The Unapologetic Mathematician

Mathematics for the interested outsider

I hear joints popping as I stretch and try to get back into the main line of my posts.

We left off defining what we mean by a matrix element of a linear transformation. Let’s see how this relates to adjoints.

We start with a linear transformation $T:V\rightarrow W$ between two inner product spaces. Given any vectors $v\in V$ and $w\in W$ we have the matrix element $\langle w,T(v)\rangle_W$, using the inner product on $W$. We can also write down the adjoint transformation $T^*:W\rightarrow V$, and its matrix element $\langle v,T^*(w)\rangle_V$, using the inner product on $V$.

But the inner product on $W$ is (conjugate) linear. That is, we know that the matrix element $\langle w,T(v)\rangle_W$ can also be written as $\overline{\langle T(v),w\rangle_W}$. And we also have the adjoint relation $\langle v,T^*(w)\rangle_V=\langle T(v),w\rangle_W$. Putting these together, we find

\displaystyle\begin{aligned}\langle v,T^*(w)\rangle_V&=\langle T(v),w\rangle_W\\&=\overline{\langle w,T(v)\rangle_W}\end{aligned}

So the matrix elements of $T$ and $T^*$ are pretty closely related.

What if we pick whole orthonormal bases $\left\{e_i\right\}$ of $V$ and $\left\{f_j\right\}$ of $W$? Now we can write out an entire matrix of $T$ as $t_i^j=\langle f_j,T(e_i)\rangle_W$. Similarly, we can write a matrix of $T^*$ as

\displaystyle\begin{aligned}\left(t^*\right)_j^i&=\langle e_i,T^*(f_j)\rangle_V\\&=\overline{\langle f_j,T(e_i)\rangle_W}\\&=\overline{t_i^j}\end{aligned}

That is, we get the matrix for the adjoint transformation by taking the original matrix, swapping the two indices, and taking the complex conjugate of each entry. This “conjugate transpose” operation on matrices reflects adjunction on transformations.

June 22, 2009 - Posted by | Algebra, Linear Algebra

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