0
$\begingroup$

We know that the improper subspaces of a vector space $V$ , $\{0\}$ and $V$ itself, are invariant under the linear operator $T$ and as we are always in search of proper invariant subspaces. (where $T \in L (V)$ and $dim (V)=n$.)

Now the eigenspace $W$ corresponding to some eigenvalue $c$ of $T$ is also invariant under $T$            $$W = \{x \in \mathbb{R}^n : Ax = cx\}$$ Now here is my question: When we talk about invariant subspaces we know that subspaces contain necessarily a zero vector but here $\{0\}$ doesn't belong to $W$ because eigenvectors are meaningful for non-zero vectors then how could be it a subspace.

$\endgroup$
  • 1
    $\begingroup$ First of all I think that you mean $T$ by $A$ in the explicit expression for $W$. Second of all $0$ does belong to $W$ since it obviously fulfils the condition $Tx=cx$ since $T\cdot 0 = 0 = c \cdot 0$. $\endgroup$ – Jakob Elias Apr 24 '17 at 6:55
  • $\begingroup$ See eigenspace contains eigenvectors and eigenvectors are non-zero i. e.,for eigenvector x not equal to Zero Ax =cx and c is the corresponding eigenvalue.Therefore, A . 0 = c.0 is not true $\endgroup$ – user437903 Apr 24 '17 at 7:04
  • 1
    $\begingroup$ @user437903 The eigenspace corresponding to the eigenvalue $c$ is defined to be $\{x \in \mathbb{R}^n : Ax = cx \}$, rather than the set of all eigenvectors with eigenvalue $c$. In particular, the eigenspace contains every eigenvector with eigenvalue $c$, but it also contains the zero vector. $\endgroup$ – lokodiz Apr 24 '17 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.