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  "body": "\n## Proof\n\nWe prove existence by induction on $n$, and uniqueness by comparing two factorisations.\n\n1. For $x, y \\in \\C^n$ let $\\ip{x}{y} = y^* x = \\sum_i x_i \\overline{y_i}$, where $^*$ is the [[conjugate transpose|0EX045E2BP]]. This is an [[inner product|0VGSYBWHV2]], since it is linear in $x$, $\\ip{y}{x} = \\overline{\\ip{x}{y}}$, and $\\ip{x}{x} = \\sum_i \\abs{x_i}^2 \u003e 0$ for $x \\ne 0$.\n2. Let $W$ be an $n \\times n$ complex matrix with columns $w_1, \\dots, w_n$. The $(i, j)$ entry of $W^* W$ is $w_i^* w_j = \\ip{w_j}{w_i}$. So $W$ is [[unitary|0D6RMJ10M9]] if and only if its columns form an [[orthonormal family|09VYHEF2VT]].\n3. For complex matrices $X$ and $Y$ with $XY$ defined, $(XY)^* = Y^* X^*$. This follows from the [[transpose of a product|09GK5PS824]] by taking complex conjugates of all entries.\n4. Let $U$ and $V$ be unitary $n \\times n$ matrices. Then $(UV)^*(UV) = V^* U^* U V = V^* V = I_n$, so $UV$ is unitary.\n5. Since $U^* U = I_n$, the [[one-sided inverse theorem|0753Y2X5K6]] gives $U U^* = I_n$. As $(U^*)^* = U$, this says that $U^*$ is unitary.\n6. Claim: for each $a \\in \\C^n$ there are a unitary $H$ and a number $\\alpha \\in \\C$ with $Ha = \\alpha e_1$, where $e_1$ is the first standard basis vector. If $a = 0$, take $H = I_n$ and $\\alpha = 0$.\n7. If $a \\ne 0$, let $\\norm{a} = \\sqrt{\\ip{a}{a}}$ and $u = a / \\norm{a}$. The family $(u)$ is orthonormal. By the [[existence of an orthonormal basis|0Y85GYEGSF]], it extends to an orthonormal basis $u = w_1, w_2, \\dots, w_n$ of $\\C^n$.\n8. Let $W$ be the matrix with columns $w_1, \\dots, w_n$. By step 2, $W$ is unitary. Also $W e_1 = w_1$, so $a = \\norm{a}\\, W e_1$.\n9. Put $H = W^*$, which is unitary by step 5, and $\\alpha = \\norm{a}$. Then $Ha = \\norm{a}\\, W^* W e_1 = \\alpha e_1$. This proves the claim.\n10. Existence for $n = 1$: take $Q = (1)$ and $R = A$. Every $1 \\times 1$ matrix is [[upper triangular|0MKABWDRNC]].\n11. Let $n \\ge 2$ and assume existence for size $n - 1$. Let $a$ be the first column of $A$, and take $H$ and $\\alpha$ as in step 6.\n12. The first column of $HA$ is $Ha = \\alpha e_1$. So $HA = \\begin{pmatrix} \\alpha \u0026 r \\\\ 0 \u0026 A' \\end{pmatrix}$, with a row $r$ of length $n - 1$ and an $(n-1) \\times (n-1)$ matrix $A'$.\n13. By the induction hypothesis, $A' = Q' R'$ with $Q'$ unitary and $R'$ upper triangular.\n14. Let $G = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 Q' \\end{pmatrix}$. Then $G^* G = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 Q'^* Q' \\end{pmatrix} = I_n$, so $G$ is unitary.\n15. Let $R = \\begin{pmatrix} \\alpha \u0026 r \\\\ 0 \u0026 R' \\end{pmatrix}$. It is upper triangular because $R'$ is. Block multiplication gives $GR = \\begin{pmatrix} \\alpha \u0026 r \\\\ 0 \u0026 Q'R' \\end{pmatrix} = HA$.\n16. Multiply step 15 on the left by $H^*$. Since $H^* H = I_n$, we get $A = H^* G R$.