Find the adjoint of this non-standard inner product space

Question

I'm really blanking out (a lot of late nights these past 10 weeks).

The point of the exercise I'm about to type up is to show that the adjoint structure may possibly change when the inner product structure changes.

We define the inner product as $\langle (v_1, v_2), (w_1, w_2)\rangle = (v_1, v_2) \begin{pmatrix} 1& \frac{1}{2} \\ \frac{1}{2} & 1 \end{pmatrix} (w_1,w_2)^t$ with $(v_1,v_2)(w_1, w_2) \in \mathbb{R}^2$.

We choose vectors $v = (1,0)$ and $w = (1,0)$ and $A= \begin{pmatrix} 1&1\\0&1 \end{pmatrix}$.

Then, we can show that $⟨Av,w⟩ = ⟨(1,0),(1,0)⟩ = (1,0)(1,1/2)^t = 1$, while $⟨v,A^\dagger w⟩ = ⟨(1,0),(1,1)⟩ =(1,0)(\frac{3}{2},\frac{3}{2})^t=\frac{3}{2}$

This shows that the usual property of adjoints in the standard inner product spaces $⟨Av, w⟩ = ⟨v, A^\dagger w⟩$ does not hold.

Now, my question is how to find the adjoint of this inner product space constructed with this "unusual" inner product.

Is there a general method to finding adjoints?

Thank you!

Answer 1

The general method is the same as usual. Let $\langle \cdot, \cdot \rangle$ be the usual inner product and let $\langle x, y \rangle_A = \langle x, Ay \rangle$, where $A$ is a symmetric positive definite matrix.

Then given a matrix $B$, the adjoint $B^\square$ with respect to $\langle \cdot, \cdot \rangle_A$ is defined by $\langle B^\square x, y \rangle_A = \langle x, By \rangle_A$. Writing this out gives $\langle B^\square x, Ay \rangle = \langle x, ABy \rangle$, and letting $Ay = z$ gives $\langle B^\square x, z \rangle = \langle x, ABA^{-1}z \rangle = \langle A^{-1}B^* A x, z \rangle$.

It follows from is that $B^\square = A^{-1}B^* A$.

Using the notation in the question, we have $A^\square = {1 \over 3} \begin{bmatrix} 1 & -1 \\ 4 & 5\end{bmatrix}$.

Answer 2

Let $M=\pmatrix{1& 1/2\\1/2&1}$ denote the matrix of the inner product. Denote by $A^M$ the adjoint matrix of $A$ with respect to the inner product: $$ \langle Av,w\rangle = \langle v, A^Mw\rangle. $$ Then writing everything with matrices gives the condition $$ w^TMAv = (A^Mw)^TMv = w^T(A^M)^TMv $$ for all $v,w$. This implies $MA=(A^M)^TM$, or $(A^M)^T = MAM^{-1}$, or $A^M = M^{-1}A^TM$.

Answer 3

Let $ B = \begin{pmatrix} 1& \frac{1}{2} \\ \frac{1}{2} & 1 \end{pmatrix} $. If there exists $A$ such that $A^T B = BA$, then $$ \forall (u,v), \, \, \langle Au,v \rangle = u^TA^TBv = u^TBAv = \langle u,Av \rangle$$ So you just need to find $A$ such that $A^T B = BA$.

If we just restrict $A$ to be symmetric as $B$ does, then the relation becomes $AB = BA$. So $A$ suffices to be a commuting matrix of $B$.

Since $B$ is symmetric. It is diagonalizable. Let $P$ be the transition matrix such that $$ B = P \cdot D_B \cdot P^{-1} $$ where $D_B$ is diagonal. If we pose $$ A = P\cdot D \cdot P^{-1}$$ with some arbitrary diagonal matrix $D$ then we can verify that $$AB = P \cdot D\cdot D_B \cdot P^{-1} P \cdot D_B \cdot D \cdot P^{-1} = BA$$ Because the diagonal matrices commute with each others.