Necessary and Sufficient Condition for Diagonal Operator to be Invertible
Theorem
Let $\mathbb F \in \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\mathbb F$.
Let $\ell^2$ be the $2$-sequence space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space on $\ell^2$.
Let $\Lambda \in \map {CL} {\ell^2}$ be the diagonal operator such that:
- $\forall \mathbf x := \tuple {x_1, x_2, \ldots} \in \ell^2 : \map \Lambda {\mathbf x} = \tuple {\lambda_1 \cdot x_1, \lambda_2 \cdot x_2, \ldots}$
Then $\Lambda$ is invertible in $\map {CL} {\ell^2}$ if and only if $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$ where $\inf$ denotes the infimum.
Proof
Necessary Condition
Suppose $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$.
Then:
\(\ds \forall k \in \N_{> 0}: \, \) | \(\ds \size {\lambda_k}\) | \(\ge\) | \(\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} }\) | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall k \in \N_{> 0}: \, \) | \(\ds \lambda_k\) | \(\ne\) | \(\ds 0\) |
Moreover:
\(\ds \sup_{n \mathop \in \N_{> 0} } \size {\frac 1 {\lambda_n} }\) | \(=\) | \(\ds \sup_{n \mathop \in \N_{> 0} } \frac 1 {\size {\lambda_n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\ds \inf_{n \mathop \in \N_{> 0} } \size {\lambda_n} }\) |
Let $V : \ell^2 \to \ell^2$ be a mapping such that:
- $\ds \forall \mathbf a := \sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, \ldots} : \map V {\mathbf a} = \tuple {\frac {a_1} {\lambda_1}, \frac {a_2} {\lambda_2}, \ldots}$
Then $V \in \map {CL} {\ell^2}$.
Furthremore:
\(\ds \forall \mathbf a \in \ell^2: \, \) | \(\ds \map V {\map \Lambda {\mathbf a} }\) | \(=\) | \(\ds \map V {\sequence {\lambda_n a_n}_{n \mathop \in \N_{> 0} } }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sequence {\frac{\lambda_n a_n}{\lambda_n} }_{n \mathop \in \N_{> 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sequence {a_n}_{n \mathop \in \N_{> 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map I { \sequence {a_n}_{n \mathop \in \N_{> 0} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sequence {\lambda_n \frac{a_n}{\lambda_n} }_{n \mathop \in \N_{> 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Lambda {\sequence {\frac{a_n}{\lambda_n} }_{n \mathop \in \N_{> 0} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Lambda {\map V {\mathbf a} }\) |
Therefore:
- $\Lambda V = V \Lambda = I$
Hence, $\Lambda$ is invertible.
Finally, the unique inverse of $\Lambda$ is $\Lambda^{-1} = V$
$\Box$
Sufficient Condition
Suppose $\Lambda$ is invertible in $\map {CL} {\ell^2}$.
Then there is a unique inverse $\Lambda^{-1}$ of $\Lambda$:
- $\ds \exists \Lambda^{-1} \in \map {CL} {\ell^2} : \Lambda \circ \Lambda^{-1} = \Lambda^{-1} \circ \Lambda = I$
Furthermore, Invertible Operator is not Zero Operator.
Thus:
- $\exists \mathbf x \in \ell^2 : \map {\Lambda^{-1} } {\mathbf x} \ne \mathbf 0$
We have that P-Sequence Space with P-Norm forms Normed Vector Space.
Hence:
\(\ds \forall \mathbf x \in \ell^2: \, \) | \(\ds \norm {\mathbf x}_2\) | \(=\) | \(\ds \norm {\map I {\mathbf x} }_2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm { \map {\Lambda^{-1} } {\map \Lambda {\mathbf x} } }_2\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {\Lambda^{-1} } \norm{\map \Lambda {\mathbf x} }_2\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {\norm {\Lambda^{-1} } } \norm {\mathbf x}_2\) | \(\le\) | \(\ds \norm {\map \Lambda {\mathbf x} }_2\) |
Let $\mathbf x := \mathbf e_k = \tuple {\underbrace{0, \ldots, 0}_{k - 1}, 1, 0, \ldots}$
Then:
\(\ds \forall k \in \N_{> 0}: \, \) | \(\ds \size {\lambda_k}\) | \(=\) | \(\ds \norm {\map \Lambda {\mathbf e_k} }_2\) | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \frac 1 {\norm {\Lambda^{-1} } } \norm {\mathbf e_k}_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\norm {\Lambda^{-1} } }\) |
Therefore:
- $\ds \inf_{k \mathop \in \N_{> 0} } \size {\lambda_k} \ge \frac 1 {\norm {\Lambda^{-1} } } > 0$
$\Box$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations