Let dim$V$ be finite. Let $T:V\rightarrow V$ be a diagonalizable operator and $W\subseteq V$ such that $T(W)\subseteq W$. Then can we write $W$ as sum of ($1$-dimensional) eigenspaces of $T$?
Invariant subspace of a diagonalizable operator
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linear-algebra
vector-spaces
eigenvalues-eigenvectors
1 Answers
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$T$ is diagonalizable iff its minimal polynomial $m$ has distinct roots in the underlying field.
Clearly $m(T) = 0$ implies $m(T|_{W}) = 0$. Therefore the minimal polynomial $n$ of $T|_{W}$ divides $m$, so that $n$ has also distinct roots, and therefore $T|_{W}$ is diagonalizable.
Now a ($1$-dimensional) eigenspace for $T|_{W}$ is also an eigenspace for $T$.