[proofplan] We prove that every nonzero closed two-sided ideal of $K(H)$ contains all rank-one operators. Starting from a nonzero [compact operator](/page/Compact%20Operator) $T$ in the ideal, rank-one multiplication on the left and right produces a nonzero rank-one operator in the ideal. Multiplying that rank-one operator again by arbitrary rank-one operators gives every rank-one operator. Since finite-rank operators are finite sums of rank-one operators and are norm dense in $K(H)$, closedness of the ideal forces it to be all of $K(H)$. [/proofplan] [step:Define the rank-one operators and compute their products] For $x,y\in H$, define the rank-one operator $\theta_{x,y}:H\to H$ by \begin{align*}\theta_{x,y}(\zeta)=(\zeta,y)_H x \qquad (\zeta\in H).\end{align*} Throughout, the [inner product](/page/Inner%20Product) on $H$ is linear in the first variable. This map is bounded because, by the [Cauchy-Schwarz inequality](/theorems/432) in $H$, \begin{align*}\|\theta_{x,y}\zeta\|_H\le \|\zeta\|_H\|y\|_H\|x\|_H\end{align*} for every $\zeta\in H$. Its range is contained in the one-dimensional subspace $\mathbb C x$, so $\theta_{x,y}$ is finite-rank and hence compact. Therefore $\theta_{x,y}\in K(H)$. Let $u,v,x,y\in H$. For every $\zeta\in H$, \begin{align*}(\theta_{u,v}\theta_{x,y})(\zeta)=\theta_{u,v}\bigl((\zeta,y)_H x\bigr)=((\zeta,y)_H x,v)_H u=(x,v)_H(\zeta,y)_H u.\end{align*} Thus \begin{align*}\theta_{u,v}\theta_{x,y}=(x,v)_H\theta_{u,y}.\end{align*} [guided] For $x,y\in H$, we introduce the basic finite-rank building block $\theta_{x,y}:H\to H$ by \begin{align*}\theta_{x,y}(\zeta)=(\zeta,y)_H x \qquad (\zeta\in H).\end{align*} The inner product is linear in the first variable, so this formula defines a [linear map](/page/Linear%20Map). The Cauchy-Schwarz inequality gives \begin{align*}\|\theta_{x,y}\zeta\|_H\le \|\zeta\|_H\|y\|_H\|x\|_H\end{align*} for every $\zeta\in H$, so $\theta_{x,y}$ is bounded. Its image is contained in $\mathbb C x$, a finite-dimensional subspace of $H$, so $\theta_{x,y}$ is compact and therefore belongs to $K(H)$. The multiplication rule is the mechanism that lets us move from one nonzero rank-one operator to all rank-one operators. Let $u,v,x,y\in H$. For $\zeta\in H$, first applying $\theta_{x,y}$ gives $(\zeta,y)_H x$. Applying $\theta_{u,v}$ afterward gives \begin{align*}(\theta_{u,v}\theta_{x,y})(\zeta)=\theta_{u,v}\bigl((\zeta,y)_H x\bigr)=((\zeta,y)_H x,v)_H u.\end{align*} Since the inner product is linear in the first variable, the scalar $(\zeta,y)_H$ factors out: \begin{align*}((\zeta,y)_H x,v)_H u=(x,v)_H(\zeta,y)_H u.\end{align*} This is exactly the action of the scalar multiple $(x,v)_H\theta_{u,y}$ on $\zeta$. Hence \begin{align*}\theta_{u,v}\theta_{x,y}=(x,v)_H\theta_{u,y}.\end{align*} [/guided] [/step] [step:Produce a nonzero rank-one operator inside a nonzero ideal] Let $I\trianglelefteq K(H)$ be a nonzero closed two-sided ideal. Choose $T\in I$ with $T\ne 0$. Since $T$ is nonzero, there exists $\xi\in H$ such that $T\xi\ne 0$. Define $x:=T\xi$. The rank-one operators $\theta_{x,x}$ and $\theta_{\xi,\xi}$ belong to $K(H)$, and $I$ is a two-sided ideal in $K(H)$. Hence \begin{align*}\theta_{x,x}T\theta_{\xi,\xi}\in I.