# When should I take conjugate transpose of a complex matrix, and when transpose of it?

I was taking the inverse of

$$A=\begin{bmatrix} 2+i &1 \\ 1&-2+i \end{bmatrix}$$

and $$\det(A)=-6$$, and cofactor matrix

$$C=\begin{bmatrix} -2+i &-1 \\ -1&2+i \end{bmatrix}$$

such that correct way to do it is

$$A^{-1}=\frac{1}{\det(A)}C^{T}$$

but I'm wondering why we are not taking conjugate transpose of C?

• It's hard to say why we don't do something unless you explain to us why you feel that we should. Commented Dec 23, 2019 at 10:09
• actually I don't know where we should take conjugate transpose of a matrix, and I thought that we take conjugate transpose instead transposing for the complex matrices. So what does it do actually and when should I use it ? Commented Dec 23, 2019 at 10:15
• The conjugate transpose arises from the standard Hermitian inner product and the usual transpose arises from the standard complex bilinear form. I’ll make that into a detailed answer if I get the chance later today. Commented Dec 23, 2019 at 11:47

Briefly: "most of the time", the correct analog for extending a matrix expression involving a transpose will be the conjugate-transpose. In the case of the inverse, however, the formula for $$A^{-1}$$ should be a "nice" (i.e. complex-differentiable) function and so we must use the entry-wise transpose.

In the case of real matrices, the transpose "usually" arises when we consider the relationship of a matrix $$A$$ to the inner-product $$\langle x, y \rangle_{\Bbb R} = y^Tx = x^Ty$$. More abstractly, this occurs when we consider how the linear transformation induced by $$A$$ interacts with the usual (Euclidean) geometry on $$\Bbb R^n$$. For instance, we have the following definitions and statements involving the transpose of a real matrix.

Definitions:

• The length of a vector $$x$$ (denoted $$\|x\|$$) is $$\sqrt{\langle x,x\rangle_{\Bbb R}}$$. Notably, $$\langle x,x \rangle_{\Bbb R} > 0$$ whenever $$x \neq 0$$.
• $$A$$ is symmetric (i.e. self-adjoint) when $$A = A^T$$, or equivalently when $$\langle Ax, y \rangle_{\Bbb R} = \langle x, Ay \rangle_{\Bbb R}$$.
• $$A$$ is orthogonal (i.e. satisfies $$\langle Ax, Ay \rangle_{\Bbb R} = \langle x, y \rangle_{\Bbb R}$$) when $$A^TA = I$$
• $$A$$ is positive definite when ($$A$$ is symmetric and) for all non-zero $$x$$, $$\langle x,Ax \rangle_{\Bbb R} > 0$$

Theorems:

• Cauchy-Schwarz: $$\langle x,y \rangle_{\Bbb R} \leq \|x\| \cdot \|y\|$$. Moreover, the angle between unit-vectors $$x,y$$ is $$\cos^{-1}(\langle x,y \rangle_{\Bbb R})$$.
• For a rectangular $$A$$, $$A^TA$$ will have the same rank as $$A$$, and $$\sqrt{\det(A^TA)}$$ is the "volume" spanned by the columns of $$A$$ when $$A$$ has linearly independent columns.
• The spectral theorem: if $$A$$ is symmetric, then $$A$$ is diagonalizable with real eigenvalues. Moreover, $$A$$ is orthogonally diagonalizable so that $$A = UDU^T$$ for some diagonal $$D$$ and orthogonal $$U$$.
• $$A$$ is positive definite if and only if it is symmetric with positive eigenvalues. This occurs if and only if $$x,y \mapsto \langle Ax, y \rangle_{\Bbb R}$$ defines an inner product.
• Every (rectangular) matrix $$A$$ has a singular-value decomposition $$A = U \Sigma V^T$$ (equivalent to the spectral theorem).

All of these statements and theorems have analogs when we consider complex matrices over the Hermitian inner product, which is defined by $$\langle x,y \rangle = y^*x$$. In the complex context, any $$A^T$$ is replaced with $$A^*$$, the conjugate-transpose of $$A$$.

Now, let's consider the entry-wise transpose for complex matrices and the corresponding bilinear form $$(x,y) = y^Tx = x^Ty$$. Here are some things that go wrong.

• It is not true that $$(x,x) > 0$$ whenever $$x \neq 0$$. For instance, the non-zero vector $$x = (1,i) \in \Bbb C^2$$ satisfies $$(x,x) = 0$$.
• Cauchy-Schwarz fails, for instance, with $$x = (1,i)$$ and $$y = (1,-i)$$.
• $$A^TA$$ no longer has the same rank as $$A$$. For instance, $$A = \pmatrix{1&i\\i&-1}$$ satisfies $$A^TA = 0$$
• $$A$$ can be symmetric without being diagonalizable. Consider for instance the $$A$$ given above, which fails to be diagonalizable.
• For any matrix $$A$$, $$x^T A x$$ is a complex polynomial on the entries of $$x$$. We can never have $$x^TAx > 0$$ for all $$x \neq 0$$.

There is, however, something gained in this case. Because $$(x,y)$$ is a polynomial on the entries of $$x$$ and $$y$$ (whereas $$\langle x, y \rangle$$ fails to be complex-differentiable), formulas involving the entry-wise transpose behave nicely with respect to computations involving complex numbers, including complex-differentiation.

For instance: the matrices satisfying $$A^T = A$$ form a complex subspace of $$\Bbb C^{n \times n}.$$ Also, the set of complex-orthogonal matrices (i.e. matrices satisfying $$A^TA = I$$) forms a smooth manifold in $$\Bbb C^{n \times n}$$.

Another consequence of all this, as you said, is that the correct choice for you determinant formula is the entry-wise transpose rather than the conjugate transpose. In this case, the formula for the cofactor matrix bears no relation to the Euclidean geometry on $$\Bbb R^n$$ or $$\Bbb C^n$$.

Over any field $$k$$ a matrix $$A\in k^{2\times 2}$$ where $$\det(A)\neq 0$$ the inverse is given by $$\frac{1}{\det(A)}\begin{pmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{pmatrix}$$ as you can easily calculate: $$\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{pmatrix}=\begin{pmatrix}\det(A)&0\\0&\det(A)\end{pmatrix}$$

• Possibly less general, but very clear! (+1) Commented Dec 23, 2019 at 10:18

I don't see why we should. We have the well-known formula, valid over any commutative ring: $$A\,^{\mathrm t\mkern-2.5mu}(\operatorname{com}A)=(\det A)I$$ where $$\;\operatorname{com}A$$ denotes the comatrix of $$A$$, a.k.a. matrix of cofactors.

This formula results directly from Laplace's formula for the expansion of a determinant.

• +1 much more general than my answer Commented Dec 23, 2019 at 10:14