Question 3 (18 marks)
a) For matrix $H=\left[\begin{array}{rrrrr}2 & -2 & 4 & -2 & 0 \ 3 & 0 & 6 & -1 & 2 \ 1 & 3 & 0 & 0 & 2\end{array} ight]$, find its rank, column space, and null space. (14 marks)
b) Let $u=\left[\begin{array}{c}-2 \ -1 \ 0\end{array} ight], v_{1}=\left[\begin{array}{l}2 \ 3 \ 1\end{array} ight], v_{2}=\left[\begin{array}{c}-2 \ 0 \ 3\end{array} ight]$, and $v_{3}=\left[\begin{array}{l}4 \ 6 \ 0\end{array} ight]$. Show that $V=\left\{v_{1}, v_{2}, v_{3} ight\}$ is a basis for $R^{3}$, and then find the coordinate vector of $u$ relative to the basis $V$. ( 4 marks)

Question

Question 3 (18 marks)
a) For matrix $H=\left[\begin{array}{rrrrr}2 & -2 & 4 & -2 & 0 \ 3 & 0 & 6 & -1 & 2 \ 1 & 3 & 0 & 0 & 2\end{array}ight]$, find its rank, column space, and null space. (14 marks)
b) Let $u=\left[\begin{array}{c}-2 \ -1 \ 0\end{array}ight], v_{1}=\left[\begin{array}{l}2 \ 3 \ 1\end{array}ight], v_{2}=\left[\begin{array}{c}-2 \ 0 \ 3\end{array}ight]$, and $v_{3}=\left[\begin{array}{l}4 \ 6 \ 0\end{array}ight]$. Show that $V=\left\{v_{1}, v_{2}, v_{3}ight\}$ is a basis for $R^{3}$, and then find the coordinate vector of $u$ relative to the basis $V$. ( 4 marks)

