4.3.5

a) Let T be a linear operator on a vector space V with matrix $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} $$ with respect to some basis $$ (v_1,v_2)$$ of V. For this problem, we will identify each vector v in V with its coordinates under the given basis: $$v = c_1v_1 + c_2v_2 \leftrightarrow \begin{bmatrix} c_1 \\ c_2 \\ \end{bmatrix} $$.

Let W be an invariant subspace of V under T, then the dimension of W is either 0,1 or 2. If W is 0-dimensional, it is simply the zero vector. If W is 2-dimensional, then it is the entire space V.

If W is 1-dimensional, then $$ W = span\{w\}$$ = $$\{cw: c \in \mathbb{F}\} $$, for some nonzero vector w, and where $$\mathbb{F} $$ is the field over which the vector space V is defined. Since W is invariant under T, $$ Tw \in W$$, so $$Tw = cw $$ for some scalar c, showing w is an eigenvector of T with eigenvalue c. Now the only eigenvectors of the matrix A are scalar multiples of $$\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} $$, with eigenvalue 1, and recalling the identification $$\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \leftrightarrow v_1 $$ which was made earlier, this shows w is a multiple of $$v_1 $$, so $$ W = span\{v_1\}$$.

Therefore, the invariant subspaces of V under T are $$\{0\}, span\{v_1\}, $$ and V.

b) Now let T have the matrix $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} $$ with respect to some basis $$(v_1,v_2,v_3) $$ of V. Observe the basis vectors are in fact eigenvectors of T, and they are also the only eigenvectors of T (up to multiplication by a scalar).

As before, if W is an invariant subspace of V under T, it is either 0,1,2, or 3 dimensional. If W is 0-dimensional it is the zero vector; if it is 3-dimensional, it is the entire space V, and if it is 1-dimensional, then the reasoning above shows W is the span of one of the eigenvectors $$ v_1,v_2$$ or $$ v_3$$. We are left with the case where W is 2-dimensional.

To address this case, we will prove the following claim: If $$v = c_1v_2 + c_2v_2 + c_3v_3$$ is a vector in V such that $$c_1,c_2,c_3 $$ are all nonzero, then the set $$\{v,Tv,T^2v\} $$ is linearly independent in V. To see this is true, note that with $$ v \leftrightarrow (c_1,c_2,c_3)^T$$, we have $$ Tv \leftrightarrow (c_2,2c_2,3c_3)^T, T^2v \leftrightarrow (c_2,4c_2,9c_3)^T$$. Letting M be the matrix with columns given by the coordinate vectors of $$v,Tv,T^2v $$, we have that $$ M = \begin{bmatrix} c_1 & c_1 & c_1 \\ c_2 & 2c_2 & 4c_2 \\ c_3 & 3c_3 & 9c_3 \\ \end{bmatrix}$$, and an explicit computation shows that $$det(M) = 2c_1c_2c_3 \ne 0 $$, since $$ c_1,c_2,c_3$$ are all nonzero. Therefore M is invertible, thus its columns are linearly independent, and so is the set $$\{v,Tv,T^2v\} $$.

Now with W being a 2-dimensional invariant subspace of V under T, given any w in W, the set $$\{w,Tw,T^2w\} $$ must be linearly dependent, and therefore writing $$w = c_1v_1 + c_2v_2 + c_3v_3 $$, one of the $$c_i $$ must be zero (if more than one was zero, the dimension of W would be less than 2). Thus, W is one of the following: $$span\{v_1,v_2\}, span\{v_1,v_3\}, span\{v_2,v_3\} $$. This fully classifies all invariant subspaces of V under T.