Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T:V \to V$ be linear transformation, $\dim V=n$. If I know that every subspace of dimension $n-1$ is $T$-invariant, then how can I prove that $T=\lambda I$ for $\lambda \in F$?

$V$ is the vector space and $F$ is the field we work in.

Thank you.

share|cite|improve this question
Presumably you're asking for a proof of this statement? It wouldn't hurt to say so. Also, please introduce symbols that you use. Presumably $F$ is the field over which $V$ is a vector space? – joriki Aug 23 '11 at 14:44
I edited it, sorry. – Jozef Aug 23 '11 at 14:54
up vote 11 down vote accepted

Assume that there is $v\in V$ such that $Tv$ is linearly independent of $v$. Then we could find a subspace of dimension $n-1$ containing $v$ but not $Tv$, which contradicts the fact that $T$ leaves such a subspace invariant. Thus there is no such $v$, and $T$ acts as a scalar on all vectors. This must be the same scalar for all vectors, since otherwise applying $T$ to the sum of two non-zero vectors multiplied by different scalars would yield a linearly independent vector.

Note that $n-1$ could be replaced by any other integer $m$ with $0<m<n$.

share|cite|improve this answer
@Jozef: "$T$ leaves such a subspace invariant" means the same as "such a subspace is $T$-invariant". – joriki Aug 23 '11 at 15:12
Ok, great answer. thanks @joriki. – Jozef Aug 23 '11 at 15:14

Every $k$-dimensional subspace of $V$ can be seen as the intersection of two $(k+1)$-dimensional subspaces (for $k < n$).

Therefore, by induction, $T$ stabilizes every subspace of $V$, in particular the $1$-dimensional ones.

Thus there exists a function $\lambda : V\setminus\{0\} \to F$ such that for all $x\in V\setminus\{0\}$ , $Tx=\lambda(x) x$. Taking $x,y$ independent, we obtain $\lambda(x)=\lambda(x+y)=\lambda(y)$, which finally shows that $\lambda$ is constant. This concludes.

share|cite|improve this answer

Pick a vector $v \in V \setminus \{ 0 \}$. You can find $v_2, \dots, v_n \in V$ such that $\{v,v_2, \dots,v_n \}$ is basis of $V$. Now consider the $(n-1)$-subspaces $W_i = \langle v, v_2, \dots, v_{i-1}, v_{i+1}, \dots, v_n \rangle$, for $i=2,\dots,n$. It is clear that the subspace $\langle v \rangle = W_2 \cap \cdots \cap W_n$ is $T$-invariant. So there exists $\lambda_v \in F$ such that $Tv = \lambda_v v$. (Pay attention: $\lambda_v$ depends on $v$.)

Now we must prove the scalars $\{ \lambda_v \}_{v \in V \setminus \{ 0 \}}$ are all the same. Choose two vectors $v, v' \neq 0$; if $v$ and $v'$ are linearly dependent then it is clear that $\lambda_v = \lambda_{v'}$. If $v$ and $v'$ are linearly independent, then $\lambda_{v+v'} (v + v') = Tv + Tv' = \lambda_v v + \lambda_{v'} v'$, so $\lambda_v = \lambda_{v+v'} = \lambda_{v'}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.