Find a Basis for each of these substances of R^4
(a) All vectors whose components are equal
(b) All vectors whose component add to zero
(c) All vectors that are perpendicular to (1,1,0,0) and (1,0,11)
(d) The column space (In R^2) and nullspace(In R^5) of U=[1 0 1 0 1]
[0 1 0 1 0]
(a) To find a basis for the subspace of vectors whose components are equal in R^4, we can follow these steps:
- Let's call this subspace S.
- Notice that any vector in S can be expressed as (x,x,x,x), where x is a scalar.
- We can rewrite this expression as x(1,1,1,1).
- Therefore, S can be generated by the vector (1,1,1,1).
- Since this single vector generates the entire subspace, it is a basis for S.
(b) To find a basis for the subspace of vectors whose components add to zero in R^4, we can follow these steps:
- Let's call this subspace S.
- Notice that any vector in S can be expressed as (x,y,z,w), where x, y, z, and w are scalars, and x + y + z + w = 0.
- We can rewrite this expression as x(-1,1,0,0) + y(0,-1,1,0) + z(0,0,-1,1).
- Therefore, S can be generated by the vectors (-1,1,0,0), (0,-1,1,0), and (0,0,-1,1).
- These three vectors form a linearly independent set, which means they span the entire subspace S. Therefore, they form a basis for S.
(c) To find a basis for the subspace of vectors perpendicular to both (1,1,0,0) and (1,0,11) in R^4, we can follow these steps:
- Let's call this subspace S.
- Notice that any vector in S will be orthogonal (perpendicular) to both (1,1,0,0) and (1,0,11). This means that the dot product of the vector and both of these vectors should be zero.
- We can set up the following system of equations:
(x,y,z,w) · (1,1,0,0) = 0,
(x,y,z,w) · (1,0,11) = 0.
- Solving this system of equations will give us the solutions for x, y, z, and w. These solutions will represent the vectors that are perpendicular to both given vectors.
- The resulting values of x, y, z, and w will give us the coefficients of the vectors that generate the subspace S.
- Therefore, the vectors obtained from the solutions are a basis for S.
(d) To find the basis for the column space of U in R^2 and the nullspace of U in R^5, we can follow these steps:
For the column space:
- The column space of U is the subspace spanned by the columns of U.
- In this case, U has two columns.
- Write down the columns of U: U = [1, 0], [0, 1], [1, 0], [0, 1], [1, 0].
- Notice that the first two columns are linearly independent, and they span a subspace in R^2.
- Therefore, the first two columns [1, 0] and [0, 1] form a basis for the column space of U in R^2.
For the nullspace:
- The nullspace of U is the set of all vectors x that satisfy the equation Ux = 0, where 0 is the zero vector in R^5.
- Write down the equation: Ux = [1, 0, 1, 0, 1; 0, 1, 0, 1, 0]x = 0.
- To find the nullspace, we need to solve the system of equations represented by this equation.
- One way to solve this system is to perform row reduction on the augmented matrix [U | 0].
- Row reducing the augmented matrix will give us the reduced row echelon form [R | 0], where R is the row echelon form of matrix U.
- Identify the pivot variables (x1 and x2) and free variables (x3, x4, and x5).
- The nullspace is then parameterized by the free variables.
- Express the solution in terms of the parameters, e.g., x3, x4, and x5.
- The nullspace is the set of all vectors that can be obtained by assigning values to the free variables.
- Write the vectors using parameterized form, e.g., [(expression in terms of x3, x4, x5)].
- These vectors form a basis for the nullspace of U in R^5.
(a) To find a basis for all vectors whose components are equal, let's consider a specific vector in R^4. We can choose the vector (1,1,1,1) as an example. Notice that any vector whose components are equal can be written as a scalar multiple of (1,1,1,1). Therefore, the vector (1,1,1,1) itself is a basis for the set of all vectors whose components are equal.
(b) To find a basis for all vectors whose components add to zero, let's consider a specific vector in R^4. We can choose the vector (1,-1,1,-1) as an example. Notice that any vector whose components add to zero can be written as a linear combination of (1,-1,1,-1). Therefore, the vector (1,-1,1,-1) itself is a basis for the set of all vectors whose components add to zero.
(c) To find a basis for all vectors that are perpendicular to (1,1,0,0) and (1,0,11), we can use the cross product. The cross product of two vectors gives a vector that is orthogonal (perpendicular) to both of the original vectors.
The cross product is calculated as follows:
(1,1,0,0) x (1,0,11) = (11,11,-1,-1)
So, the vector (11,11,-1,-1) is perpendicular to both (1,1,0,0) and (1,0,11). Therefore, it is a basis for the set of all vectors that are perpendicular to (1,1,0,0) and (1,0,11).
(d) For the column space of U=[1 0 1 0 1; 0 1 0 1 0] in R^2, we can simply take the columns of U as a basis. So, the basis for the column space of U is {(1, 0), (0, 1)}.
For the null space of U in R^5, we need to find all vectors x in R^5 such that Ux = 0. This means that the columns of U are linearly dependent, and the null space will contain all linear combinations of the vectors in U that equal the zero vector.
To find the null space, we can row-reduce the augmented matrix [U | 0] to get the reduced row-echelon form.
[RREF] => [1 0 1 0 1 | 0
0 1 0 1 0 | 0]
From the RREF, we can see that the first and third columns are pivot columns, while the second and fourth columns are free (non-pivot) columns.
So, a general form of a vector x in the null space of U can be written as x = t(-1, -1, 1, 1, 0), where t is any scalar.
Therefore, the basis for the null space of U in R^5 is {(-1, -1, 1, 1, 0)}.