The Matrix of the Adjoint of a Linear Map

# The Matrix of the Adjoint of a Linear Map

Recall from the The Conjugate Transpose of a Matrix page that if $A$ is an $m \times n$ matrix then the conjugate transpose of $A$ is the matrix obtained by taking the complex conjugate of each entry in $A$ and then transposing $A$.

Now let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $T \in \mathcal L (V, W)$. Furthermore, let $B_V = \{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and let $B_W = \{ w_1, w_2, ..., w_m \}$ be a basis of $W$. From The Matrix of a Linear Map page we saw that the matrix of $T$ with respect to the bases $B_V$ of $V$ and the basis $B_W$ of $W$ is, once again, denoted $\mathcal M (T, B_V, B_W)$ and:

(1)
\begin{align} \quad \mathcal M (T, B_V, B_W) = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \end{align}

The $a$'s are obtained by $T(v_k) = a_{1,k}w_1 + a_{2,k}w_2 + ... + a_{m,k}w_m$ for each $k = 1, 2, ..., n$, that is by applying the transformation $T$ to each basis vector in $B_V$ to get a linear combination of the basis vectors in $B_W$. The coefficients in this linear combination form $\mathcal M (T, B_V, B_W)$.

In the following theorem we will see that if we have orthonormal bases for $V$ and $W$ then finding the matrix of the adjoint of $T$, $T^*$ is very simple.

 Theorem 1: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces and let $T \in \mathcal L (V, W)$. Let $B_V = \{ e_1, e_2, ..., e_n \}$ be an orthonormal basis of $V$ and let $B_W = \{ f_1, f_2, ..., f_m \}$ be an orthonormal basis of $W$. Then $\mathcal M (T^*, B_W, B_V)$ can be obtained by taking the conjugate transpose of $\mathcal M (T, B_V, B_W)$.
• Consider the $k^{\mathrm{th}}$ column of $\mathcal M (T, B_V, B_W)$. Then we will have that for each basis vector $e_k$, $k = 1, 2, ... n$ that:
(2)
\begin{align} \quad T(e_k) = <T(e_k), f_1>f_1 + <T(e_k), f_2>f_2 + ... + <T(e_k), f_m>f_m \end{align}
• Therefore the entry in row $k$ column $j$, $j = 1, 2, ..., m$ of $\mathcal M (T, B_V, B_W)$ is $<T(e_k), f_j>$
• Now consider the $j^{\mathrm{th}}$ column of $\mathcal M(T^*, B_W, B_V)$. Then we will have that for each basis vector $f_j$, $j = 1, 2, ..., m$ that:
(3)
\begin{align} \quad T^*(f_j) = <T^*(f_j), e_1>e_1 + <T^*(f_j), e_2>e_2 + ... + <T(f_j), e_n>e_n \\ \end{align}
• Therefore the entry in row $j$ column $k$ of $\mathcal M (T^*, B_W, B_V)$ is $<T^*(f_j), e_k>$. If we take the complex conjugate of this inner product and we get $\overline{<e_k, T^*(f_j)>}$.
• However, $<T(e_k), f_j> = <e_k, T^*(f_j)>$, so indeed $\mathcal M (T^*, B_W, B_V)$ is the conjugate transpose of $\mathcal M (T, B_V, B_W)$. $\blacksquare$