Proof: |detA| = 1 if A is unitary (that is, if $AA^* = I_n).$ Does it make sense?

Rafael Vergnaud just a moment. 1 answers, 0 views
linear-algebra proof-verification determinant

A matrix $A \in Mat(n, \mathbb{F})$ is called unitary if $AA^* = I_n,$ where $A^* = \overline{A^T} = \operatorname{conj}(A^T)$ is the conjugate transpose of $A$ (where the complex conjugate is taken for every entry in $A$). Prove that $\left\vert \det{A} \right\vert = 1$ only if $A$ is unitary.

If $A$ is orthogonal, then $AA^* = I_n$ and $$\det{(AA^*)} = \det{A}\det{A^*} = \det{I_n} = 1.$$ It will be shown that $$\det{A^*} = (\det{A})^*.$$ It is established that for $x, y \in \mathbb{C}$ and $\lambda \in \mathbb{R}$ $$\overline{\lambda} = \lambda \\ \overline{\lambda x} = \lambda \overline{x} \\ \overline{\overline{x}} = x \\ \overline{x+y} = \overline{x} + \overline{y} \\ \overline{x \cdot y} = \overline{x} \cdot \overline{y} \\$$

Consider the arbitrary $2 \times 2$ matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ and its conjugate transpose $$M^* = \begin{pmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{pmatrix}.$$

As $$\det{M} = a\det{d} - b\det{c} = ad - bc$$ and $$\det{M^*} = \overline{a} \cdot \overline{\det{d}} - \overline{b} \cdot \overline{\det{c}} \\ = \overline{a}\overline{d} - \overline{b}\overline{c} \\ = \overline{a}\overline{d} - \overline{b}\overline{c} \\ = \overline{ad} - \overline{bc} = \overline{ad - bc},$$ and $$\overline{\overline{ad - bc}} = ad - bc,$$ it follows that $$\det{M} = \det{M^T} = \overline{\det{M^*}} = (\det{M^*})^*$$ and $$\overline{(\det{M})} = (\det{M})^* = \det{\overline{M^T}} = \det{M^*}.$$

Now, suppose the equality holds for any $n \times n$ matrix $M.$ Consider the $(n+1) \times (n+1)$ matrix $M'.$ By the cofactor expansion of determinants, $$ \det{M'} = \sum_{j=1}^{n+1} A_{ij} \cdot \operatorname{cof}(A)_{ji} $$

So, given $$\det{M} = \det{M^T}$$ and by the properties of conjugates, $$\overline{\det{M'}} = (\det{M'})^* = \operatorname{conj}{(\sum_{j=1}^{n+1}A_{ij}\cdot\operatorname{cof}(A)_{ji})} \\ =\sum_{j=1}^{n+1}\overline{(A_{ij}\cdot\operatorname{cof}(A)_{ji})} \\ =\sum_{j=1}^{n+1}\overline{A_{ij}}\cdot\overline{\operatorname{cof}(A)_{ji}} \\ =\sum_{j=1}^{n+1}\overline{A_{ji}}\cdot\overline{\operatorname{cof}(A)^T_{ji}} \\ = \det{\overline{(M')^T}} = \det{M'^*}.$$

Hence, by induction, $$\det{A^*} = (\det{A})^*,$$ and therefore $$\det{A}\det{A^*} = (\det{A})(\det{A})^* = 1.$$ Suppose $$\det{A} = x + iy.$$ Therefore, $$(\det{A})^* = x - iy,$$ and $$(\det{A})(\det{A})^* = (x + iy)(x - iy) = x^2 + y^2 = 1.$$ So, $$|\det{A}|^2 = 1$$ and $$\left\vert \det{A} \right\vert \geq 0 \implies \left\vert \det{A} \right\vert = 1.$$

1 Answers


Aweygan 04/05/2018.

Everything seems fine. Since $|\det A|\geq0$, the $-1$ in $\pm1$ doesn't occur.


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