Solution 3d 2

Suppose $V$ is finite-dimensional and $\dim V > 1$ . Prove that the set of noninvertible operators on $V$ is not a subspace of $\mathcal L(V)$ .

It isn't closed under addition, the sum of noninvertible operators can be invertible. For an example let $v_1,v_2$ be independent vectors in $V$ and define

T(c_1v_1 + c_2v_2) = c_1v_1 \quad\text{and}\quad S(c_1v_1 + c_2v_2) = c_2v_2

$T$ and $S$ are clearly not invertible, but $T + S$ is the identity, which is clearly invertible!

Another way to view this is in terms of matrices

\underbrace{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}}_{T} + \underbrace{\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}}_{S} = \underbrace{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}_{I}