consider an orthogonal basis for $\mathbb C^n$, $\tilde e=(\tilde e_1, \tilde e_2,...,\tilde e_n)$ under inner product $<.,.>$. Then the matrix of $T$ under this basis is $[m_{ij}]_{n\times n}$, where $m_{ij}=<T\tilde e_j,\tilde e_i>$, which we denoted as $M(T,\tilde e)$.

On the other hand, since $$ T(x)=Ax $$ Then the matrix of $T$ under the trivial basis $e=(e_1, e_2,...,e_n)$ where $e_1=[1,0,...,0]^T$, $e_2=[0,1,...,0]^T$,..., i.e., $M(T, e)=A$. We use $A^H$ to denote the conjugate transpose of $A$. Then we can show the matrix of the linear operator $T^*$ under basis $\tilde e$ (well, we need to prove it's linear first, which is not very hard) is the conjugate transpose of $M(T,\tilde e)$, i.e., $$ M(T^*,\tilde e)=M(T,\tilde e)^H\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ Now change the basis from $\tilde e$ to $e$, we get $$ \tilde e^{-1} M(T^*,e) \tilde e= (\tilde e^{-1} A \tilde e)^H=(\tilde e)^H A^H (\tilde e^{-1})^H $$ If we want $M(T^*,e)=A^H$ holds for all $A$, then $$ \tilde e^{-1} A^H \tilde e=(\tilde e)^H A^H (\tilde e^{-1})^H $$ which requires $\tilde e$ being a unitary matrix: $$ (\tilde e)^H \tilde e = I_n $$


To see why (1) holds, consider the $j^{th}$ column of $M \equiv M(T,\tilde e)$, call it $M_j$. It's obvious that $$ M_j=M\tilde e_j $$ on the other hand, $$ M\tilde e_j=T\tilde e_j=<T\tilde e_j,\tilde e_1>\tilde e_1+<T\tilde e_j,\tilde e_2>\tilde e_2+...+<Te_j,\tilde e_n>\tilde e_n $$ so $m_{ij}=<T\tilde e_j,\tilde e_i>$

Similarly, let $M^* \equiv M(T^*,\tilde e)$, then $$ m^*_{ji}=<T^* \tilde e_i,\tilde e_j>=< \tilde e_i,(T^*)^*\tilde e_j>=< \tilde e_i,T\tilde e_j>= \overline {< T\tilde e_i,\tilde e_j>} $$ this holds for each $(i,j)$, which means $M^*=M^H$

$\endgroup$
-2
$\begingroup$

$〈Ax,y⟩=〈x,A^*y⟩ $ this holds for any inner product ! when you assume $A^*$ is adjoint of $A$ in space X equipped with inner product of $〈.,.⟩$

This is how adjoint operators defined in finite dimension.

$\endgroup$
  • $\begingroup$ Hey Down voter(s): never vote down bcz of your lack of knowledge ! $\endgroup$ – Red shoes Oct 5 '17 at 4:08
  • $\begingroup$ I think the downvotes are because the question states specifically that $A^*$ is the conjugate transpose of the matrix $A$. $\endgroup$ – littleO Oct 5 '17 at 6:30
  • $\begingroup$ @littleO I was just trying to convey a message to OP that what he has seen in many material is correct ... but his/her interpretation of adjoint as simple transpose is not always true ! adjoint = transpose if one works with Euclidean inner product ... And I bet those who down vote thinks adjoint is same as transpose ... $\endgroup$ – Red shoes Oct 5 '17 at 6:39
  • $\begingroup$ In general, if you want to convey a message, it's best to convey that message, rather than to redefine the notation explicitly defined in the question. You had a good point to make, and a slight rewriting would probably get folks to remove their downvotes. As for "I bet those downvoters are idiots,"... well...that's your choice, but a lot of experience with the folks who tend to read multiple answers and think about their quality here suggests otherwise, at least to me. $\endgroup$ – John Hughes Oct 5 '17 at 11:11
  • $\begingroup$ @JohnHughes I believe I didn't redefine anything. Notice that I used the word "WHEN you assume ..." since the notation $A^*$ generally refers to adjoint which in the particular case ,the dot product, the notation $\bar{A^T}$ is used for conjugate transpose. And I thought that's the where confusion come from. I posted my answer after you, so I already read yours and voted it up, but I felt still confusion exists or maybe your answer make it worse! then I posted this answer. BTW Someone who plays with words it is you not me! I never used the word "idiots". $\endgroup$ – Red shoes Oct 5 '17 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.