Why the Complex Gram Matrix Matters
For a real matrix, the product \(\mathbf{A}^{\top}\mathbf{A}\) records how the columns of \(\mathbf{A}\) relate to one another and produces a symmetric positive semidefinite matrix. With complex entries, ordinary transpose does not account for complex conjugation, so the corresponding matrix is \(\mathbf{A}^{*}\mathbf{A}\). Its key property is an exact identity: for every input \(\mathbf{x}\), the quadratic expression \(\mathbf{x}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{x}\) equals the squared length of \(\mathbf{A}\mathbf{x}\). This makes the product a useful bridge between a matrix and the lengths of its outputs.
Throughout, let \(\mathbf{A}\in\mathbb{C}^{m\times n}\), where \(m,n\in\mathbb{N}\), and let \(\mathbf{x}\in\mathbb{C}^{n}\) be a column vector. For a complex number \(z=a+bi\), its conjugate is \(\overline{z}=a-bi\). The squared Euclidean norm of \(\mathbf{y}\in\mathbb{C}^{m}\) is \(\|\mathbf{y}\|^{2}=\sum_{k=1}^{m}|y_k|^{2}=\mathbf{y}^{*}\mathbf{y}\), where \(|y_k|^{2}=\overline{y_k}y_k\). This is a real number, and it is nonnegative.
We use the course convention that the complex inner product is linear in its first slot: \(\langle\mathbf{u},\mathbf{v}\rangle=\mathbf{v}^{*}\mathbf{u}\). In particular, \(\|\mathbf{u}\|^{2}=\langle\mathbf{u},\mathbf{u}\rangle=\mathbf{u}^{*}\mathbf{u}\). The conjugate transpose is consistent with this convention: for compatible \(\mathbf{A},\mathbf{x},\mathbf{y}\), \(\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{A}^{*}\mathbf{y}\rangle\). The norm identity below can also be established directly from matrix multiplication, without relying on inner-product notation.
Dimensions and Entries
If \(\mathbf{A}\) has \(m\) rows and \(n\) columns, then \(\mathbf{A}^{*}\) has \(n\) rows and \(m\) columns, so \(\mathbf{A}^{*}\mathbf{A}\) is square with size \(n\times n\). The order matters: \(\mathbf{A}\mathbf{A}^{*}\) is instead \(m\times m\). If the columns of \(\mathbf{A}\) are \(\mathbf{a}_1,\dots,\mathbf{a}_n\), then the entry in row \(i\), column \(j\) of \(\mathbf{A}^{*}\mathbf{A}\) is \(\mathbf{a}_i^{*}\mathbf{a}_j\). Thus the diagonal entries are squared column lengths, and the off-diagonal entries record conjugated dot products of pairs of columns.
A complex matrix \(\mathbf{H}\) is Hermitian if \(\mathbf{H}^{*}=\mathbf{H}\). It is positive semidefinite if it is Hermitian and \(\mathbf{x}^{*}\mathbf{H}\mathbf{x}\ge 0\) for every complex column vector \(\mathbf{x}\) of the appropriate size. Here the inequality means that the scalar is real and nonnegative. It is important to require reality as well as nonnegativity; a general complex scalar is not ordered.
Proof. Write the columns of \(\mathbf{A}\) as \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The entry of \(\mathbf{A}^{*}\mathbf{A}\) in position \((i,j)\) is \(\mathbf{a}_i^{*}\mathbf{a}_j\). The corresponding entry in position \((j,i)\) is \(\mathbf{a}_j^{*}\mathbf{a}_i\), whose conjugate is \(\mathbf{a}_i^{*}\mathbf{a}_j\). Therefore each entry in position \((i,j)\) equals the conjugate of the entry in position \((j,i)\), so \((\mathbf{A}^{*}\mathbf{A})^{*}=\mathbf{A}^{*}\mathbf{A}\).
For the identity, associativity gives \(\mathbf{x}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{x}=(\mathbf{A}\mathbf{x})^{*}(\mathbf{A}\mathbf{x})\), since \((\mathbf{A}\mathbf{x})^{*}=\mathbf{x}^{*}\mathbf{A}^{*}\). Writing \(\mathbf{y}=\mathbf{A}\mathbf{x}\), the right side is \(\mathbf{y}^{*}\mathbf{y}=\sum_{k=1}^{m}\overline{y_k}y_k=\sum_{k=1}^{m}|y_k|^{2}=\|\mathbf{y}\|^{2}\). Every term is a nonnegative real number, so the sum is real and nonnegative. This proves both the identity and positive semidefiniteness. \(\square\)
Computing a Complex Gram Matrix
Worked Example: A Two-By-Two Complex Matrix
Let \(\mathbf{A}=\begin{bmatrix}1&i\\2&0\end{bmatrix}\) and \(\mathbf{x}=\begin{bmatrix}1\\i\end{bmatrix}\). Compute the Gram matrix and check the quadratic-form identity.
Transpose the matrix and conjugate each entry. Since \(\overline{i}=-i\),
Multiplying gives
The off-diagonal entries are conjugates, as Hermitian symmetry requires. To check the norm identity, first compute
Now calculate using the Gram matrix:
The two calculations agree. Also, \((\mathbf{A}^{*}\mathbf{A})^{*}=\begin{bmatrix}5&i\\-i&1\end{bmatrix}\), so the computed matrix is indeed Hermitian.
The Null Space Is Preserved
The norm identity gives more than nonnegativity: it identifies exactly which vectors the Gram matrix sends to zero. For a matrix \(\mathbf{A}\in\mathbb{C}^{m\times n}\), define \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{C}^{n}:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). This is the set of inputs mapped to the zero vector.
