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A=[1 a -1 1 0 2a 0 2 0 0 0 a]for some a ∈ R. Determine bases for...
Mar 23, 2024
A=[1 a -1 1 0 2a 0 2 0 0 0 a]for some a ∈ R. Determine bases for Nul(A) for all possible values of a. You should set out your working clearly and with precise use of terminology.
Solution by Steps
step 1
To determine the bases for Nul(A) \text{Nul}(A) , we need to find the null space of the matrix A A
step 2
The matrix A A is given by A=[1amp;aamp;−1amp;10amp;2aamp;0amp;20amp;0amp;0amp;a] A = \begin{bmatrix} 1 & a & -1 & 1 \\ 0 & 2a & 0 & 2 \\ 0 & 0 & 0 & a \end{bmatrix} . We need to consider different cases for a a
step 3
Case 1: If a≠0 a \neq 0 , the matrix A A has three pivot positions, and the null space is one-dimensional
step 4
The null space in this case is spanned by the vector (1,0,1,0) (1, 0, 1, 0) , which can be normalized to (12,0,12,0) \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}, 0\right)
step 5
Case 2: If a=0 a = 0 , the matrix A A has only one pivot position, and the null space is three-dimensional
step 6
The null space in this case is spanned by the vectors (0,1,0,0) (0, 1, 0, 0) , (1,0,0,0) (1, 0, 0, 0) , and (0,0,1,0) (0, 0, 1, 0)
Answer
For a≠0 a \neq 0 , the basis for Nul(A) \text{Nul}(A) is {(12,0,12,0)} \left\{ \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}, 0\right) \right\} . For a=0 a = 0 , the basis for Nul(A) \text{Nul}(A) is {(0,1,0,0),(1,0,0,0),(0,0,1,0)} \left\{ (0, 1, 0, 0), (1, 0, 0, 0), (0, 0, 1, 0) \right\} .
Key Concept
Null Space and Basis
Explanation
The null space of a matrix consists of all vectors that result in the zero vector when multiplied by the matrix. The basis for the null space is a set of linearly independent vectors that span this space. The dimension of the null space depends on the number of pivot positions in the matrix.
To find a basis for the row space we row reduce and then take the non zero rows in reduced row echelon form, and to find a basis for the column space we row reduce to find the pivot columns and then take the corresponding columns from the original matrix. In this question we consider what happens if we mix up these techniques. (a) Suppose that you take the pivot columns from the row reduced matrix, will this in general give a basis for the column space? (b) Suppose that you identify the non-zero rows of the row reduced matrix, but then take the corresponding rows of the original matrix, will this in general give a basis for the row space? In each case you must give arguments and specific examples where applicable to support your answers.
Solution by Steps
step 1
To address part (a), we consider the process of finding a basis for the column space of a matrix
step 2
The column space of a matrix is spanned by the pivot columns of the original matrix, not the row-reduced matrix
step 3
Taking the pivot columns from the row-reduced matrix does not, in general, give a basis for the column space of the original matrix
step 4
For example, if we row reduce the matrix [1amp;2amp;34amp;5amp;6]\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, we get [1amp;0amp;−10amp;1amp;2]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix}. The pivot columns of the row-reduced matrix are the first and second columns
step 5
However, the pivot columns of the original matrix are also the first and second columns. In this specific case, taking the pivot columns from the row-reduced matrix would give the correct basis for the column space, but this is not true in general
Answer
No, taking the pivot columns from the row-reduced matrix will not in general give a basis for the column space.
Key Concept
Column Space Basis
Explanation
The basis for the column space of a matrix is given by the pivot columns of the original matrix, not the row-reduced version.
step 1
To address part (b), we consider the process of finding a basis for the row space of a matrix
step 2
The row space of a matrix is spanned by the non-zero rows of its row-reduced echelon form
step 3
If we identify the non-zero rows of the row-reduced matrix and then take the corresponding rows of the original matrix, this does not in general give a basis for the row space
step 4
The non-zero rows of the row-reduced matrix represent the leading terms that span the row space, but the corresponding rows in the original matrix may not be linearly independent
step 5
For example, if we row reduce the matrix [1amp;2amp;34amp;5amp;6]\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, we get [1amp;0amp;−10amp;1amp;2]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix}. The non-zero rows of the row-reduced matrix are both rows, but taking the corresponding rows of the original matrix would give [1amp;2amp;34amp;5amp;6]\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, which are not linearly independent
Answer
No, identifying the non-zero rows of the row-reduced matrix and then taking the corresponding rows of the original matrix will not in general give a basis for the row space.
Key Concept
Row Space Basis
Explanation
The basis for the row space is given by the non-zero rows of the row-reduced echelon form, not by taking the corresponding rows from the original matrix.
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