Exercise 3.B.4

Question

Show that
$$
\{T \in \mathcal{L}(R^5,R^4): 	ext{dim null } T > 2\}
$$
is not a subspace of $\mathcal{L}(R^5,R^4)$

Arden Shi · Accepted Answer

Let $T_1, T_2 \in \mathcal{L}(R^5,R^4)$ and dim null $T_1$ and dim null $T_2 > 2$ where
$$
T_1(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, 0)
$$
$$
T_2(x_1, x_2, x_3, x_4, x_5) = (0, x_2, 0, 0)
$$
However, dim null $(T_1 + T_2) = 2$, so it is not a subspace.

Exercise 3.B.4

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Exercise 3.B.4

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