Question

Math

Posted 3 months ago

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

Posted 3 months ago

Solution by Steps

step 1

To find the rank of the matrix $H = \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 3 & 0 & 6 & -1 & 2 \\ 1 & 3 & 0 & 0 & 2 \end{bmatrix}$, we perform row reduction to echelon form

step 2

After row reduction, the matrix $H$ is transformed into echelon form, revealing that there are 3 non-zero rows. Therefore, the rank of $H$ is 3

step 3

To find the column space of $H$, we identify the pivot columns in the row-reduced form of $H$. The pivot columns correspond to the original columns of $H$

step 4

The column space of $H$ is spanned by the vectors $\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix}$. Therefore, the column space is $R^3$

step 5

To find the null space of $H$, we solve the equation $H \mathbf{x} = \mathbf{0}$. This involves finding the solutions to the homogeneous system of linear equations represented by $H$

step 6

The null space of $H$ is given by the set of all vectors $\mathbf{x}$ such that $H \mathbf{x} = \mathbf{0}$. The null space can be expressed as $\{(12x, -4x - 2y, -5x - y, 6x, 3y) : x, y \in \mathbb{R}\}$

step 7

To show that $V = \left\{ \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix} \right\}$ is a basis for $R^3$, we need to show that these vectors are linearly independent and span $R^3$

step 8

We form the matrix $A = \begin{bmatrix} 2 & -2 & 4 \\ 3 & 0 & 6 \\ 1 & 3 & 0 \end{bmatrix}$ and perform row reduction. The matrix $A$ reduces to the identity matrix, indicating that the vectors are linearly independent and span $R^3$. Therefore, $V$ is a basis for $R^3$

step 9

To find the coordinate vector of $u = \begin{bmatrix} -2 \\ -1 \\ 0 \end{bmatrix}$ relative to the basis $V$, we express $u$ as a linear combination of the basis vectors: $u = c_1 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix} + c_3 \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix}$

step 10

Solving the system of equations, we find the coordinates $c_1$, $c_2$, and $c_3$. The coordinate vector of $u$ relative to the basis $V$ is $\begin{bmatrix} -0.668965 \\ 0.594635 \\ -0.445976 \end{bmatrix}$

Answer

The rank of $H$ is 3. The column space of $H$ is $R^3$. The null space of $H$ is $\{(12x, -4x - 2y, -5x - y, 6x, 3y) : x, y \in \mathbb{R}\}$. The set $V$ is a basis for $R^3$. The coordinate vector of $u$ relative to the basis $V$ is $\begin{bmatrix} -0.668965 \\ 0.594635 \\ -0.445976 \end{bmatrix}$.

Key Concept

Matrix rank, column space, null space, basis, and coordinate vector

Explanation

The rank of a matrix is the number of linearly independent rows or columns. The column space is the span of the column vectors. The null space is the set of solutions to the homogeneous system. A basis for $R^3$ is a set of three linearly independent vectors that span $R^3$. The coordinate vector expresses a vector in terms of a given basis.

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