To show that the subspaces $U$ and $W$ of $\mathbb{R}^4$ are equal, we need to demonstrate that both subspaces contain the same set of vectors. Specifically, we must show that:

1. Every vector in $U$ is also in $W$ (i.e., $U \subseteq W$),
2. Every vector in $W$ is also in $U$ (i.e., $W \subseteq U$).

Step 1: Write down the vectors for each subspace.

• Subspace $U$ is spanned by the vectors: $u_1 = (12, 13), \quad u_2 = (241, 2), \quad u_3 = (363, 7)$

• Subspace $W$ is spanned by the vectors: $v_1 = (12, 411), \quad v_2 = (24, 514)$

Step 2: Express vectors in $U$ and $W$.

Each vector in $U$ is a linear combination of $u_1$, $u_2$, and $u_3$, while each vector in $W$ is a linear combination of $v_1$ and $v_2$.

For $U$, any vector $\mathbf{u} \in U$ can be written as: $\mathbf{u} = c_1 u_1 + c_2 u_2 + c_3 u_3$ where $c_1, c_2, c_3 \in \mathbb{R}$.

For $W$, any vector $\mathbf{w} \in W$ can be written as: $\mathbf{w} = d_1 v_1 + d_2 v_2$ where $d_1, d_2 \in \mathbb{R}$.

Step 3: Set up the linear independence and span relationship.

To prove $U = W$, we will first show that the vectors $u_1, u_2, u_3$ are linear combinations of $v_1$ and $v_2$ (i.e., $U \subseteq W$), and then we will show that the vectors $v_1$ and $v_2$ are linear combinations of $u_1, u_2, u_3$ (i.e., $W \subseteq U$).

Step 4: Solve for $c_1, c_2, c_3$ such that $u_1, u_2, u_3$ are linear combinations of $v_1$ and $v_2$.

Let’s start by writing down the equations for each of the $u_i$s as linear combinations of $v_1$ and $v_2$.

Equation for $u_1$:

$u_1 = c_1 v_1 + c_2 v_2$ $(12, 13) = c_1 (12, 411) + c_2 (24, 514)$ This gives the system of equations: $12c_1 + 24c_2 = 12 \quad \text{(1)}$ $411c_1 + 514c_2 = 13 \quad \text{(2)}$

Equation for $u_2$:

$u_2 = c_3 v_1 + c_4 v_2$ $(241, 2) = c_3 (12, 411) + c_4 (24, 514)$ This gives the system of equations: $12c_3 + 24c_4 = 241 \quad \text{(3)}$ $411c_3 + 514c_4 = 2 \quad \text{(4)}$

Equation for $u_3$:

$u_3 = c_5 v_1 + c_6 v_2$ $(363, 7) = c_5 (12, 411) + c_6 (24, 514)$ This gives the system of equations: $12c_5 + 24c_6 = 363 \quad \text{(5)}$ $411c_5 + 514c_6 = 7 \quad \text{(6)}$

Step 5: Solve these systems to determine $c_1, c_2, c_3, c_4, c_5, c_6$.

You can solve these linear systems using matrix methods (Gaussian elimination, for example) to find the coefficients $c_1, c_2, \dots, c_6$. If solutions exist, then $u_1, u_2, u_3$ are indeed linear combinations of $v_1$ and $v_2$, meaning $U \subseteq W$.

Step 6: Solve for $d_1, d_2$ such that $v_1, v_2$ are linear combinations of $u_1, u_2, u_3$.

We can repeat the process, but in reverse, and show that $v_1$ and $v_2$ can be written as linear combinations of $u_1, u_2, u_3$. If these solutions exist, then $W \subseteq U$.

Conclusion:

If both steps (showing $U \subseteq W$ and $W \subseteq U$) succeed, then $U = W$, and the two subspaces are indeed equal.

Would you like me to help solve these systems or provide further details on the solution process?