1. **Problem Statement:** Given matrix $$ A = \begin{bmatrix}1 & 2 & 3 & 4 \\ 2 & 2 & 1 & 3 \\ 4 & 6 & 7 & 11 \end{bmatrix} $$ Find bases and dimensions for these four fundamental subspaces: - Null space of $$A$$, $$N(A)$$ - Row space of $$A$$, $$R(A)$$ - Null space of transpose $$A^T$$, $$N(A^T)$$ - Row space of transpose $$A^T$$, $$R(A^T)$$ (which is the column space of $$A$$) --- 2. **Concepts Explanation:** - **Null space \(N(A)\)** is the set of all vectors $$x$$ with $$Ax=0$$. - **Row space \(R(A)\)** is the span of the rows of $$A$$. - **Column space \(R(A^T)\) is the span of columns of $$A$$ (also called image or range). - **Null space of transpose \(N(A^T)\)** consists of vectors orthogonal to all rows of $$A$$. The dimensions are called the rank (for row/column spaces) and nullity (for null spaces), and satisfy the Rank-Nullity theorem: $$\text{rank}(A) + \text{nullity}(A) = \text{number of columns of } A = 4$$ --- 3. **Step 1: Find the Reduced Row Echelon Form (RREF) of $$A$$.** We perform elementary row operations to simplify $$A$$: Start: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 2 & 1 & 3 \\ 4 & 6 & 7 & 11 \end{bmatrix}$$ - Replace row 2 by $$R_2 - 2R_1$$: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -2 & -5 & -5 \\ 4 & 6 & 7 & 11 \end{bmatrix}$$ - Replace row 3 by $$R_3 - 4R_1$$: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -2 & -5 & -5 \\ 0 & -2 & -5 & -5 \end{bmatrix}$$ - Replace row 3 by $$R_3 - R_2$$ to eliminate duplicate: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ - Multiply row 2 by $$-\frac{1}{2}$$: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ - Eliminate above second row pivot: replace row 1 by $$R_1 - 2R_2$$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 0 & 1 & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ This is the RREF. --- 4. **Step 2: Determine rank and free variables.** - Rank = number of pivot columns = 2 (columns 1 and 2) - Number of columns = 4 - Nullity = 4 - 2 = 2 So, the null space dimension is 2. --- 5. **Step 3: Find basis for Null space $$N(A)$$.** The system $$Ax=0$$ in RREF: $$\begin{cases} x_1 - 2x_3 - x_4 = 0 \\ x_2 + \frac{5}{2} x_3 + \frac{5}{2} x_4 = 0 \end{cases}$$ Let $$x_3 = s$$ and $$x_4 = t$$ be free variables. Then, $$x_1 = 2s + t$$ $$x_2 = -\frac{5}{2}s - \frac{5}{2}t$$ Null space vector form: $$x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = s \begin{bmatrix} 2 \\ -\frac{5}{2} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ -\frac{5}{2} \\ 0 \\ 1 \end{bmatrix}$$ Basis for $$N(A)$$ is the set of these two vectors. --- 6. **Step 4: Find basis for Row space $$R(A)$$.** The row space basis corresponds to the nonzero rows of RREF: $$\left\{ \begin{bmatrix}1 & 0 & -2 & -1\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{5}{2} & \frac{5}{2} \end{bmatrix} \right\}$$ --- 7. **Step 5: Find basis for Null space of $$A^T$$, $$N(A^T)$$.** Vectors in $$N(A^T)$$ satisfy $$A^T y = 0$$. Since $$\text{dim}(N(A^T)) = m - \text{rank}(A)$$ where $$m$$ is row count = 3, Dimension of $$N(A^T) = 3 - 2 = 1$$. Find vector $$y = \begin{bmatrix}y_1 & y_2 & y_3\end{bmatrix}^T$$ with $$A^T y = 0$$: $$ \begin{bmatrix}1 & 2 & 4 \\ 2 & 2 & 6 \\ 3 & 1 & 7 \\ 4 & 3 & 11\end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = 0 $$ Equations: $$\begin{cases}1y_1 + 2y_2 + 4y_3 = 0 \\ 2y_1 + 2y_2 + 6y_3 = 0 \\ 3y_1 + y_2 + 7y_3 = 0 \\ 4y_1 + 3y_2 + 11y_3 = 0 \end{cases}$$ Only 3 variables but 4 equations; last is dependent. Row reduce coefficient matrix: Start with: $$\begin{bmatrix}1 & 2 & 4 \\ 2 & 2 & 6 \\ 3 & 1 & 7 \end{bmatrix}$$ - $$R_2 - 2R_1 \to R_2: 0, -2, -2$$ - $$R_3 - 3R_1 \to R_3: 0, -5, -5$$ - $$R_3 - \frac{5}{2}R_2 \to R_3: 0, 0, 0$$ From $$R_2$$: $$-2 y_2 - 2 y_3 = 0 \Rightarrow y_2 = -y_3$$ From $$R_1$$: $$y_1 + 2 y_2 + 4 y_3 = 0 \Rightarrow y_1 + 2(-y_3) + 4 y_3 = 0 \Rightarrow y_1 + 2 y_3 = 0 \Rightarrow y_1 = -2 y_3$$ Let $$y_3 = t$$ free. So, $$ y = t \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} $$ Basis for $$N(A^T)$$ is $$\{ \begin{bmatrix} -2, -1, 1 \end{bmatrix}^T \}$$. --- 8. **Step 6: Basis for $$R(A^T)$$ (Column space of $$A$$).** Column space basis is columns of $$A$$ corresponding to pivots in RREF (columns 1 and 2): $$ \left\{ \begin{bmatrix}1 \\ 2 \\ 4 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ 6 \end{bmatrix} \right\} $$ --- 9. **Summary of dimensions and bases:** - $$\dim N(A) = 2$$; basis: $$ \left\{ \begin{bmatrix}2 \\ -\frac{5}{2} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}1 \\ -\frac{5}{2} \\ 0 \\ 1 \end{bmatrix} \right\} $$ - $$\dim R(A) = 2$$; basis rows of RREF: $$ \left\{ \begin{bmatrix}1 & 0 & -2 & -1\end{bmatrix}, \begin{bmatrix}0 & 1 & \frac{5}{2} & \frac{5}{2}\end{bmatrix} \right\} $$ - $$\dim N(A^T) = 1$$; basis: $$ \left\{ \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} \right\}$$ - $$\dim R(A^T) = 2$$; basis columns 1 and 2 of $$A$$: $$ \left\{ \begin{bmatrix}1 \\ 2 \\ 4 \end{bmatrix} , \begin{bmatrix}2 \\ 2 \\ 6 \end{bmatrix} \right\} $$ This completes the assignment with detailed explanation and stepwise row reduction.