Let $F$ be an infinite field, $V/F$ is a finite dimensional vector space. Let $T\in \mathcal L(V,V)$. Prove that there exists $\lambda\in F$ such that $T-\lambda I$ is injective.
I saw this. I am little confuse how to start it?
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1Do you know about determinants? – paul blart math cop Jun 07 '21 at 16:43
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@paulblartmathcop Yes, I know – Unknown Jun 07 '21 at 16:43
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1In that case consider $det(T - \lambda I)$ as a function of $\lambda$. What sort of function is it and what do you know about its roots? – paul blart math cop Jun 07 '21 at 16:45
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@paulblartmathcop I didn't get could you please elaborate – Unknown Jun 07 '21 at 16:47
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its rhymes with matrimonial – C Squared Jun 07 '21 at 16:58
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its a polynomial – C Squared Jun 07 '21 at 17:01
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1Yes it is a polynomial in $\lambda$. In fact, it is a nonzero polynomial. @AmanPandey do you see then what to do? Try to recall how determinants relate to injectivity. – paul blart math cop Jun 07 '21 at 17:26
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$T- \lambda I$ is injective iff ker$(T-\lambda I)= \{0\}$ so if there is NO vector $v \ne 0$ with $(T-\lambda I)v =0 \Leftrightarrow Tv= \lambda v$.
From the definition we know that this means: $T- \lambda I$ is injective iff $\lambda$ is no eigenvalue of $T$.
Now since $V$ is finite-dimensional we can only have finitely many eigenvalues (because each eigenvalue brings with it at least one linear independent eigenvector and eigenvectors corresponding to distinct eigenvalues are linearly independent - therefore we can only have dim$(V)$ eigenvalues). Since $F$ is given as an infinite field we can find $\lambda \in F$ which is not an eigenvalue and that gives us that $T- \lambda I$ is injective for that $\lambda$.
Lukas
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