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Complex Eigenvalues and Complex Scalars · Tutorial 549 of 1000

Conjugation And Conjugate Transpose

Compute conjugates and conjugate transposes of complex matrices, and justify the identities that govern sums, products, scalar multiples, and transposes.

Intermediate 9 min read

What You'll Learn

  • Find the entrywise conjugate and conjugate transpose of a complex matrix.
  • Distinguish ordinary transpose from conjugate transpose.
  • Apply conjugation to sums and scalar multiples.
  • Verify why conjugate transpose reverses the order of a matrix product.
  • Check dimensions and confirm identities with exact calculations.
  • Recognize and correct the error of keeping product order unchanged.

Two Operations on a Complex Matrix

For a real matrix, transposing is the familiar operation that switches rows and columns. With complex entries, there is another operation to keep track of: complex conjugation changes the sign of each imaginary part. Combining these operations gives the conjugate transpose. Careful notation matters because transposing alone does not conjugate entries, and conjugating a product does not reverse its factors in the same way as taking a conjugate transpose.

For a complex number \(z=a+bi\), where \(a,b\in\mathbb{R}\) and \(i^2=-1\), its conjugate is \(\overline{z}=a-bi\). The scalar identities used below follow by multiplying these expressions: \(\overline{z+w}=\overline z+\overline w\), \(\overline{zw}=\overline z\,\overline w\), and \(\overline{cz}=\overline c\,\overline z\). For example, if \(z=a+bi\) and \(w=d+ei\), then \(zw=(ad-be)+(ae+bd)i\), whose conjugate is \((ad-be)-(ae+bd)i=(a-bi)(d-ei)\).

Definition: If \(\mathbf{A}=[a_{ij}]\in\mathbb{C}^{m\times n}\), its entrywise conjugate \(\overline{\mathbf{A}}\) is the \(m\times n\) matrix whose \((i,j)\) entry is \(\overline{a_{ij}}\). Its transpose \(\mathbf{A}^{\top}\) is the \(n\times m\) matrix with \((i,j)\) entry \(a_{ji}\). Its conjugate transpose is \(\mathbf{A}^{*}=(\overline{\mathbf{A}})^{\top}\), an \(n\times m\) matrix.

Thus, taking a conjugate changes entries but not the matrix shape; transposing changes the shape but not the entries; and taking a conjugate transpose does both. For a real matrix, every entry is unchanged by conjugation, so \(\mathbf{A}^{*}=\mathbf{A}^{\top}\). For a complex matrix, that equality need not hold.

Worked Example: Conjugating and Transposing a Matrix

Let \(\mathbf{A}=\begin{bmatrix}1+2i&-i&3-i\\2i&4&-1+i\end{bmatrix}\). Conjugate each entry to obtain

$$ \overline{\mathbf{A}}= \begin{bmatrix} 1-2i&i&3+i\\ -2i&4&-1-i \end{bmatrix}. $$

This is still a \(2\times 3\) matrix. To take the conjugate transpose, first use those conjugated entries and then switch rows and columns:

$$ \mathbf{A}^{*}= \begin{bmatrix} 1-2i&-2i\\ i&4\\ 3+i&-1-i \end{bmatrix}. $$

The result has size \(3\times 2\), as required. Applying the same operation again conjugates every displayed entry and transposes again:

$$ (\mathbf{A}^{*})^{*}= \begin{bmatrix} 1+2i&-i&3-i\\ 2i&4&-1+i \end{bmatrix} =\mathbf{A}. $$

This calculation illustrates why both parts of the definition matter: transposing alone would leave the entry \(1+2i\) unchanged, whereas the conjugate transpose places \(1-2i\) in the corresponding position.

Identities for Sums, Scalars, and Transposes

The next identities apply to complex matrices of the indicated sizes. In each sum, the matrices must have the same size; in each product, the number of columns of the first factor must equal the number of rows of the second. These conditions ensure that both sides of an identity are defined and have the same size.

