$\textbf{Exercise 27.}$ Suppose that $V$ is finite-dimensional and $k \in \{1, \ldots, \dim V - 1\}$. Suppose $T \in \mathcal{L}(V)$ is such that every subspace of $V$ of dimension $k$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator. $\textbf{Solution 27.}$ Suppose $T$ is not a scalar multiple of the identity operator. By Exercise 26, there exists $v_1 \in V \setminus \{0\}$ such that $v_1$ is not an eigenvector of $T$. Thus $v_1, Tv_1$ is linearly independent. Extend $v_1, Tv_1$ to a basis $v_1, Tv_1, v_2, \ldots, v_{\dim V - 1}$ of $V$. Let $\begin{equation*} U = \operatorname{span}(v_1, \ldots, v_k). \end{equation*}$ Then $U$ is a subspace of $V$ and $\dim U = k$. However, $U$ is not invariant under $T$ because $v_1 \in U$ but $Tv_1 \notin U$. This contradiction to our hypothesis about $T$ shows that our assumption that $T$ is not a scalar multiple of the identity was false. ---  --- Commentary: $\textbf{Exercise 27.}$ Suppose that $V$ is finite-dimensional and $k \in \{1, \ldots, \dim V - 1\}$. Suppose $T \in \mathcal{L}(V)$ is such that every subspace of $V$ of dimension $k$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator. $\textbf{Solution 27.}$ Suppose $T$ is not a scalar multiple of the identity operator. By Exercise 26, there exists $v_1 \in V \setminus \{0\}$ such that $v_1$ is not an eigenvector of $T$. Thus $v_1, Tv_1$ is linearly independent. Extend $v_1, Tv_1$ to a basis $v_1, Tv_1, v_2, \ldots, v_{\dim V - 1}$ of $V$. Let $\begin{equation*} U = \operatorname{span}(v_1, \ldots, v_k). \end{equation*}$ Then $U$ is a subspace of $V$ and $\dim U = k$. However, $U$ is not invariant under $T$ because $v_1 \in U$ but $Tv_1 \notin U$. This contradiction to our hypothesis about $T$ shows that our assumption that $T$ is not a scalar multiple of the identity was false. $\textit{Commentary:}$ This exercise extends the previous one to a more general setting. If every subspace of a specific dimension $k$ is invariant under $T$, then $T$ must be a scalar multiple of the identity. The proof is by contradiction. If $T$ is not a scalar multiple of the identity, then by the previous exercise, there must be a vector $v_1$ that is not an eigenvector. Then $v_1$ and $Tv_1$ are linearly independent, and can be extended to a basis of $V$. The subspace spanned by the first $k$ basis vectors is then a $k$-dimensional subspace that is not invariant under $T$, contradicting the hypothesis. This result has applications in the study of group representations and invariant theory. It shows that the only linear operators that preserve all subspaces of a given dimension are the scalar multiples of the identity. $\textit{Example:}$ Let $V = \mathbb{R}^3$ and define $T: \mathbb{R}^3 \to \mathbb{R}^3$ by $T(x, y, z) = (x, y, 0)$. Let $k = 2$. Then every 2-dimensional subspace of $\mathbb{R}^3$ is invariant under $T$. For example, the $xy$-plane ${(x, y, 0) : x, y \in \mathbb{R}}$ is invariant because if $(x, y, 0)$ is in the $xy$-plane, then $T(x, y, 0) = (x, y, 0)$ is also in the $xy$-plane. Similarly, any other plane that passes through the origin is invariant under $T$. However, $T$ is not a scalar multiple of the identity because $T(0, 0, 1) = (0, 0, 0) \neq \lambda(0, 0, 1)$ for any scalar $\lambda$. Now let's define $S: \mathbb{R}^3 \to \mathbb{R}^3$ by $S(x, y, z) = (2x, 2y, 2z)$. Then every subspace of $\mathbb{R}^3$ of any dimension is invariant under $S$, because for any vector $(x, y, z)$, we have $S(x, y, z) = 2(x, y, z)$, which is a scalar multiple of $(x, y, z)$. So $S$ is a scalar multiple of the identity with $\lambda = 2$. ---