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Chapter 1: Problem 40

Let \(A\) be an \(m \times n\) matrix, and let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\mathbb{R}^{n}\) with the property that \(A \mathbf{u}=\mathbf{0}\) and \(A \mathbf{v}=\mathbf{0} .\) Explain why \(A(\mathbf{u}+\mathbf{v})\) must be the zero vector. Then explain why \(A(c \mathbf{u}+d \mathbf{v})=\mathbf{0}\) for each pair of scalars \(c\) and \(d\)

Short Answer

Expert verified
\(A(\mathbf{u} + \mathbf{v}) = \mathbf{0}\) by linearity, and \(A(c\mathbf{u} + d\mathbf{v}) = \mathbf{0}\) since the null space is closed under linear combinations.

Step by step solution

01

Understand what it means for vectors to be in the null space

The vectors \(\mathbf{u}\) and \(\mathbf{v}\) are in the null space of the matrix \(A\). This means that when \(A\) multiplies \(\mathbf{u}\) or \(\mathbf{v}\), the result is the zero vector: \(A \mathbf{u} = \mathbf{0}\) and \(A \mathbf{v} = \mathbf{0}\). The null space is a subspace consisting of all the vectors that \(A\) maps to the zero vector.
02

Show that A(u+v) = 0

Using the linearity (or distributive property) of matrix multiplication, we calculate \(A(\mathbf{u} + \mathbf{v})\). Expanding this gives: \[A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}\]Since both \(A\mathbf{u}\) and \(A\mathbf{v}\) are zero vectors, their sum is also the zero vector:\[A\mathbf{u} + A\mathbf{v} = \mathbf{0} + \mathbf{0} = \mathbf{0}\]
03

Justify why A(cu+dv) = 0 for scalars c and d

Matrix multiplication is also linear with respect to scalar multiplication. This means for scalars \(c\) and \(d\), we have:\[A(c\mathbf{u} + d\mathbf{v}) = cA\mathbf{u} + dA\mathbf{v}\]Again, since \(A\mathbf{u} = \mathbf{0}\) and \(A\mathbf{v} = \mathbf{0}\), we can substitute:\[cA\mathbf{u} + dA\mathbf{v} = c \cdot \mathbf{0} + d \cdot \mathbf{0} = \mathbf{0}\]
04

Conclude about the null space property

The properties just discussed about the matrix \(A\), when operating on vectors within its null space, show that any linear combination of these vectors will also map to the zero vector. Thus, the null space is closed under addition and scalar multiplication, making it a vector subspace.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearity of Matrix Multiplication
Matrix multiplication follows specific rules, one of which is linearity. This means that matrices distribute over vector addition and are compatible with scalar multiplication.

In the original exercise, the matrix \(A\) and vectors \(\mathbf{u}\) and \(\mathbf{v}\) demonstrated this property:
  • When you compute \(A(\mathbf{u} + \mathbf{v})\), you apply the matrix to the addition of the vectors. Linearity allows us to say that \(A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}\).
  • This decomposition into parts makes calculations simpler and assures that you can handle complex operations step-by-step.
  • Both \(A\mathbf{u}\) and \(A\mathbf{v}\) resulting in the zero vector nicely confirms that \(A(\mathbf{u} + \mathbf{v})\) is also zero. This is a clear example of how linearity helps in problem-solving.
Vector Subspace
A vector subspace is a fundamental concept in linear algebra, especially when discussing null spaces. It is essentially a set of vectors closed under addition and scalar multiplication.

In context, the null space of a matrix \(A\) is a subspace of \(\mathbb{R}^n\). For any vectors \(\mathbf{u}\) and \(\mathbf{v}\) within this null space, their sum \(\mathbf{u} + \mathbf{v}\) is also within it.
  • The step-by-step exercise outlined how the sum of two vectors that return zero when multiplied with \(A\), \(A\mathbf{u} = \mathbf{0}\) and \(A\mathbf{v} = \mathbf{0}\), results in \(A(\mathbf{u} + \mathbf{v}) = \mathbf{0}\).
  • This property alludes to the definition of a vector subspace鈥攃losure under addition.
  • Importantly, the null space contains the zero vector and is closed under scalar multiplication as well, reinforcing that it is indeed a subspace.
Scalar Multiplication in Linear Algebra
Scalar multiplication involves multiplying each component of a vector by a scalar, and linear algebra ensures this operation behaves predictably with matrices.

The exercise showed that if \(c\) and \(d\) are scalars and \(\mathbf{u}, \mathbf{v}\) are vectors in the null space of a matrix \(A\), then multiplying components of the vectors by \(c\) and \(d\) still results in the zero vector.
  • Equation: \(A(c\mathbf{u} + d\mathbf{v}) = cA\mathbf{u} + dA\mathbf{v} = c \cdot \mathbf{0} + d \cdot \mathbf{0} = \mathbf{0}\).
  • The combination of these scalars with vectors from a null space always adheres to the properties of null space, that is, resulting in \(\mathbf{0}\).
This reliance on scalar multiplication further emphasizes the structured nature of vector spaces and their interactions with matrices.

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Most popular questions from this chapter

Suppose \(A\) is a \(4 \times 3\) matrix and \(\mathbf{b}\) is a vector in \(\mathbb{R}^{4}\) with the property that \(A \mathbf{x}=\mathbf{b}\) has a unique solution. What can you say about the reduced echelon form of \(A ?\) Justify your answer.

Construct a \(3 \times 3\) nonzero matrix \(A\) such that the vector \(\left[\begin{array}{r}{1} \\ {-2} \\ {1}\end{array}\right]\) is a solution of \(A \mathbf{x}=\mathbf{0}\) .

In Exercises \(29-32,\) (a) does the equation \(A \mathbf{x}=0\) have a nontrivial solution and (b) does the equation \(A \mathbf{x}=\mathbf{b}\) have at least one solution for every possible \(\mathbf{b} ?\) \(A\) is a \(3 \times 3\) matrix with three pivot positions.

[M] Budget庐 Rent A Car in Wichita, Kansas, has a fleet of about 500 cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix below. Suppose that on Monday there are 295 cars at the airport (or rented from there), 55 cars at the east side office, and 150 cars at the west side office. What will be the approximate distribution of cars on Wednesday? \(\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {.00} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right]\)

In Exercises 21 and \(22,\) find a parametric equation of the line \(M\) through \(\mathbf{p}\) and \(\mathbf{q} \cdot[\text { Hint: } M \text { is parallel to the vector } \mathbf{q}-\mathbf{p} . \text { See the }\) figure below. $$ \mathbf{p}=\left[\begin{array}{r}{2} \\ {-5}\end{array}\right], \mathbf{q}=\left[\begin{array}{r}{-3} \\ {1}\end{array}\right] $$
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