The Matrix of the Adjoint of a Linear Map

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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 $\mathcal M (T, B_V, B_W)$ . $\blacksquare$

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The Matrix of the Adjoint of a Linear Map