\n17. Put $Q = H^* G$. It is unitary by steps 4 and 5. So $A = QR$, which completes the induction.\n18. Real case: let $A$ be real. Repeat steps 6 to 17 in $\\R^n$ with the inner product $\\ip{x}{y} = y^T x$. The existence theorem of step 7 also holds over $\\R$, so $W$ is real.\n19. For real matrices, $X^* = X^T$, so a real unitary matrix is an [[orthogonal matrix|0QRBVFJ50X]]. All matrices built in steps 6 to 17 are then real, so $Q$ is orthogonal and $R$ is real.\n20. Uniqueness, first a lemma. Let $T$ be an invertible upper triangular $n \\times n$ matrix with entries $t_{ij}$. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $\\det T = t_{11} \\cdots t_{nn}$.\n21. By the [[invertibility criterion|FBSAPTP1PR]], $\\det T \\ne 0$. So $t_{ii} \\ne 0$ for all $i$.\n22. Fix $j$ and let $x$ be column $j$ of $T^{-1}$, so $Tx = e_j$. For $i \u003e j$, row $i$ of this system reads $\\sum_{k \\ge i} t_{ik} x_k = 0$.\n23. For $i = n, n-1, \\dots, j+1$ in turn, this gives $t_{ii} x_i = -\\sum_{k \u003e i} t_{ik} x_k = 0$, since the later $x_k$ already vanish. So $x_i = 0$ for all $i \u003e j$.\n24. Row $j$ then reads $t_{jj} x_j = 1$. So $T^{-1}$ is upper triangular, with diagonal entries $1/t_{jj}$.\n25. If $S$ and $T$ are upper triangular, then $(ST)_{ij} = \\sum_k s_{ik} t_{kj}$, and a term is nonzero only if $i \\le k \\le j$. So $ST$ is upper triangular with diagonal entries $s_{ii} t_{ii}$.\n26. Now let $A$ be invertible, and let $A = Q_1 R_1 = Q_2 R_2$ with $Q_1, Q_2$ unitary and $R_1, R_2$ upper triangular with positive diagonal entries.\n27. Since the [[determinant is multiplicative|04CPJ6AF00]], $0 \\ne \\det A = \\det Q_1 \\det R_1$. So $\\det R_1 \\ne 0$, and $R_1$ is invertible.\n28. From $Q_1 R_1 = Q_2 R_2$ we get $Q_2^* Q_1 R_1 = R_2$, so $T = Q_2^* Q_1$ equals $R_2 R_1^{-1}$.\n29. By steps 24 and 25, $T$ is upper triangular with positive diagonal entries. By steps 4 and 5, $T$ is unitary.\n30. Since $T^* T = I_n$, the one-sided inverse theorem gives $T^{-1} = T^*$. By step 24, $T^*$ is upper triangular.\n31. But $T^*$ is the conjugate transpose of an upper triangular matrix, so it is lower triangular. Hence $T^*$ is diagonal, and so is $T$.\n32. The diagonal entries $t_{jj}$ are positive reals with $t_{jj}^2 = (T^* T)_{jj} = 1$. So $t_{jj} = 1$ and $T = I_n$.\n33. Then $R_2 = T R_1 = R_1$. Also $Q_2^* Q_1 = I_n$, so $Q_1 = Q_2 Q_2^* Q_1 = Q_2$, using $Q_2 Q_2^* = I_n$ from step 5.\n34. So every complex $n \\times n$ matrix is $QR$, with real orthogonal factors for real $A$. The factorisation is unique for invertible $A$ with positive diagonal in $R$.\n\n## Idea\n\nA unitary matrix can turn any vector into a multiple of $e_1$. Doing this to the first column of $A$ leaves a smaller block, which we factor by induction. For uniqueness, two factorisations differ by a matrix that is both unitary and triangular with positive diagonal, and only the identity is both.\n",
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