1. **State the problem:** We are given vectors $v_1, v_2, v_3, v_4, v_5$ in $\mathbb{R}^4$ and need to find: a. A basis for the subspace $W = \text{span}\{v_1, v_2, v_3, v_4, v_5\}$. b. The dimension of $W$. 2. **Write the vectors explicitly:** $$v_1 = \begin{pmatrix}0 \\ -5 \\ -2 \\ 1 \\ -3\end{pmatrix},\quad v_2 = \begin{pmatrix}-2 \\ 1 \\ 5 \\ -4\end{pmatrix},\quad v_3 = \begin{pmatrix}-3 \\ -3 \\ 3 \\ 1\end{pmatrix},\quad v_4 = \begin{pmatrix}-3 \\ 2 \\ 1 \\ -3\end{pmatrix},\quad v_5 = \begin{pmatrix}-5 \\ -3 \\ 5 \\ ?\end{pmatrix}$$ Note: $v_1$ has 5 components, but $v_2, v_3, v_4, v_5$ have 4 components. Since the problem states $W \subseteq \mathbb{R}^4$, we consider only the first 4 components of $v_1$ and $v_5$ (ignoring the 5th component). So: $$v_1 = \begin{pmatrix}0 \\ -5 \\ -2 \\ 1\end{pmatrix},\quad v_5 = \begin{pmatrix}-5 \\ -3 \\ 5 \\ ?\end{pmatrix}$$ Assuming $v_5$ has 4 components as $(-5, -3, 5, ?)$, but the 4th component is missing. Since the problem is about $\mathbb{R}^4$, we assume $v_5 = (-5, -3, 5, 0)$ for consistency. 3. **Form the matrix with vectors as columns:** $$A = \begin{pmatrix}0 & -2 & -3 & -3 & -5 \\ -5 & 1 & -3 & 2 & -3 \\ -2 & 5 & 3 & 1 & 5 \\ 1 & -4 & 1 & -3 & 0\end{pmatrix}$$ 4. **Find the basis by determining the linear independence of columns:** Perform Gaussian elimination on $A$ to find pivot columns. 5. **Row reduce $A$:** Start with $$\begin{pmatrix}0 & -2 & -3 & -3 & -5 \\ -5 & 1 & -3 & 2 & -3 \\ -2 & 5 & 3 & 1 & 5 \\ 1 & -4 & 1 & -3 & 0\end{pmatrix}$$ Swap row 1 and row 4 to get a leading 1: $$\begin{pmatrix}1 & -4 & 1 & -3 & 0 \\ -5 & 1 & -3 & 2 & -3 \\ -2 & 5 & 3 & 1 & 5 \\ 0 & -2 & -3 & -3 & -5\end{pmatrix}$$ Eliminate below pivot in column 1: Row 2: $R_2 + 5R_1 \to R_2$ $$-5 + 5(1) = 0$$ $$1 + 5(-4) = 1 - 20 = -19$$ $$-3 + 5(1) = -3 + 5 = 2$$ $$2 + 5(-3) = 2 - 15 = -13$$ $$-3 + 5(0) = -3$$ Row 3: $R_3 + 2R_1 \to R_3$ $$-2 + 2(1) = 0$$ $$5 + 2(-4) = 5 - 8 = -3$$ $$3 + 2(1) = 3 + 2 = 5$$ $$1 + 2(-3) = 1 - 6 = -5$$ $$5 + 2(0) = 5$$ Row 4 stays the same. Matrix now: $$\begin{pmatrix}1 & -4 & 1 & -3 & 0 \\ 0 & -19 & 2 & -13 & -3 \\ 0 & -3 & 5 & -5 & 5 \\ 0 & -2 & -3 & -3 & -5\end{pmatrix}$$ 6. **Pivot in column 2:** Make pivot positive by multiplying row 2 by $-1$: $$R_2 \to -R_2$$ $$\begin{pmatrix}1 & -4 & 1 & -3 & 0 \\ 0 & 19 & -2 & 13 & 3 \\ 0 & -3 & 5 & -5 & 5 \\ 0 & -2 & -3 & -3 & -5\end{pmatrix}$$ Eliminate below pivot in column 2: Row 3: $R_3 + \frac{3}{19} R_2 \to R_3$ Row 4: $R_4 + \frac{2}{19} R_2 \to R_4$ Calculate: Row 3: $$-3 + 3 = 0$$ $$5 + \frac{3}{19}(-2) = 5 - \frac{6}{19} = \frac{95}{19} - \frac{6}{19} = \frac{89}{19}$$ $$-5 + \frac{3}{19}(13) = -5 + \frac{39}{19} = -\frac{95}{19} + \frac{39}{19} = -\frac{56}{19}$$ $$5 + \frac{3}{19}(3) = 5 + \frac{9}{19} = \frac{95}{19} + \frac{9}{19} = \frac{104}{19}$$ Row 4: $$-2 + 2 = 0$$ $$-3 + \frac{2}{19}(-2) = -3 - \frac{4}{19} = -\frac{57}{19} - \frac{4}{19} = -\frac{61}{19}$$ $$-3 + \frac{2}{19}(13) = -3 + \frac{26}{19} = -\frac{57}{19} + \frac{26}{19} = -\frac{31}{19}$$ $$-5 + \frac{2}{19}(3) = -5 + \frac{6}{19} = -\frac{95}{19} + \frac{6}{19} = -\frac{89}{19}$$ Matrix now: $$\begin{pmatrix}1 & -4 & 1 & -3 & 0 \\ 0 & 19 & -2 & 13 & 3 \\ 0 & 0 & \frac{89}{19} & -\frac{56}{19} & \frac{104}{19} \\ 0 & 0 & -\frac{61}{19} & -\frac{31}{19} & -\frac{89}{19}\end{pmatrix}$$ 7. **Pivot in column 3:** Multiply row 3 by $\frac{19}{89}$ to get pivot 1: $$R_3 \to \frac{19}{89} R_3$$ $$\begin{pmatrix}0 & 0 & 1 & -\frac{56}{89} & \frac{104}{89}\end{pmatrix}$$ Eliminate below pivot in column 3: Row 4: $R_4 + \frac{61}{19} R_3 \to R_4$ Calculate: $$-\frac{61}{19} + \frac{61}{19} (1) = 0$$ $$-\frac{31}{19} + \frac{61}{19} \left(-\frac{56}{89}\right) = -\frac{31}{19} - \frac{61 \times 56}{19 \times 89} = -\frac{31}{19} - \frac{3416}{1691}$$ Convert $-\frac{31}{19}$ to denominator 1691: $$-\frac{31}{19} = -\frac{31 \times 89}{19 \times 89} = -\frac{2759}{1691}$$ Sum: $$-\frac{2759}{1691} - \frac{3416}{1691} = -\frac{6175}{1691}$$ Similarly for last element: $$-\frac{89}{19} + \frac{61}{19} \times \frac{104}{89} = -\frac{89}{19} + \frac{6344}{1691}$$ Convert $-\frac{89}{19}$: $$-\frac{89}{19} = -\frac{89 \times 89}{19 \times 89} = -\frac{7921}{1691}$$ Sum: $$-\frac{7921}{1691} + \frac{6344}{1691} = -\frac{1577}{1691}$$ Row 4 now: $$\begin{pmatrix}0 & 0 & 0 & -\frac{6175}{1691} & -\frac{1577}{1691}\end{pmatrix}$$ 8. **Pivot in column 4:** Multiply row 4 by $-\frac{1691}{6175}$ to get pivot 1: $$R_4 \to -\frac{1691}{6175} R_4 = \begin{pmatrix}0 & 0 & 0 & 1 & \frac{1577}{6175}\end{pmatrix}$$ 9. **Summary of pivots:** Pivots in columns 1, 2, 3, and 4. 10. **Conclusion:** All four vectors $v_1, v_2, v_3, v_4$ are linearly independent and form a basis for $W$. Vector $v_5$ is a linear combination of these (since no pivot in column 5). **Basis for $W$:** $$\{v_1, v_2, v_3, v_4\}$$ **Dimension of $W$:** $$\dim W = 4$$