‎Banach ‎algebra

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1: We know that ‎$M‎_n‎(\mathbb{‎C}) $ ‎is a‎ ‎Banach ‎algebra.

‎ a) Can ‎we ‎say ‎the ‎set ‎of ‎all ‎upper ‎triangular(down ‎triangular) ‎is a‎ ‎subalgebra ‎of‎ ‎$ M‎_{n}‎(\mathbb{‎C}) ‎$, ‎but ‎not ‎an ideal?‎

b) ‎suppose‎ ‎$ D=\{ (a‎_{ij} ) ‎\in ‎‎‎ M‎_{n} ‎(\mathbb{‎C}) ‎:\ (a‎_{ij} )‎ =‎ 0‎ ,‎ i‎ ‎‎\neq j‎ ‎\}‎ ‎‎‎‎‎$, ‎so ‎is‎ ‎$ D $ a‎ ‎subalgebra ‎of‎‎ ‎‎‎‎$ M‎_{n} ‎(\mathbb{‎C}) ‎$, ‎but ‎not ‎ideal?‎

2: Let‎ $\mathbb D‎ = ‎\{‎‎‎ ‎‎\lambda ‎‎\in‎‎ ‎‎\mathbb{‎C‎} : | ‎‎\lambda |‎ <‎ ‎1\}‎ $‎‎, $ ‎\mathbb T‎ = ‎\{‎‎‎ ‎‎\lambda ‎‎\in‎‎ ‎‎\mathbb{‎C‎} : | ‎‎\lambda |‎ =‎ ‎1\}‎ $‎‎, and‎‎ ‎$ K = \{ f \in C (‎\overline{\mathbb D}) : ‎‎\quad‎ ‎‎f(‎\lambda) =‎ 0 ,‎ ‎‎\forall‎‎ ‎\lambda ‎\in\mathbb ‎‎ ‎T‎\}. $ ‎‎ Show ‎that‎ ‎$ K $ ‎is a‎ ‎closed ‎ideal ‎of‎ ‎$C (‎\overline {‎\mathbb D})‎ $‎?

‎ (if‎ ‎$ ‎\Omega‎ $ is a compact Hausdorff space, the ‎set‎ ‎$ C(‎\Omega) =‎ ‎\{ f‎ :‎ ‎‎\Omega‎‎ ‎\longrightarrow‎ ‎‎\mathbb{‎C} :\ f\ \text{ is continuous on} \Omega‎‎\}$ ‎is a‎ ‎algebra)‎

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a) Sums and products of upper triangular matrices are again upper triangular, so the set of upper triangular matrices is indeed a subalgebra of $M_n(\mathbb C)$. It is not an ideal.

b) Indeed, $D$ is also a subalgebra, and also not an ideal. In fact, $M_n(\mathbb C)$ has no nontrivial two-sided ideals.

The argument for 2) is straightforward: if $f\in K$ and $g\in C(\overline{\mathbb D})$, then $(fg)(\lambda)=f(\lambda)g(\lambda)=0$ for any $\lambda\in\mathbb T$, since $f(\lambda)=0$.