Existence of inverses of linear combinations of bounded operators

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If $V$ is a banach space and $T$ is a bounded linear operator on $V$. If $|T|< a$ where $a>0$, then is $A=T-aI$ invertible? If so why? We know that if $|T|<1$ then $(I-T)$ is invertible.






functional-analysis operator-theory linear-transformations







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    If $V$ is a banach space and $T$ is a bounded linear operator on $V$. If $|T|< a$ where $a>0$, then is $A=T-aI$ invertible? If so why? We know that if $|T|<1$ then $(I-T)$ is invertible.






    functional-analysis operator-theory linear-transformations







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      If $V$ is a banach space and $T$ is a bounded linear operator on $V$. If $|T|< a$ where $a>0$, then is $A=T-aI$ invertible? If so why? We know that if $|T|<1$ then $(I-T)$ is invertible.






      functional-analysis operator-theory linear-transformations







      share|cite|improve this question















      If $V$ is a banach space and $T$ is a bounded linear operator on $V$. If $|T|< a$ where $a>0$, then is $A=T-aI$ invertible? If so why? We know that if $|T|<1$ then $(I-T)$ is invertible.






      functional-analysis operator-theory linear-transformations




      functional-analysis operator-theory linear-transformations






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      edited 12 mins ago









      qbert

      21.2k32356




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      asked 22 mins ago









      user593295

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          1 Answer
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          Note
          $$
          T-aI=-a(I-frac1aT)
          $$
          and by assumption
          $$
          left|left| frac1aT right|right|=frac1a||T||<1
          $$
          So $(I-frac1aT)^-1$ is invertible and the inverse of $T-aI$ is then
          $$
          -frac1a(I-frac1aT)^-1
          $$






          share|cite|improve this answer




















          • I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
            – user593295
            11 mins ago










          • yes, and you can explicitly give its inverse, as in the above
            – qbert
            1 min ago











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          1 Answer
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          active

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          1 Answer
          1






          active

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          active

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          up vote
          3
          down vote













          Note
          $$
          T-aI=-a(I-frac1aT)
          $$
          and by assumption
          $$
          left|left| frac1aT right|right|=frac1a||T||<1
          $$
          So $(I-frac1aT)^-1$ is invertible and the inverse of $T-aI$ is then
          $$
          -frac1a(I-frac1aT)^-1
          $$






          share|cite|improve this answer




















          • I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
            – user593295
            11 mins ago










          • yes, and you can explicitly give its inverse, as in the above
            – qbert
            1 min ago















          up vote
          3
          down vote













          Note
          $$
          T-aI=-a(I-frac1aT)
          $$
          and by assumption
          $$
          left|left| frac1aT right|right|=frac1a||T||<1
          $$
          So $(I-frac1aT)^-1$ is invertible and the inverse of $T-aI$ is then
          $$
          -frac1a(I-frac1aT)^-1
          $$






          share|cite|improve this answer




















          • I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
            – user593295
            11 mins ago










          • yes, and you can explicitly give its inverse, as in the above
            – qbert
            1 min ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          Note
          $$
          T-aI=-a(I-frac1aT)
          $$
          and by assumption
          $$
          left|left| frac1aT right|right|=frac1a||T||<1
          $$
          So $(I-frac1aT)^-1$ is invertible and the inverse of $T-aI$ is then
          $$
          -frac1a(I-frac1aT)^-1
          $$






          share|cite|improve this answer












          Note
          $$
          T-aI=-a(I-frac1aT)
          $$
          and by assumption
          $$
          left|left| frac1aT right|right|=frac1a||T||<1
          $$
          So $(I-frac1aT)^-1$ is invertible and the inverse of $T-aI$ is then
          $$
          -frac1a(I-frac1aT)^-1
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 16 mins ago









          qbert

          21.2k32356




          21.2k32356











          • I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
            – user593295
            11 mins ago










          • yes, and you can explicitly give its inverse, as in the above
            – qbert
            1 min ago

















          • I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
            – user593295
            11 mins ago










          • yes, and you can explicitly give its inverse, as in the above
            – qbert
            1 min ago
















          I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
          – user593295
          11 mins ago




          I assume if $(1-1/aT)$ is invertible, then any scale of that operator is invertible? (i.e we can freely multiply by $-a$ and its fine)
          – user593295
          11 mins ago












          yes, and you can explicitly give its inverse, as in the above
          – qbert
          1 min ago





          yes, and you can explicitly give its inverse, as in the above
          – qbert
          1 min ago


















           

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