Let V be the subspace of R3 defined by the equation a1 + 2a2 + 3a3 = 0. Find a basis of V and give a proof that it is indeed a basis. --- To find a basis of the subspace V = {(a1, a2, a3) ∈ ℝ^3 | a1 + 2a2 + 3a3 = 0}, we need to find a set of linearly independent vectors in V that span the subspace. One way to do this is by considering the null space of the matrix [1 2 3], which corresponds to the equation a1 + 2a2 + 3a3 = 0. The null space of [1 2 3] is the set of all vectors (x, y, z) such that [1 2 3] [x, y, z]^T = 0. This equation can be rewritten as: x + 2y + 3z = 0 By equating the coefficients of x, y, and z to zero, we get: x = -2y - 3z This means that any vector (x, y, z) in the null space can be expressed as a linear combination of the vectors (-2, 1, 0) and (-3, 0, 1). Therefore, a basis for V is B = {(-2, 1, 0), (-3, 0, 1)}. Proof that B is a basis: 1. Vectors in B are in V: (-2, 1, 0) satisfies -2 + 1 + 0 = 0 (-3, 0, 1) satisfies -3 + 0 + 3 = 0 2. Vectors in B are linearly independent: Suppose α(-2, 1, 0) + β(-3, 0, 1) = (0, 0, 0) Then, -2α - 3β = 0, α + 0 = 0, 0 + β = 0 Solving this system, we get α = 0 and β = 0. This means that the only way to express the zero vector as a linear combination of the vectors in B is by using the trivial coefficients (α = 0, β = 0). 3. Span of B is V: Any vector (a1, a2, a3) in V satisfies a1 + 2a2 + 3a3 = 0. We can express (a1, a2, a3) as a linear combination of (-2, 1, 0) and (-3, 0, 1) as follows: (a1, a2, a3) = a2(-2, 1, 0) + (a3 + 2a2)(-3, 0, 1) Therefore, the set B = {(-2, 1, 0), (-3, 0, 1)} is a basis for the subspace V. --- student: please explain how you found the basis for the vectors (x,y,z) in the null space of the matrix [1 2 3 ] --- Sure, let me explain the reasoning step-by-step: 1) We start with the equation x + 2y + 3z = 0, which represents the null space of the matrix [1 2 3]. 2) Rearranging the terms, we get x = -2y - 3z. 3) This equation tells us that for any vector (x, y, z) in the null space, the value of x is completely determined by the values of y and z. In other words, we can express x as a linear combination of two vectors: (-2, 1, 0) and (-3, 0, 1). 4) To see this, let's substitute the values of y and z in the equation x = -2y - 3z: a) If y = 1 and z = 0, then x = -2(1) - 3(0) = -2. This means that when y = 1 and z = 0, the vector (x, y, z) is (-2, 1, 0). b) If y = 0 and z = 1, then x = -2(0) - 3(1) = -3. This means that when y = 0 and z = 1, the vector (x, y, z) is (-3, 0, 1). 5) Now, for any other vector (x, y, z) in the null space, we can express it as a linear combination of (-2, 1, 0) and (-3, 0, 1) as follows: (x, y, z) = y(-2, 1, 0) + z(-3, 0, 1) = (-2y, y, 0) + (-3z, 0, z) = (-2y - 3z, y, z) 6) Substituting the expression for x = -2y - 3z, we get: (x, y, z) = (x, y, z) This shows that any vector (x, y, z) in the null space can indeed be expressed as a linear combination of the vectors (-2, 1, 0) and (-3, 0, 1). The key idea is that once we have the equation x = -2y - 3z, we can express x in terms of y and z, which allows us to represent any vector (x, y, z) in the null space using the two vectors (-2, 1, 0) and (-3, 0, 1). --- Sure, let me explain how I found the basis for the null space of the matrix [1 2 3]. The null space of a matrix A is the set of all vectors x such that Ax = 0. In our case, we want to find the vectors (x, y, z) that satisfy the equation [1 2 3] [x, y, z]^T = 0. To find these vectors, we can set up the matrix equation and solve for x, y, and z: [1 2 3] [x, y, z]^T = 0 [x + 2y + 3z] = 0 x + 2y + 3z = 0 This equation represents the null space of the matrix [1 2 3]. To find a basis for this null space, we need to find a set of linearly independent vectors that span the null space. From the equation x + 2y + 3z = 0, we can express x in terms of y and z: x = -2y - 3z Now, we can substitute this expression for x into the vector (x, y, z) to get: (x, y, z) = (-2y - 3z, y, z) This shows that any vector (x, y, z) in the null space can be expressed as a linear combination of the vectors (-2, 1, 0) and (-3, 0, 1). To verify that these two vectors (-2, 1, 0) and (-3, 0, 1) are linearly independent, we can check that the only way to express the zero vector (0, 0, 0) as a linear combination of these two vectors is by using the trivial coefficients (α = 0, β = 0). Let α(-2, 1, 0) + β(-3, 0, 1) = (0, 0, 0) -2α - 3β = 0 α = 0 β = 0 Since the only solution to this system is α = 0 and β = 0, the vectors (-2, 1, 0) and (-3, 0, 1) are linearly independent. Furthermore, since any vector (x, y, z) in the null space can be expressed as a linear combination of these two vectors, they span the null space. Therefore, the set B = {(-2, 1, 0), (-3, 0, 1)} forms a basis for the null space of the matrix [1 2 3]. --- You're right, the statement "(x, y, z) = (-2y - 3z, y, z)" alone does not directly show that any vector in the null space can be expressed as a linear combination of the vectors (-2, 1, 0) and (-3, 0, 1). Let me explain the reasoning more clearly. 1) We start with the equation x + 2y + 3z = 0, which represents the null space of the matrix [1 2 3]. 2) Rearranging the terms, we get x = -2y - 3z. 3) Substituting this expression for x into the vector (x, y, z), we get: (x, y, z) = (-2y - 3z, y, z) 4) Now, let's consider the vectors (-2, 1, 0) and (-3, 0, 1). We can express these vectors as: (-2, 1, 0) = (-2(1) - 3(0), 1, 0) (-3, 0, 1) = (-2(0) - 3(1), 0, 1) 5) Observe that any vector (x, y, z) in the null space can be expressed as a linear combination of (-2, 1, 0) and (-3, 0, 1) by using the appropriate scalar multiples of y and z: (x, y, z) = (-2y - 3z, y, z) = y(-2, 1, 0) + z(-3, 0, 1) 6) This shows that any vector (x, y, z) in the null space can be written as a linear combination of the vectors (-2, 1, 0) and (-3, 0, 1) by choosing the appropriate values of y and z. The key step is recognizing that the expression (-2y - 3z, y, z) allows us to express any vector (x, y, z) in the null space using scalar multiples of the vectors (-2, 1, 0) and (-3, 0, 1). The substitution alone does not directly show this, but it sets up the representation that allows us to express any null space vector as a linear combination of these two vectors.