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  "body": "\n## Proof\n\nThe converse is a direct computation. For the main direction, we argue by induction on $n$, splitting off one eigenvector at a time.\n\n1. Write $M^*$ for the [[conjugate transpose|0EX045E2BP]]. For matrices of matching sizes, $(MN)^* = N^* M^*$ and $(M^*)^* = M$, by the formula $(M^*)_{ij} = \\overline{m_{ji}}$.\n2. Converse. Let $A = U D U^*$ with $U$ [[unitary|0D6RMJ10M9]] and $D$ [[diagonal|0Q9PGXD9E4]]. Then $U^* U = I$, so $U U^* = I$ by the [[one-sided inverse theorem|0753Y2X5K6]].\n3. Diagonal matrices commute, so $D D^* = D^* D$. Hence $A A^* = U D U^* U D^* U^* = U D D^* U^* = U D^* D U^* = A^* A$. So $A$ is [[normal|0XYKE8BE7X]].\n4. For the main direction, let $P(n)$ say: every normal $n \\times n$ complex matrix is $U D U^*$ with $U$ unitary and $D$ diagonal. For $n = 1$, take $U = (1)$ and $D = A$.\n5. Let $n \\ge 2$, assume $P(n-1)$, and let $A$ be normal of size $n$. On $\\C^n$, use the inner product $\\ip{x}{y} = y^* x$, with norm $\\norm{x}^2 = x^* x = \\sum_i \\abs{x_i}^2$.\n6. $\\C$ is [[algebraically closed|02SJG1HC32]] by the [[fundamental theorem of algebra|0JP0W07Y9A]]. So $A$ has an [[eigenvalue|08CJB2NQ92]] $\\lambda$, with $A x = \\lambda x$ for some $x \\ne 0$. Set $u_1 = x / \\norm{x}$, so $u_1^* u_1 = 1$.\n7. Let $N = A - \\lambda I$. Then $N^* = A^* - \\overline{\\lambda} I$. Expanding gives $N N^* = A A^* - \\overline{\\lambda} A - \\lambda A^* + \\abs{\\lambda}^2 I$, and $N^* N$ is the same with $A^* A$ in place of $A A^*$.\n8. Since $A A^* = A^* A$, step 7 gives $N N^* = N^* N$. Then $\\norm{N^* u_1}^2 = u_1^* N N^* u_1 = u_1^* N^* N u_1 = \\norm{N u_1}^2$. This is $0$, since $A u_1 = \\lambda u_1$ gives $N u_1 = 0$.\n9. So $N^* u_1 = 0$, that is, $A^* u_1 = \\overline{\\lambda} u_1$.\n10. The map $\\ip{\\cdot}{\\cdot}$ of step 5 is an [[inner product|0VGSYBWHV2]]. The family $(u_1)$ is orthonormal, so it [[extends to an orthonormal basis|0Y85GYEGSF]] $u_1, \\dots, u_n$ of $\\C^n$.\n11. Let $U_1$ have columns $u_1, \\dots, u_n$. The $(i, j)$ entry of $U_1^* U_1$ is $u_i^* u_j = \\delta_{ij}$. So $U_1$ is unitary, and $U_1 U_1^* = I$ by the one-sided inverse theorem.\n12. Let $B = U_1^* A U_1$. By [[matrix multiplication|076JGP6R3T]], its $(i, j)$ entry is $b_{ij} = u_i^* A u_j$.\n13. First column: $b_{i1} = u_i^* (\\lambda u_1) = \\lambda \\delta_{i1}$. First row: $b_{1j} = (A^* u_1)^* u_j = (\\overline{\\lambda} u_1)^* u_j = \\lambda \\delta_{1j}$, by steps 1 and 9.\n14. So $B$ is the [[block matrix|01AXG9KZQF]] $\\begin{pmatrix} \\lambda \u0026 0 \\\\ 0 \u0026 C \\end{pmatrix}$ for some $(n-1) \\times (n-1)$ complex matrix $C$.\n15. By steps 1 and 11, $B B^* = U_1^* A U_1 U_1^* A^* U_1 = U_1^* A A^* U_1$. Likewise $B^* B = U_1^* A^* A U_1$. So $B$ is normal.\n16. Multiplying blocks gives $B B^* = \\begin{pmatrix} \\abs{\\lambda}^2 \u0026 0 \\\\ 0 \u0026 C C^* \\end{pmatrix}$ and $B^* B = \\begin{pmatrix} \\abs{\\lambda}^2 \u0026 0 \\\\ 0 \u0026 C^* C \\end{pmatrix}$. So $C C^* = C^* C$, and $C$ is normal.\n17. By $P(n-1)$, $C = V D' V^*$ with $V$ unitary of size $n - 1$ and $D'$ diagonal.\n18. Let $U_2 = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 V \\end{pmatrix}$ and $D = \\begin{pmatrix} \\lambda \u0026 0 \\\\ 0 \u0026 D' \\end{pmatrix}$. Then $D$ is diagonal, and $U_2^* U_2 = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 V^* V \\end{pmatrix} = I$.\n19. Multiplying blocks gives $U_2 D U_2^* = \\begin{pmatrix} \\lambda \u0026 0 \\\\ 0 \u0026 V D' V^* \\end{pmatrix} = B$.\n20. By step 11, $A = U_1 U_1^* A U_1 U_1^* = U_1 B U_1^*$. With step 19, $A = (U_1 U_2) D (U_1 U_2)^*$.\n21. Let $U = U_1 U_2$. Then $U^* U = U_2^* U_1^* U_1 U_2 = U_2^* U_2 = I$, so $U$ is unitary. Thus $A = U D U^*$, which proves $P(n)$.\n22. By induction, $P(n)$ holds for all $n \\ge 1$. With steps 2 and 3, $A$ is normal if and only if $A = U D U^*$ with $U$ unitary and $D$ diagonal.\n\n## Idea\n\nFor a normal matrix, $\\norm{A^* x} = \\norm{A x}$ for every $x$, and the same holds for $A - \\lambda I$. So an eigenvector of $A$ for $\\lambda$ is an eigenvector of $A^*$ for $\\overline{\\lambda}$. In an orthonormal basis that starts with this vector, both the first row and the first column of $A$ vanish off the diagonal. The rest is a smaller normal matrix, and induction finishes.\n",
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