1. **Problem Statement:** Given subspaces $$V_1 = \left\{\begin{bmatrix} a & 0 \\ c & d \end{bmatrix} : a,c,d \in \mathbb{R} \right\}$$ and $$V_2 = \left\{\begin{bmatrix} u & v \\ 0 & r \end{bmatrix} : u,v,r \in \mathbb{R} \right\}$$ show that $$M_2(\mathbb{R}) = V_1 + V_2$$ and determine if $$M_2(\mathbb{R}) = V_1 \oplus V_2$$ 2. **Recall:** - The sum of subspaces $V_1 + V_2$ is the set of all sums $v_1 + v_2$ where $v_1 \in V_1$ and $v_2 \in V_2$. - The direct sum $V_1 \oplus V_2$ means $V_1 + V_2 = M_2(\mathbb{R})$ and $V_1 \cap V_2 = \{0\}$. 3. **Show $M_2(\mathbb{R}) = V_1 + V_2$: ** Take any matrix $$A = \begin{bmatrix} x & y \\ z & w \end{bmatrix} \in M_2(\mathbb{R})$$ We want to write $A = A_1 + A_2$ with $$A_1 = \begin{bmatrix} a & 0 \\ c & d \end{bmatrix} \in V_1, \quad A_2 = \begin{bmatrix} u & v \\ 0 & r \end{bmatrix} \in V_2$$ Equate entries: $$\begin{cases} a + u = x \\ c + 0 = z \\ 0 + v = y \\ d + r = w \end{cases}$$ Choose $$c = z, \quad v = y$$ Then $$a + u = x, \quad d + r = w$$ Since $a,u,d,r$ are arbitrary real numbers, pick for example $$a = 0, \quad u = x, \quad d = 0, \quad r = w$$ Thus, $$A_1 = \begin{bmatrix} 0 & 0 \\ z & 0 \end{bmatrix}, \quad A_2 = \begin{bmatrix} x & y \\ 0 & w \end{bmatrix}$$ which are in $V_1$ and $V_2$ respectively. Therefore, $$M_2(\mathbb{R}) = V_1 + V_2$$ 4. **Check if the sum is direct:** We need to check if $$V_1 \cap V_2 = \{0\}$$ Suppose $$X = \begin{bmatrix} a & 0 \\ c & d \end{bmatrix} = \begin{bmatrix} u & v \\ 0 & r \end{bmatrix}$$ with $X \in V_1 \cap V_2$. Equate entries: $$a = u, \quad 0 = v, \quad c = 0, \quad d = r$$ From $0 = v$ and $c=0$, the matrix is $$X = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$$ which is diagonal. Since $a,d$ can be any real numbers, the intersection contains all diagonal matrices. Hence, $$V_1 \cap V_2 \neq \{0\}$$ 5. **Conclusion:** - $M_2(\mathbb{R}) = V_1 + V_2$ - But $M_2(\mathbb{R}) \neq V_1 \oplus V_2$ because the intersection is not trivial. **Final answer:** $$M_2(\mathbb{R}) = V_1 + V_2$$ but $$M_2(\mathbb{R}) \neq V_1 \oplus V_2$$ because $$V_1 \cap V_2 = \left\{ \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} : a,d \in \mathbb{R} \right\} \neq \{0\}$$