 stop2dance3l

2022-01-06

Label the following statements as being true or false.
(a) There exists a linear operator T with no T-invariant subspace.
(b) If T is a linear operator on a finite-dimensional vector space V, and W is a T-invariant subspace of V, then the characteristic polynomial of Tw divides the characteristic polynomial of T.
(c) Let T be a linear operator on a finite-dimensional vector space V, and let x and y be elements of V. If W is the T-cyclic subspace generated by x, W Anzante2m

Expert

a) Given statement: there exists a linear operator T with no T-invariant subspace. Since, subspace {0} is a T-invariant for every linear operator T, that means, for linear operator T(0)=0. Hence, the given statement is false.
b) If T is a linear operator on a finite-dimensional vector space V, and W is a T-invariant subspace of V, then the characteristics polynomial of ${T}_{W}$ divides the characteristics polynomial of T.
The provided statement is the direct theorem.
So, the given statement is true. Corgnatiui

Expert

(c)
Let T be a linear operator on a finite-dimensional vector space V, and let x and y be elements of V. If W is the T-cyclic subspace generated by x, W’ is the T-cyclic subspace generated by y, and W = W’, then x = y.
Let $W=\left\{x,T\left(x\right),{T}^{2}\left(x\right),\dots \right\}=\left\{x,y\right\}$
Since, $T\left(y\right)=-x$
$T\left(-x\right)=-y$
Which means that,
${W}^{\prime }=\left\{y,T\left(y\right),{T}^{2}\left(y\right),\dots \right\}=\left\{x,y\right\}$
Thus, $W={W}^{\prime }$ but $x\ne y$Hence, the given statement is false

Do you have a similar question?