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Let $T:V\to V$ be a linear operator on vector space $V$. Let $W$ be a $T$-invariant subspace of $V$ and denote $V(\lambda)$ the $\lambda$-eigenspace of $T$. Is it true that $(V/W)(\lambda)\cong V(\lambda)/W(\lambda)$? Given the dimensions of $(V/W)(\lambda)$ and $W(\lambda)$, what can I say about the dimension of $V(\lambda)$?

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A vector space $V$ endowed with an endomorphism is the same thing as a $k[X]$-module, so let's talk about the latter, which are nicer.

If $V$ is a $k[X]$-module and $\lambda$ is a scalar, then what you write $V(\lambda)$ can be identified with $\hom_{k[X]}(Q_\lambda,V)$, with $Q_\lambda$ the $k[X]$-module which as a vector space is simply $k$ and on which $X$ acts by multiplication by $\lambda$ (in other words, $Q_\lambda=k[X]/(X-\lambda)$).

Now suppose that $V$ is a module and that $W$ is a submodule of $V$ (which is the same thing as an invariant subspace), so that we have a shoret exact sequence $$0\to W\to V\to V/W\to 0$$ If we apply $\hom_{k[X]}(Q_\lambda,\mathord-)$, we obtain an exact sequence $$0\to\hom_{k[X]}(Q_\lambda,W)\to\hom_{k[X]}(Q_\lambda,V)\to\hom_{k[X]}(Q_\lambda,V/W)$$ which amounts to a sequence $$0\to W(\lambda)\to V(\lambda)\to (V/W)(\lambda)$$

Your question is essentially if we can extend these exact sequences by adding a zero on the right. Homological algebra tells us that what we can certainly do is add an $\def\Ext{\operatorname{Ext}}\Ext$, so get an exact sequence

$$0\to\hom_{k[X]}(Q_\lambda,W)\to\hom_{k[X]}(Q_\lambda,V)\to\hom_{k[X]}(Q_\lambda,V/W)\to\Ext^1_{k[X]}(Q_\lambda,W)$$ and therefore this vector space $\Ext^1_{k[X]}(Q_\lambda,W)$ measures in some sense how close we are, and if you know how to compute it tells you where to look for examples.

An easy way to get an example is to pick the matrix of $f$ on $V$ to be a Jordan block of eigenvalue $\lambda$, $W$ any proper nonzero invariant subspace. In that case you have $V(\lambda)$, $W(\lambda)$ and $(V/W)(\lambda)$ all one-dimensional.