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

Good morning,

Let T be a linear transformation, which acts on a vector space V, over a field F.

Let W be a sub-space which invariants of T, and f,g polynomials from the same field F.

I need to prove that W is invariant of g(T) and f(T)(W) invariant of g(T).

obviously, g(T) is also a linear mapping and therefore is also invariants of W as T does, I just don't know how to prove it correctly.

Have a good day.

share|cite|improve this question
up vote 5 down vote accepted

Because $W$ is a subspace of $V$, it is invariant under all the maps $m_c:V\rightarrow V$, where for each $c\in F$ the map $m_c$ is defined by $m_c(v)=cv$. By assumption, $W$ is invariant under the map $T:V\rightarrow V$.

Hint: Show that if $W$ is invariant under two maps $C,D:V\rightarrow V$, then $W$ is invariant under $C\circ D$ and $C+D$. Then you will have that $W$ is invariant under all

$$f(T)=a_nT^n+\cdots+a_1T+a_0=(m_{a_n}\circ T\circ\cdots\circ T)+\cdots +(m_{a_1}\circ T)+m_{a_0}.$$

share|cite|improve this answer
thank you. helpful answer. – user6163 Apr 3 '11 at 9:17
Can I claim that f(T)(W) is a sub-space of W? and if so, why will it be invariant of g(T)? Sub-spaces of vector space who is invariant of a linear mapping T, don't have to invariant to T. – user6163 Apr 3 '11 at 9:28
@Nir: The image of $f(T)(W)$ is certainly a subspace of $V$ (image of a subspace under a linear transformation). Because W is invariant under f and under scalar multiplication, the image of $f(T)(W)$ is necessarily contained in $W$, so it is a subspace of $W$. To show it is invariant under $g(T)$, I suggest showing that $g(T)\circ f(T) = f(T)\circ g(T)$ and think about what that tells you. – Arturo Magidin Apr 3 '11 at 19:02

Your Answer


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