Proof. First suppose \(\mathbf{x}\in\operatorname{Nul}\mathbf{A}\). Then \(\mathbf{A}\mathbf{x}=\mathbf{0}\), so \(\mathbf{A}^{*}\mathbf{A}\mathbf{x}=\mathbf{A}^{*}\mathbf{0}=\mathbf{0}\), and hence \(\mathbf{x}\in\operatorname{Nul}(\mathbf{A}^{*}\mathbf{A})\). Conversely, suppose \(\mathbf{x}\in\operatorname{Nul}(\mathbf{A}^{*}\mathbf{A})\). Then \(\mathbf{A}^{*}\mathbf{A}\mathbf{x}=\mathbf{0}\), so multiplying on the left by \(\mathbf{x}^{*}\) gives \(\mathbf{x}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{x}=0\). By the norm identity, \(\|\mathbf{A}\mathbf{x}\|^{2}=0\). This squared norm is the sum \(\sum_{k=1}^{m}|(\mathbf{A}\mathbf{x})_k|^{2}\) of nonnegative real numbers. A finite sum of nonnegative numbers equals zero only if each term is zero, so every component of \(\mathbf{A}\mathbf{x}\) is zero. Thus \(\mathbf{x}\in\operatorname{Nul}\mathbf{A}\), proving the reverse inclusion and the equality. \(\square\)
Worked Example: A Nonzero Vector in the Null Space
Let \(\mathbf{A}=\begin{bmatrix}1&i\\2&2i\end{bmatrix}\) and \(\mathbf{x}=\begin{bmatrix}-i\\1\end{bmatrix}\). This example checks the null-space result and shows why a Gram matrix need not be positive definite.
First form the conjugate transpose and multiply:
For instance, the upper-right entry is \(1\cdot i+2\cdot2i=5i\), and the lower-right entry is \((-i)i+(-2i)(2i)=1+4=5\). Now multiply \(\mathbf{A}\) by \(\mathbf{x}\):
For a direct check with the Gram matrix,
The vector \(\mathbf{x}\) is nonzero, yet its quadratic-form value is zero. Thus this Gram matrix is positive semidefinite, but it does not have a strictly positive quadratic-form value for every nonzero input.
A Rectangular Example
Worked Example: A Two-By-Three Complex Matrix
Let \(\mathbf{A}=\begin{bmatrix}1&i&0\\0&1&1-i\end{bmatrix}\) and \(\mathbf{x}=\begin{bmatrix}1\\0\\i\end{bmatrix}\). Compute \(\mathbf{A}^{*}\mathbf{A}\) and verify the identity, paying attention to its dimensions.
Here \(\mathbf{A}\) is \(2\times3\), so \(\mathbf{A}^{*}\) is \(3\times2\), and the Gram matrix is \(3\times3\):
For example, the entry in position \((2,3)\) is \((-i)\cdot0+1\cdot(1-i)=1-i\), and the entry in position \((3,3)\) is \((1+i)(1-i)=2\). The entry in position \((3,2)\) is \(1+i\), the conjugate of the entry in position \((2,3)\).
Compute the output and its squared norm:
Using the Gram matrix instead gives
Since \(\mathbf{x}^{*}=\begin{bmatrix}1&0&-i\end{bmatrix}\), the quadratic-form value is \(1\cdot1+0\cdot1+(-i)(2i)=1+2=3\), matching the squared norm. The dimensions also agree: a \(3\times3\) Gram matrix acts on the original three-component input \(\mathbf{x}\).
Common Pitfall: Using Transpose Instead
For complex matrices, the ordinary transpose \(\mathbf{A}^{\top}\) does not generally replace the conjugate transpose \(\mathbf{A}^{*}\). The difference can change even the sign of a quadratic-form value. Reuse \(\mathbf{A}=\begin{bmatrix}1&i\\2&0\end{bmatrix}\) from the first example, but incorrectly use \(\mathbf{A}^{\top}\mathbf{A}\). The erroneous product is
For \(\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}\), this gives \(\mathbf{e}_2^{*}(\mathbf{A}^{\top}\mathbf{A})\mathbf{e}_2=-1\), which is negative. But \(\mathbf{A}\mathbf{e}_2=\begin{bmatrix}i\\0\end{bmatrix}\), so \(\|\mathbf{A}\mathbf{e}_2\|^{2}=|i|^{2}=1\). The false negative value exposes the mistake. With the correct product, \(\mathbf{A}^{*}\mathbf{A}=\begin{bmatrix}5&i\\-i&1\end{bmatrix}\), the value is \(1\), exactly as the norm identity requires.
When checking a complex matrix calculation, conjugate every entry after transposing, and check that the off-diagonal entries of \(\mathbf{A}^{*}\mathbf{A}\) are conjugates of one another. Then test the quadratic form against \(\|\mathbf{A}\mathbf{x}\|^{2}\) for a chosen input. These checks do not replace the proof, but they often reveal a missing conjugation or a dimension error.
Check Your Understanding
Use conjugate transposes throughout, and verify quadratic-form calculations by computing the matrix-vector product first.
- Find \(\mathbf{A}^{*}\) for \(\mathbf{A}=\begin{bmatrix}1&-i\\2&3i\end{bmatrix}\).
- For \(\mathbf{A}=\begin{bmatrix}1\\i\end{bmatrix}\), compute \(\mathbf{A}^{*}\mathbf{A}\) and check that it is Hermitian.
- Explain why \(\mathbf{x}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{x}\) must be real and nonnegative for every complex vector \(\mathbf{x}\).
- Prove that \(\operatorname{Nul}(\mathbf{A}^{*}\mathbf{A})\subseteq\operatorname{Nul}\mathbf{A}\) using the squared-norm identity.
- Give a reason that \(\mathbf{A}^{\top}\mathbf{A}\) is not the appropriate product for a general complex matrix.