Theorem: Let \(\mathbf{A},\mathbf{B}\in\mathbb{C}^{m\times n}\), let \(c\in\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{C}^{m\times n}\). Then \(\overline{\mathbf{A}+\mathbf{B}}=\overline{\mathbf{A}}+\overline{\mathbf{B}}\), \(\overline{c\mathbf{A}}=\overline c\,\overline{\mathbf{A}}\), \(\mathbf{A}^{*}=(\overline{\mathbf{A}})^{\top}=\overline{\mathbf{A}^{\top}}\), \((\mathbf{A}^{\top})^{*}=\overline{\mathbf{A}}\), and \((\mathbf{A}^{*})^{*}=\mathbf{A}\). In particular, \((\mathbf{A}+\mathbf{B})^{*}=\mathbf{A}^{*}+\mathbf{B}^{*}\) and \((c\mathbf{A})^{*}=\overline c\,\mathbf{A}^{*}\).

Proof. Write \(\mathbf{A}=[a_{ij}]\) and \(\mathbf{B}=[b_{ij}]\). The \((i,j)\) entry of \(\overline{\mathbf{A}+\mathbf{B}}\) is \(\overline{a_{ij}+b_{ij}}=\overline{a_{ij}}+\overline{b_{ij}}\), which is the \((i,j)\) entry of \(\overline{\mathbf{A}}+\overline{\mathbf{B}}\). The \((i,j)\) entry of \(\overline{c\mathbf{A}}\) is \(\overline{ca_{ij}}=\overline c\,\overline{a_{ij}}\), giving the scalar identity. By definition, the \((i,j)\) entry of \(\mathbf{A}^{*}\) is \(\overline{a_{ji}}\), which is also the \((i,j)\) entry of both \((\overline{\mathbf{A}})^{\top}\) and \(\overline{\mathbf{A}^{\top}}\). The \((i,j)\) entry of \((\mathbf{A}^{\top})^{*}\) is \(\overline{(\mathbf{A}^{\top})_{ji}}=\overline{a_{ij}}\), giving \((\mathbf{A}^{\top})^{*}=\overline{\mathbf{A}}\). Finally, the \((i,j)\) entry of \((\mathbf{A}^{*})^{*}\) is \(\overline{(\mathbf{A}^{*})_{ji}}=\overline{\overline{a_{ij}}}=a_{ij}\), so \((\mathbf{A}^{*})^{*}=\mathbf{A}\). Applying the definition to a sum or scalar multiple, together with the entrywise identities already proved, gives the last two identities. \(\square\)

The same entry-by-entry method proves the product rules. Their proof is worth examining because it explains the difference between conjugating a product and taking its conjugate transpose.

Theorem: Let \(\mathbf{A}\in\mathbb{C}^{m\times n}\) and \(\mathbf{B}\in\mathbb{C}^{n\times p}\). Then \(\overline{\mathbf{A}\mathbf{B}}=\overline{\mathbf{A}}\,\overline{\mathbf{B}}\) and \((\mathbf{A}\mathbf{B})^{*}=\mathbf{B}^{*}\mathbf{A}^{*}\).

Proof. By the definition of matrix multiplication, the \((i,j)\) entry of \(\mathbf{A}\mathbf{B}\) is \(\sum_{k=1}^{n}a_{ik}b_{kj}\). Its conjugate is \(\sum_{k=1}^{n}\overline{a_{ik}}\,\overline{b_{kj}}\), which is the \((i,j)\) entry of \(\overline{\mathbf{A}}\,\overline{\mathbf{B}}\). This proves the first identity. For the second, the \((i,j)\) entry of \((\mathbf{A}\mathbf{B})^{*}\) is the conjugate of the \((j,i)\) entry of \(\mathbf{A}\mathbf{B}\), namely \[ \overline{\sum_{k=1}^{n}a_{jk}b_{ki}} =\sum_{k=1}^{n}\overline{b_{ki}}\,\overline{a_{jk}}. \] The \((i,j)\) entry of \(\mathbf{B}^{*}\mathbf{A}^{*}\) is \(\sum_{k=1}^{n}(\mathbf{B}^{*})_{ik}(\mathbf{A}^{*})_{kj} =\sum_{k=1}^{n}\overline{b_{ki}}\,\overline{a_{jk}}\). The entries agree for every \(i,j\). The product \(\mathbf{B}^{*}\mathbf{A}^{*}\) has size \(p\times m\), the same size as \((\mathbf{A}\mathbf{B})^{*}\), so the identity is defined with matching dimensions. \(\square\)