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the rank of matrix $H$, we first need to reduce it to its row echelon form (REF). ‖ ‖ step 2 ⋮ The given matrix $H$ is: $$ H = \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 3 & 0 & 6 & -1 & 2 \ 1 & 3 & 0 & 0 & 2 \end{bmatrix} $$ We perform row operations to convert it to REF. ‖ ‖ step 3 ⋮ Subtract $\frac{3}{2}$ times the first row from the second row: $$ R_2 \leftarrow R_2 - \frac{3}{2}R_1 $$ $$ \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 0 & 3 & 0 & 2 & 2 \ 1 & 3 & 0 & 0 & 2 \end{bmatrix} $$ ‖ ‖ step 4 ⋮ Subtract $\frac{1}{2}$ times the first row from the third row: $$ R_3 \leftarrow R_3 - \frac{1}{2}R_1 $$ $$ \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 0 & 3 & 0 & 2 & 2 \ 0 & 4 & -2 & 1 & 2 \end{bmatrix} $$ ‖ ‖ step 5 ⋮ Subtract $\frac{4}{3}$ times the second row from the third row: $$ R_3 \leftarrow R_3 - \frac{4}{3}R_2 $$ $$ \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 0 & 3 & 0 & 2 & 2 \ 0 & 0 & -2 & -\frac{5}{3} & -\frac{2}{3} \end{bmatrix} $$ ‖ ‖ step 6 ⋮ The matrix is now in row echelon form. The non-zero rows are: $$ \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 0 & 3 & 0 & 2 & 2 \ 0 & 0 & -2 & -\frac{5}{3} & -\frac{2}{3} \end{bmatrix} $$ Thus, the rank of $H$ is 3. ‖ ‖ step 7 ⋮ The column space of $H$ is the span of the pivot columns in the original matrix: $$ \text{Column space} = \text{span} \left\{ \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}, \begin{bmatrix} -2 \ 0 \ 3 \end{bmatrix}, \begin{bmatrix} 4 \ 6 \ 0 \end{bmatrix} \right\} $$ ‖ ‖ step 8 ⋮ To find the null space, we solve $H \mathbf{x} = \mathbf{0}$: $$ \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \ 3 & 0 & 6 & -1 & 2 \ 1 & 3 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \ x_5 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} $$ From the REF, we get: $$ x_3 = \frac{5}{6}x_4 + \frac{1}{3}x_5 $$ $$ x_2 = -\frac{2}{3}x_4 - \frac{2}{3}x_5 $$ $$ x_1 = \frac{1}{3}x_4 + \frac{1}{3}x_5 $$ Thus, the null space is: $$ \text{Null space} = \text{span} \left\{ \begin{bmatrix} \frac{1}{3} \ -\frac{2}{3} \ \frac{5}{6} \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} \frac{1}{3} \ -\frac{2}{3} \ \frac{1}{3} \ 0 \ 1 \end{bmatrix} \right\} $$ ‖ ∻Answer∻ ⚹ Rank: 3, Column space: span of $\left\{ \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}, \begin{bmatrix} -2 \ 0 \ 3 \end{bmatrix}, \begin{bmatrix} 4 \ 6 \ 0 \end{bmatrix} \right\}$, Null space: span of $\left\{ \begin{bmatrix} \frac{1}{3} \ -\frac{2}{3} \ \frac{5}{6} \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} \frac{1}{3} \ -\frac{2}{3} \ \frac{1}{3} \ 0 \ 1 \end{bmatrix} \right\}$ ⚹ ∻Key Concept∻ ⚹Rank, Column Space, and Null Space of a Matrix⚹ ∻Explanation∻ ⚹The rank of a matrix is the number of linearly independent rows or columns. The column space is the span of the pivot columns, and the null space is the set of solutions to the homogeneous equation $H \mathbf{x} = \mathbf{0}$.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To show that $V = \{v_1, v_2, v_3\}$ is a basis for $\mathbb{R}^3$, we need to show that $v_1, v_2, v_3$ are linearly independent and span $\mathbb{R}^3$. ‖ ‖ step 2 ⋮ The vectors are: $$ v_1 = \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}, v_2 = \begin{bmatrix} -2 \ 0 \ 3 \end{bmatrix}, v_3 = \begin{bmatrix} 4 \ 6 \ 0 \end{bmatrix} $$ We form the matrix $A$ with these vectors as columns: $$ A = \begin{bmatrix} 2 & -2 & 4 \ 3 & 0 & 6 \ 1 & 3 & 0 \end{bmatrix} $$ ‖ ‖ step 3 ⋮ We find the determinant of $A$ to check for linear independence: $$ \det(A) = \begin{vmatrix} 2 & -2 & 4 \ 3 & 0 & 6 \ 1 & 3 & 0 \end{vmatrix} $$ Expanding along the first row: $$ \det(A) = 2 \begin{vmatrix} 0 & 6 \ 3 & 0 \end{vmatrix} - (-2) \begin{vmatrix} 3 & 6 \ 1 & 0 \end{vmatrix} + 4 \begin{vmatrix} 3 & 0 \ 1 & 3 \end{vmatrix} $$ $$ = 2(0 - 18) + 2(0 - 6) + 4(9 - 0) $$ $$ = -36 - 12 + 36 = -12 $$ Since $\det(A) \neq 0$, the vectors are linearly independent. ‖ ‖ step 4 ⋮ Since $v_1, v_2, v_3$ are linearly independent and there are 3 vectors in $\mathbb{R}^3$, they span $\mathbb{R}^3$. Thus, $V$ is a basis for $\mathbb{R}^3$. ‖ ‖ step 5 ⋮ To find the coordinate vector of $u$ relative to the basis $V$, we express $u$ as a linear combination of $v_1, v_2, v_3$: $$ u = c_1 v_1 + c_2 v_2 + c_3 v_3 $$ $$ \begin{bmatrix} -2 \ -1 \ 0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} -2 \ 0 \ 3 \end{bmatrix} + c_3 \begin{bmatrix} 4 \ 6 \ 0 \end{bmatrix} $$ ‖ ‖ step 6 ⋮ This gives us the system of equations: $$ 2c_1 - 2c_2 + 4c_3 = -2 $$ $$ 3c_1 + 6c_3 = -1 $$ $$ c_1 + 3c_2 = 0 $$ Solving this system, we get: $$ c_1 = -\frac{1}{3}, c_2 = -\frac{1}{9}, c_3 = -\frac{1}{18} $$ Thus, the coordinate vector of $u$ relative to the basis $V$ is: $$ [u]_V = \begin{bmatrix} -\frac{1}{3} \ -\frac{1}{9} \ -\frac{1}{18} \end{bmatrix} $$ ‖ ∻Answer∻ ⚹ $V$ is a basis for $\mathbb{R}^3$, and the coordinate vector of $u$ relative to $V$ is $\begin{bmatrix} -\frac{1}{3} \ -\frac{1}{9} \ -\frac{1}{18} \end{bmatrix}$ ⚹ ∻Key Concept∻ ⚹Basis and Coordinate Vectors⚹ ∻Explanation∻ ⚹A set of vectors forms a basis for a vector space if they are linearly independent and span the space. The coordinate vector of a vector relative to a basis is the unique combination of basis vectors that equals the vector.⚹◊What are the steps to determine the rank of a matrix using elementary row operations?⍭ Generate me a similar question◊

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