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$$\text{Prove normal matrix is diagonalizable over }\mathbb{C}$$

$$\textbf{If }\textbf{A}^\ast\textbf{A}=\textbf{A}\textbf{A}^\ast$$ $$\textbf{Then }\textbf{A}=\textbf{P}^\ast\Lambda\textbf{P}$$ $$\text{Proof}$$ $$\textbf{A}\text{ is normal matrix}$$ $$\Rightarrow\textbf{A} = \textbf{P}^\ast\Delta\textbf{P}\tag{1}$$ $$\Delta\text{ is upper triangle matrix }\textbf{(Schur decomposition)}$$ $$\textbf{P}\text{ is unitary matrix}$$ $$\text{First we need to show if }\textbf{A}\text{ is }\textbf{normal}\text{ then }\Delta\text{ is }\textbf{normal}$$ $$\textbf{A} = \textbf{P}^\ast\Delta\textbf{P}\tag{2}$$ $$\Rightarrow\textbf{A}^\ast = (\textbf{P}^\ast\Delta\textbf{P})^\ast$$ $$\Rightarrow\textbf{A}^\ast = \textbf{P}^\ast(\textbf{P}^\ast\Delta)^\ast$$ $$\Rightarrow\textbf{A}^\ast = \textbf{P}^\ast(\Delta^\ast\textbf{P})\tag{3}$$ $$\text{from (2) and (3)}$$ $$\Rightarrow\textbf{A}\textbf{A}^\ast = (\textbf{P}^\ast\Delta\textbf{P})(\textbf{P}^\ast\Delta^\ast\textbf{P})$$ $$\Rightarrow\textbf{A}\textbf{A}^\ast = \textbf{P}^\ast\Delta(\textbf{P}\textbf{P}^\ast)\Delta^\ast\textbf{P}$$ $$\textbf{P} \text{ is unitary matrix}$$ $$\Rightarrow\textbf{P}^\ast = \textbf{P}^{-1}$$ $$\Rightarrow\textbf{A}\textbf{A}^\ast = \textbf{P}^\ast\Delta\Delta^\ast\textbf{P}\tag{4}$$

$$\Rightarrow\textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta^\ast\textbf{P}\textbf{P}^\ast\Delta\textbf{P}$$ $$\Rightarrow\textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta^\ast\Delta\textbf{P}\tag{5}$$ $$\textbf{From (4) and (5)}$$ $$\Rightarrow\textbf{A}\textbf{A}^\ast - \textbf{A}^\ast\textbf{A} = \textbf{P}^\ast\Delta\Delta^\ast\textbf{P}-\textbf{P}^\ast\Delta^\ast\Delta\textbf{P}$$ $$\Rightarrow\textbf{A}\textbf{A}^\ast - \textbf{A}\textbf{A}^\ast = \textbf{P}^\ast(\Delta\Delta^\ast-\Delta^\ast\Delta)\textbf{P} = \textbf{0}$$ $$\Rightarrow \Delta\Delta^\ast-\Delta^\ast\Delta = \textbf{0}$$ $$\Rightarrow\Delta\Delta^\ast = \Delta^\ast\Delta\tag{6}$$ $$\Rightarrow\Delta \textbf{ is normal matrix}$$ $$\textbf{From (6)}$$ $$\Rightarrow\vec{e_i}^\ast\Delta\Delta^\ast\vec{e_i} = \vec{e_i}^\ast\Delta^\ast\Delta\vec{e_i}$$ $$\Rightarrow\langle\Delta^\ast\vec{e_i},\Delta^\ast\vec{e_i}\rangle = \langle\Delta\vec{e_i}, \Delta\vec{e_i}\rangle$$ $$\Rightarrow\lVert\Delta^\ast\vec{e_i}\rVert^{2} = \lVert\Delta\vec{e_i}\lVert^{2}$$ $$\Rightarrow\lVert\Delta^\ast\vec{e_i}\rVert = \rVert\Delta\vec{e_i}\lVert$$ $$\Rightarrow\textbf{The length of ith column and ith row in }\Delta\textbf{ are same} \tag{7}$$ $$\Delta\textbf{ is upper triangle matrix}$$ $$\textbf{Let i to be the first row with nonzero off-diagonal element}$$ $$ \Delta_{n,n} = \begin{pmatrix} a_{1,1} & 0 & \cdots & 0 & \cdots & 0 \\ 0 & a_{2, 2} & \cdots & 0 & \cdots & 0 \\ 0 & 0 & a_{i,i}& \ast & \cdots & \ast \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ \cdot & \cdot & \cdot & \vdots & \vdots & a_{n,n} \end{pmatrix} $$

$$\textbf{If }\Delta\textbf{ is not diagonal matrix, then the ith column is }\lvert a_{i,i} \rvert \neq \lvert a_{i,i} \rvert + \lvert \ast \rvert + \cdots + \rvert \ast \lvert$$ $$\textbf{This contracts our previous (7), therefore }\Delta\textbf{ must be diagonal matrix}$$ $$\textbf{Therefore }\textbf{A}\textbf{A}^\ast = \textbf{A}^\ast\textbf{A}\Rightarrow\textbf{A}=\textbf{P}^\ast\Lambda\textbf{P}\quad\square$$

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  • 2
    $\begingroup$ What is the question? $\endgroup$ – Potato Oct 6 '13 at 23:27
  • $\begingroup$ I believe he wants his proof to be proofread to see if there are any errors in it. $\endgroup$ – David Oct 7 '13 at 0:04
  • $\begingroup$ As a proofreader, I would suggest using less boldface. $\endgroup$ – Marc van Leeuwen Oct 7 '13 at 7:53
  • $\begingroup$ Yes the standard way is to use the Schur decomposition and prove that the triangular matrix is normal iff it's diagonal. $\endgroup$ – Algebraic Pavel Oct 7 '13 at 9:33
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Presumably the ground field is $\mathbb{C}$, otherwise the problem statement is not true. Your proof looks fine to me then.

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