Checking the Product Rules by Calculation

Worked Example: Conjugating a Product and Reversing Its Factors

Take \(\mathbf{A}=\begin{bmatrix}1+i&2\\i&1-i\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}i&1\\1+i&-i\end{bmatrix}\). First multiply in the given order, showing each entry:

$$ \mathbf{A}\mathbf{B} = \begin{bmatrix} (1+i)i+2(1+i)&(1+i)\cdot 1+2(-i)\\ i\cdot i+(1-i)(1+i)&i+(1-i)(-i) \end{bmatrix} = \begin{bmatrix} 1+3i&1-i\\ 1&-1 \end{bmatrix}. $$

Conjugating the product gives \(\overline{\mathbf{A}\mathbf{B}}=\begin{bmatrix}1-3i&1+i\\1&-1\end{bmatrix}\). Now conjugate each factor and multiply them in the same order:

$$ \overline{\mathbf{A}}\,\overline{\mathbf{B}} = \begin{bmatrix}1-i&2\\-i&1+i\end{bmatrix} \begin{bmatrix}-i&1\\1-i&i\end{bmatrix} = \begin{bmatrix} (-i-1)+(2-2i)&1-i+2i\\ -1+2&-i+(1+i)i \end{bmatrix} = \begin{bmatrix}1-3i&1+i\\1&-1\end{bmatrix}. $$

The two results match. For the conjugate-transpose identity, the individual conjugate transposes are \(\mathbf{A}^{*}=\begin{bmatrix}1-i&-i\\2&1+i\end{bmatrix}\) and \(\mathbf{B}^{*}=\begin{bmatrix}-i&1-i\\1&i\end{bmatrix}\). Multiplying in reverse order gives

$$ \mathbf{B}^{*}\mathbf{A}^{*} = \begin{bmatrix} (-i)(1-i)+2(1-i)&(-i)(-i)+(1-i)(1+i)\\ 1-i+2i&-i+(1+i)i \end{bmatrix} = \begin{bmatrix}1-3i&1\\1+i&-1\end{bmatrix}. $$

On the other hand, transposing the computed product and conjugating its entries gives \((\mathbf{A}\mathbf{B})^{*}=\begin{bmatrix}1-3i&1\\1+i&-1\end{bmatrix}\), in agreement with \(\mathbf{B}^{*}\mathbf{A}^{*}\).

Check: the product \(\mathbf{A}\mathbf{B}\) is \(2\times2\), so its conjugate transpose is also \(2\times2\). Both calculations of that matrix give the same four entries. The conjugated product keeps the factor order, while the conjugate transpose reverses it.

Rectangular Matrices and the Role of Dimensions

The identities do not require square matrices. Dimensions are particularly helpful for detecting mistakes: if \(\mathbf{A}\) is \(m\times n\), then \(\mathbf{A}^{*}\) is \(n\times m\). If \(\mathbf{A}\) is \(m\times n\) and \(\mathbf{B}\) is \(n\times p\), then \(\mathbf{B}^{*}\mathbf{A}^{*}\) has size \(p\times m\), just like \((\mathbf{A}\mathbf{B})^{*}\). By contrast, \(\mathbf{A}^{*}\mathbf{B}^{*}\) may not even be defined.

Worked Example: A Rectangular Matrix and Its Transposes

Let \(\mathbf{C}=\begin{bmatrix}i&1-i&2\\1+i&-i&0\end{bmatrix}\), a \(2\times3\) matrix. Its ordinary transpose switches rows and columns without changing entries:

$$ \mathbf{C}^{\top} = \begin{bmatrix} i&1+i\\ 1-i&-i\\ 2&0 \end{bmatrix}. $$

Its entrywise conjugate keeps the original shape, whereas its conjugate transpose changes shape and conjugates the entries:

$$ \overline{\mathbf{C}}= \begin{bmatrix} -i&1+i&2\\ 1-i&i&0 \end{bmatrix}, \qquad \mathbf{C}^{*}= \begin{bmatrix} -i&1-i\\ 1+i&i\\ 2&0 \end{bmatrix}. $$

Taking the conjugate transpose of \(\mathbf{C}^{\top}\) gives \(\begin{bmatrix}-i&1+i&2\\1-i&i&0\end{bmatrix}\), which is exactly \(\overline{\mathbf{C}}\). This verifies \((\mathbf{C}^{\top})^{*}=\overline{\mathbf{C}}\) in this rectangular case. The shapes also confirm the identity: \(\mathbf{C}^{\top}\) is \(3\times2\), so its conjugate transpose is \(2\times3\), matching \(\overline{\mathbf{C}}\).

