Necessary and Sufficient Condition for Diagonal Operator to be Invertible

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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

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