\end{align*} For every $\zeta\in H$, \begin{align*}(\theta_{x,x}T\theta_{\xi,\xi})(\zeta)=\theta_{x,x}\bigl((\zeta,\xi)_H T\xi\bigr)=\theta_{x,x}\bigl((\zeta,\xi)_H x\bigr)=(x,x)_H(\zeta,\xi)_H x.\end{align*} Therefore \begin{align*}\theta_{x,x}T\theta_{\xi,\xi}=\|x\|_H^2\theta_{x,\xi}.\end{align*} Since $x\ne 0$, the scalar $\|x\|_H^2$ is nonzero, so $\theta_{x,\xi}\in I$ and $\theta_{x,\xi}\ne 0$. [/step] [step:Generate every rank-one operator from one nonzero rank-one operator] Let $a,b\in H$ be arbitrary. Since $x\ne 0$, define \begin{align*}c:=\frac{x}{\|x\|_H^2}\in H.\end{align*} Using the rank-one multiplication formula twice, \begin{align*}\theta_{a,x}\theta_{x,\xi}=\|x\|_H^2\theta_{a,\xi}.\end{align*} Then \begin{align*}\theta_{a,\xi}\theta_{c,b}=(c,\xi)_H\theta_{a,b}\end{align*} would not in general give a nonzero scalar, so instead choose \begin{align*}d:=\frac{\xi}{\|\xi\|_H^2}\in H.\end{align*} This is well-defined because $T\xi=x\ne 0$ implies $\xi\ne 0$. Applying the multiplication formula gives \begin{align*}\theta_{a,x}\theta_{x,\xi}\theta_{d,b}=\|x\|_H^2\theta_{a,\xi}\theta_{d,b}=\|x\|_H^2(d,\xi)_H\theta_{a,b}=\|x\|_H^2\theta_{a,b}.\end{align*} Thus \begin{align*}\theta_{a,b}=\frac{1}{\|x\|_H^2}\theta_{a,x}\theta_{x,\xi}\theta_{d,b}.\end{align*} The operators $\theta_{a,x}$ and $\theta_{d,b}$ lie in $K(H)$, and $\theta_{x,\xi}\in I$. Since $I$ is a two-sided ideal, the product belongs to $I$. Hence $\theta_{a,b}\in I$ for all $a,b\in H$. [/step] [step:Pass from rank-one operators to all compact operators] Every finite-rank operator on $H$ is a finite sum of rank-one operators, because its range is contained in a finite-dimensional subspace and expansion in a basis of that subspace writes the operator as such a sum. Since $I$ is a linear subspace and contains every rank-one operator, it contains every finite-rank operator. We now recall the density argument. Let $S\in K(H)$ and let $\varepsilon>0$. The image of the closed unit ball \begin{align*}B_H:=\{\zeta\in H:\|\zeta\|_H\le 1\}\end{align*} under $S$ has compact closure in $H$. Since compact metric spaces are totally bounded, $S(B_H)$ is totally bounded as a subset of $H$. Hence there exist vectors $\eta_1,\dots,\eta_n\in B_H$ such that for every $\zeta\in B_H$ there is an index $j\in\{1,\dots,n\}$ with \begin{align*}\|S\zeta-S\eta_j\|_H<\varepsilon.\end{align*} Let $M:=\operatorname{span}\{S\eta_1,\dots,S\eta_n\}$, and let $P_M:H\to H$ be the [orthogonal projection](/theorems/437) onto $M$. Then $P_MS$ has finite rank. For every $\zeta\in B_H$, choose $j$ as above. Since $S\eta_j\in M$ and $P_M$ is the nearest-point projection onto $M$, \begin{align*}\|S\zeta-P_MS\zeta\|_H\le \|S\zeta-S\eta_j\|_H<\varepsilon.\end{align*} Taking the supremum over $\zeta\in B_H$ gives \begin{align*}\|S-P_MS\|\le \varepsilon.\end{align*} Thus finite-rank operators are norm dense in $K(H)$. Since $I$ contains every finite-rank operator and is closed in the operator norm, every $S\in K(H)$ belongs to $I$. Therefore $I=K(H)$. We have shown that every nonzero closed two-sided ideal of $K(H)$ is all of $K(H)$, so $K(H)$ is simple. [/step]