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Math
Posted 9 months ago

a) For matrix H=[224203061213002]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=[210],v1=[231],v2=[203]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 v3=[460]v_{3}=\left[\begin{array}{l}4 \\ 6 \\ 0\end{array}\right]. Show that V={v1,v2,v3}V=\left\{v_{1}, v_{2}, v_{3}\right\} is a basis for R3R^{3}, and then find the coordinate vector of uu relative to the basis VV. ( 4 marks)
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Answer from Sia
Posted 9 months ago
Solution by Steps
step 1
To find the rank of the matrix H=[224203061213002]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 HH is transformed into echelon form, revealing that there are 3 non-zero rows. Therefore, the rank of HH is 3
step 3
To find the column space of HH, we identify the pivot columns in the row-reduced form of HH. The pivot columns correspond to the original columns of HH
step 4
The column space of HH is spanned by the vectors [231]\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, [203]\begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, and [460]\begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix}. Therefore, the column space is R3R^3
step 5
To find the null space of HH, we solve the equation Hx=0H \mathbf{x} = \mathbf{0}. This involves finding the solutions to the homogeneous system of linear equations represented by HH
step 6
The null space of HH is given by the set of all vectors x\mathbf{x} such that Hx=0H \mathbf{x} = \mathbf{0}. The null space can be expressed as {(12x,4x2y,5xy,6x,3y):x,yR}\{(12x, -4x - 2y, -5x - y, 6x, 3y) : x, y \in \mathbb{R}\}
step 7
To show that V={[231],[203],[460]}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 R3R^3, we need to show that these vectors are linearly independent and span R3R^3
step 8
We form the matrix A=[224306130]A = \begin{bmatrix} 2 & -2 & 4 \\ 3 & 0 & 6 \\ 1 & 3 & 0 \end{bmatrix} and perform row reduction. The matrix AA reduces to the identity matrix, indicating that the vectors are linearly independent and span R3R^3. Therefore, VV is a basis for R3R^3
step 9
To find the coordinate vector of u=[210]u = \begin{bmatrix} -2 \\ -1 \\ 0 \end{bmatrix} relative to the basis VV, we express uu as a linear combination of the basis vectors: u=c1[231]+c2[203]+c3[460]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 c1c_1, c2c_2, and c3c_3. The coordinate vector of uu relative to the basis VV is [0.6689650.5946350.445976]\begin{bmatrix} -0.668965 \\ 0.594635 \\ -0.445976 \end{bmatrix}
Answer
The rank of HH is 3. The column space of HH is R3R^3. The null space of HH is {(12x,4x2y,5xy,6x,3y):x,yR}\{(12x, -4x - 2y, -5x - y, 6x, 3y) : x, y \in \mathbb{R}\}. The set VV is a basis for R3R^3. The coordinate vector of uu relative to the basis VV is [0.6689650.5946350.445976]\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 R3R^3 is a set of three linearly independent vectors that span R3R^3. The coordinate vector expresses a vector in terms of a given basis.

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