In Rickart, page 50 Theorem 2.2.1, the statement is made: A linear subspace $\mathfrak{M}$ of the algebra $\mathfrak{A}-\mathfrak{L}$ is invariant with respect to the representation $a{\rightarrow}A_a^{\mathfrak{A}-\mathfrak{L}}$. And, from which, it is concluded that $\mathfrak{L}_1$ (defined below) is a left ideal in $\mathfrak{A}$ that contains $\mathfrak{L}$.

I try to arrive at the same conclusion in the following manner. First the representation $a{\rightarrow}A_a^{\mathfrak{A}-\mathfrak{L}}$ means that $\mathfrak{A}-\mathfrak{L}$ is a left ideal in the algebra $\mathfrak{A}$ and that the left regular representation $a{\rightarrow}A_a$ of $\mathfrak{A}$ is restricted to the $\mathfrak{A}-\mathfrak{L}$ left ideal (definition according to Rickart). Second, all images of operators in the restricted left representation (image of $x$ denoted by $x^{\prime}$) are left ideal images. Third, since $\mathfrak{M}$ is an invariant subspace of $\mathfrak{A}-\mathfrak{L}$ then $\mathfrak{L}_1=\{x\,{:}\,x^{\prime}\,{\in}\,\mathfrak{M}\}$ must be a left ideal in $\mathfrak{A}-\mathfrak{L}$ and hence, $\mathfrak{L}_1$ is also a left ideal in $\mathfrak{A}$. This result is not quite right. According to Rickart I should have, in addition, concluded that $\mathfrak{L}_1$ contains $\mathfrak{L}$ but this does not seem possible since we're constrained to $\mathfrak{A}-\mathfrak{L}$. Can anyone point out what I missed?

Additional question. Rickart makes the statement that $\mathfrak{M}$ is a linear subspace of the algebra $\mathfrak{A}-\mathfrak{L}$. Is he saying that $\mathfrak{M}$ is $\textit{NOT}$ to be taken as a subalgebra of $\mathfrak{A}-\mathfrak{L}$ ? Or does he really mean to say that $\mathfrak{M}$ is a subalgebra of $\mathfrak{A}-\mathfrak{L}$ ?

(Sidebar: definition of an invariant subspace. Take $\mathfrak{M}$ as a subalgebra of $\mathfrak{A}-\mathfrak{L}$. If $Tx\:{\in}\:M$ for every $T\:{\in}\:\mathfrak{M}$ where $M$ is the set of vectors that generate the algebra $\mathfrak{M}$, then $M$ is said to be invariant with respect to the algebra $\mathfrak{M}$.)

Rickart: Theorem 2.2.1 page 50 of "$\textit{General Theory of Banach Algebras,}$" 1960.

In Rickart, page 50 Theorem 2.2.1, the statement is made: A linear subspace $\mathfrak{M}$ of the algebra $\mathfrak{A}-\mathfrak{L}$ is invariant with respect to the representation $a{\rightarrow}A_a^{\mathfrak{A}-\mathfrak{L}}$. And, from which, it is concluded that $\mathfrak{L}_1$ (defined below) is a left ideal in $\mathfrak{A}$ that contains $\mathfrak{L}$.

I try to arrive at the same conclusion in the following manner. First the representation $a{\rightarrow}A_a^{\mathfrak{A}-\mathfrak{L}}$ means that $\mathfrak{A}-\mathfrak{L}$ is a left ideal in the algebra $\mathfrak{A}$ and that the left regular representation $a{\rightarrow}A_a$ of $\mathfrak{A}$ is restricted to the $\mathfrak{A}-\mathfrak{L}$ left ideal (definition according to Rickart). Second, all images of operators in the restricted left representation (image of $x$ denoted by $x^{\prime}$) are left ideal images. Third, since $\mathfrak{M}$ is an invariant subspace of $\mathfrak{A}-\mathfrak{L}$ then $\mathfrak{L}_1=\{x\,{:}\,x^{\prime}\,{\in}\,\mathfrak{M}\}$ must be a left ideal in $\mathfrak{A}-\mathfrak{L}$ and hence, $\mathfrak{L}_1$ is also a left ideal in $\mathfrak{A}$. This result is not quite right. According to Rickart I should have, in addition, concluded that $\mathfrak{L}_1$ contains $\mathfrak{L}$ but this does not seem possible since we're constrained to $\mathfrak{A}-\mathfrak{L}$. Can anyone point out what I missed?

Additional question. Rickart makes the statement that $\mathfrak{M}$ is a linear subspace of the algebra $\mathfrak{A}-\mathfrak{L}$. Is he saying that $\mathfrak{M}$ is $\textit{NOT}$ to be taken as a subalgebra of $\mathfrak{A}-\mathfrak{L}$ ? Or does he really mean to say that $\mathfrak{M}$ is a subalgebra of $\mathfrak{A}-\mathfrak{L}$ ?

(Sidebar: definition of an invariant subspace. Take $\mathfrak{M}$ as a subalgebra of $\mathfrak{A}-\mathfrak{L}$. If $Tx\:{\in}\:M$ for every $T\:{\in}\:\mathfrak{M}$ where $M$ is the set of vectors that generate the algebra $\mathfrak{M}$, then $M$ is said to be invariant with respect to the algebra $\mathfrak{M}$.)

Rickart: Theorem 2.2.1 page 50 of "General theory of Banach algebras," 1960.