Check: The entry in position \((1,2)\) of \(\mathbf{C}\) is \(1-i\). It remains \(1-i\) in \(\mathbf{C}^{\top}\), becomes \(1+i\) in \(\overline{\mathbf{C}}\), and appears as \(1-i\) in position \((2,1)\) of \(\mathbf{C}^{*}\). Each operation follows its definition.

Common Mistake: Forgetting to Reverse the Product

A common incorrect calculation is \((\mathbf{A}\mathbf{B})^{*}=\mathbf{A}^{*}\mathbf{B}^{*}\). The conjugate transpose does conjugate and transpose, but transposing a product reverses the order of its factors. The correct identity is \((\mathbf{A}\mathbf{B})^{*}=\mathbf{B}^{*}\mathbf{A}^{*}\), as the entrywise proof established.

For the matrices in the previous product example, the incorrect order gives a concrete contradiction:

$$ \mathbf{A}^{*}\mathbf{B}^{*} = \begin{bmatrix} (1-i)(-i)-i&(1-i)(1-i)+(-i)i\\ 2(-i)+(1+i)&2(1-i)+(1+i)i \end{bmatrix} = \begin{bmatrix} -1-2i&1-2i\\ 1-i&1-i \end{bmatrix}. $$

But the correct result is \((\mathbf{A}\mathbf{B})^{*}=\begin{bmatrix}1-3i&1\\1+i&-1\end{bmatrix}\). These matrices are unequal, so the claimed identity fails. The correction is not to change the conjugation of entries, but to reverse the factor order: compute \(\mathbf{B}^{*}\mathbf{A}^{*}\).

Another useful distinction is between \(\mathbf{A}^{\top}\) and \(\mathbf{A}^{*}\). For the complex matrix \(\mathbf{C}\) above, the entry \(i\) stays \(i\) in \(\mathbf{C}^{\top}\), but becomes \(-i\) in \(\mathbf{C}^{*}\). Using a transpose where a conjugate transpose is required loses that sign change; conjugating without transposing loses the required change in shape.

Why These Identities Are Useful

Conjugation and conjugate transpose let complex matrix calculations be organized using familiar entrywise operations. The product identities are especially important: \(\overline{\mathbf{A}\mathbf{B}}=\overline{\mathbf{A}}\,\overline{\mathbf{B}}\) preserves the order, but \((\mathbf{A}\mathbf{B})^{*}=\mathbf{B}^{*}\mathbf{A}^{*}\) reverses it. The distinction comes from the transpose, not from scalar conjugation. Keeping the definitions and dimensions visible provides a reliable check before applying these operations in more involved calculations.

When checking an identity, a useful routine is to record the dimensions first, then compute entries using the definitions. For an equality involving a product, compare a typical \((i,j)\) entry on each side; for a conjugate transpose, remember that the entry comes from the transposed position and is then conjugated. This method works for square and rectangular complex matrices alike.

Check Your Understanding

For each question, use the definitions and verify dimensions before calculating.

  1. Find \(\overline{\mathbf{D}}\) and \(\mathbf{D}^{*}\) for \(\mathbf{D}=\begin{bmatrix}2-i&1+3i\\-i&4\end{bmatrix}\).
  2. State the correct product identity for \((\mathbf{X}\mathbf{Y})^{*}\), including compatible sizes for \(\mathbf{X}\) and \(\mathbf{Y}\).
  3. Explain why \(\overline{\mathbf{X}\mathbf{Y}}\) keeps the factor order, while \((\mathbf{X}\mathbf{Y})^{*}\) reverses it.
  4. If \(\mathbf{E}\) is \(2\times4\), what are the sizes of \(\overline{\mathbf{E}}\), \(\mathbf{E}^{\top}\), and \(\mathbf{E}^{*}\)?
  5. For a complex matrix \(\mathbf{E}\), which operation changes the sign of each imaginary part without switching rows and columns: transpose, entrywise conjugation, or conjugate transpose?

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