Hello, thank you for your question. The statement is indeed true and can be proven using induction on the dimension of the vector space. Let's first define what a T-invariant subspace is.
A subspace W of a vector space V is called T-invariant if T(W) is a subset of W. In other words, the linear operator T maps vectors in W to other vectors in W.
Now, let's consider the base case n=2. For a vector space V with dimension 2, we can have two T-invariant subspaces: the zero subspace {0} and the entire vector space V.
Now, assume that for some integer k>=2, the statement is true for all vector spaces with dimension k. That is, for any linear operator T on a vector space V with dimension k, there exists exactly k T-invariant subspaces.
We will now prove that the statement is also true for dimension k+1. Consider a vector space V with dimension k+1 and a linear operator T on V. By the induction hypothesis, we know that there exist k T-invariant subspaces W1, W2, ..., Wk of V.
Now, consider the subspace W = W1 + W2 + ... + Wk. This subspace is also T-invariant since for any vector w in W, we can write it as a linear combination of vectors in W1, W2, ..., Wk and T(w) will also be a linear combination of vectors in W1, W2, ..., Wk, thus remaining in W.
Additionally, we know that V/W is also a vector space with dimension 1. Therefore, by the base case, there exists exactly one T-invariant subspace U of V/W.
Combining these two subspaces, we have a total of k+1 T-invariant subspaces of V, satisfying the statement for dimension k+1.
Thus, by induction, the statement is true for all n>=2. I hope this helps.
