Number of invariant subspaces A is symmetric linear map $R^{2n} \rightarrow R^{2n}$. Characteristic polynomial of A is $(\lambda -1)^2 \cdots (\lambda - n)^2$. How many invariant subspaces does A have?
Choices: $2n; 2^n; 2^{2n}$; infinitely many.
What I deduced: A has n different eigenvalues equal {1 ... n}, each twice. Also, all of them are real due to the fact that A is symmetric. What to do next?
 A: If you take the standard basis and represent $A$ as a $2n \times 2n$ matrix with respect to this basis, then $A$ is a real symmetric matrix. It is a fact that such matrices are diagonalizable. In other words, there is a change of basis so that $A$ becomes a diagonal matrix with respect to this new basis. Now, if you think about what multiplication does to a basis vector in this diagonal basis, you can see that the basis vectors are eigenvectors. Thus, we have 2n different eigenvectors. The 1-dimensional subspace that is generated by a given eigenvector is an invariant subspace and so we have 2n invariant subspaces.
But wait! If you take the subspace spanned by any combination of eigenvectors, then you can show that this will also be invariant. So the total number of invariant subspaces is at least $2^{2n}$.
But wait! Lets consider the two eigenvectors $b_1, b_2$ with eigenvalue 1 in the diagonal basis. Then $A(b_1)=b_1, A(b_2)=b_2$. But also, for any $r, s \in \mathbb{R}$, $A(rb_1+sb_2)=A(rb_1)+A(sb_2)=rb_1+sb_2$. Thus, any linear combination of $b_1, b_2$ is an eigenvector with eigenvalue $1$. Then the subspace generated by any eigenvector gives an invariant subspace. We can easily pick infinitely many of these so that no two give the same subspace. For instance $b_1+mb_2$ for $m \in \mathbb{N}$ so that we have infinitely many invariant subspaces.
Hopefully this gives some intuition as to the answer to the problem and helps you think about why the other answer choices were given.
A: Hints: 


*

*Consider the case $n = 1$. 

*What is $A$ in this case? 

*What are its invariant subspaces?

*How may invariant subspaces are there?

