Show that

{T∈L(R5,R4):dim⁡null T>2}\{ T\in \mathcal L(\mathbf R^5, \mathbf R^4) : \dim \text{null }T > 2 \}

Is not a subspace of L(R5,R4)\mathcal L(\mathbf R^5, \mathbf R^4).

It isn't closed under addition, consider

T(x1,…,x5)=(x1,x2,0,0,0)S(x1,…,x5)=(0,0,x3,x4,0)\begin{aligned} T(x_1,\dots,x_5) &= (x_1,x_2,0,0,0) \\ S(x_1,\dots,x_5) &= (0,0,x_3,x_4,0) \end{aligned}

Both S,TS,T are in the subspace since their nullspaces are three dimensional, but

(S+T)=(x1,x2,x3,x4,0)(S + T) = (x_1,x_2,x_3,x_4,0)

Is not as it's nullspace is only one dimensional.