What does $V(\lambda)$ mean?

Question

I came across this question:

Let $\lambda_1$, $\lambda_2$, ..., $\lambda_m$ be $m$ distinct eigenvalues of linear operator $T:V \rightarrow V$. Show that $V(\lambda_1) + V(\lambda_2) + ... + V(\lambda_m)$ is an invariant subspace under $T$.

I assume $V$ is a vector space. Can anyone explain what $V(\lambda_1), V(\lambda_2),..., V(\lambda_m)$ is? What does it mean to multiply a vector space by a scalar?

Answer 1

Probably, $V(\lambda)$ denotes the eigenspace of the eigenvalue $\lambda$, which is defined as

$$V(\lambda) = \{x\in V \mid Tx = \lambda x\}$$

i.e. it is the space of all eigenvectors belonging to an eigenvalue (and the zero vector).

You can also write it as $V(\lambda)=\ker(T-\lambda I)$

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There is no multiplication of a vector space by a scalar, $V(\lambda)$ does not mean "$V$ times $\lambda$", it means "$V$ of $\lambda$" (i.e., like $f(x)=x^2$ is read as $f$ of $x$ equals $x^2$).

Answer 2

$$\forall k=1,2,...,m\;,\;\;V_{\lambda_k}:=\left\{\,v\in V\;/\;Tv=\lambda_k v\,\right\}$$

and that alreayd includes the zero vector, which by definition is not an eigenvector.

Well, now take

$$w:=w_1+w_2+\ldots+w_m\in \sum_{k=1}^mV_{\lambda_k}\implies Tw=\sum_{k=1}^mTw_k=\sum_{k=1}^m\lambda_kv_k\in \sum_{k=1}^mV_{\lambda_k}$$

Answer 3

$V(\lambda)$ might also refer to the generalized eigenspace of the eigenvalue $\lambda$, in which case we would have:

$$ V(\lambda) = \{ x \in V | x \in \ker(T-\lambda)^N \text{ for some N} \} .$$

You should be warned that the generalized eigenspaces and eigenspaces do not agree in general. Generalized eigenspaces are also preserved by linear operators, so both definitions of $V(\lambda)$ are possible and you will need to check the context to figure out what is meant.