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Chapter 3: Problem 19

Let \(A\) be an \(m \times n\) matrix with rank equal to \(n\). Show that if \(\mathbf{x} \neq \mathbf{0}\) and \(\mathbf{y}=A \mathbf{x},\) then \(\mathbf{y} \neq \mathbf{0}\).

Short Answer

Expert verified
Since the rank of the matrix \(A\) is equal to \(n\), it implies that the null space of \(A\) only contains the zero vector. If we assume that \(\mathbf{y} = A\mathbf{x} = \mathbf{0}\), it contradicts the given condition \(\mathbf{x} \neq \mathbf{0}\) as, in this case, \(A\mathbf{x} = \mathbf{0}\) would mean \(\mathbf{x}\) is in the null space of \(A\). Therefore, if \(\mathbf{y} = A\mathbf{x}\) and \(\mathbf{x} \neq \mathbf{0}\), then \(\mathbf{y} \neq \mathbf{0}\).

Step by step solution

01

Determine the null space of matrix A

Using the given information that the rank of the matrix \(A\) is equal to \(n\), we can infer that the null space only contains the zero vector. This is because, for an \(m \times n\) matrix, the rank-nullity theorem states that the rank of the matrix plus the nullity (dimension of the null space) must equal the number of columns n. Here, since rank \(A = n\), it implies that there are no non-zero vectors in the null space of the matrix \(A\). Thus, the nullity is zero.
02

Use the null space of A to show that when \(\mathbf{x} \neq \mathbf{0}\), \(\mathbf{y} \neq \mathbf{0}\)

Now, we are given that \(\mathbf{x} \neq \mathbf{0}\), and we want to show that \(\mathbf{y} = A \mathbf{x} \neq \mathbf{0}\). Suppose, for contradiction, that \(\mathbf{y} = A \mathbf{x} = \mathbf{0}\). Then, \(A \mathbf{x}\) should be in the null space of matrix \(A\) as \(A \mathbf{x} = \mathbf{0}\). Since the null space of \(A\) only contains the zero vector, it implies that \(\mathbf{x}\) must be equal to the zero vector. However, this contradicts our given condition that \(\mathbf{x} \neq \mathbf{0}\). Hence, our assumption that \(\mathbf{y} = \mathbf{0}\) must be incorrect. Therefore, if \(A\mathbf{x} = \mathbf{y}\) and \(\mathbf{x} \neq \mathbf{0}\), then \(\mathbf{y} \neq \mathbf{0}\).

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Most popular questions from this chapter

Let \(\mathbb{R}^{+}\) denote the set of positive real numbers. Define the operation of scalar multiplication, denoted ?, by $$\alpha \circ x=x^{\alpha}$$ for each \(x \in \mathbb{R}^{+}\) and for any real number \(\alpha\). Define the operation of addition, denoted \(\oplus,\) by $$x \oplus y=x \cdot y \quad \text { for all } \quad x, y \in \mathbb{R}^{+}$$ Thus, for this system, the scalar product of -3 \(\operatorname{times} \frac{1}{2}\) is given by $$-3 \circ \frac{1}{2}=\left(\frac{1}{2}\right)^{-3}=8$$ and the sum of 2 and 5 is given by $$2 \oplus 5=2 \cdot 5=10$$ Is \(\mathbb{R}^{+}\) a vector space with these operations? Prove your answer.

Let \(A\) and \(B\) be \(6 \times 5\) matrices. If \(\operatorname{dim} N(A)=2\) what is the rank of \(A\) ? If the rank of \(B\) is \(4,\) what is the dimension of \(N(B) ?\)

Which of the sets that follow are spanning sets for \(\mathbb{R}^{3} ?\) Justify your answers. (a) \(\left\\{(1,0,0)^{T},(0,1,1)^{T},(1,0,1)^{T}\right\\}\) (b) \(\left\\{(1,0,0)^{T},(0,1,1)^{T},(1,0,1)^{T},(1,2,3)^{T}\right\\}\) (c) \(\left\\{(2,1,-2)^{T},(3,2,-2)^{T},(2,2,0)^{T}\right\\}\) (d) \(\left\\{(2,1,-2)^{T},(-2,-1,2)^{T},(4,2,-4)^{T}\right\\}\) (e) \(\left\\{(1,1,3)^{T},(0,2,1)^{T}\right\\}\)

Is it possible to find a pair of two-dimensional subspaces \(U\) and \(V\) of \(\mathbb{R}^{3}\) such that \(U \cap V=\\{0\\} ?\) Prove your answer. Give a geometrical interpretation of your conclusion. [Hint: Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) and \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) be bases for \(U\) and \(V,\) respectively. Show that \(\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}_{1}, \mathbf{v}_{2}\) are linearly dependent.

Let \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) be linearly independent vectors in \(\mathbb{R}^{n}\) and let $$\mathbf{y}_{1}=\mathbf{x}_{1}+\mathbf{x}_{2}, \quad \mathbf{y}_{2}=\mathbf{x}_{2}+\mathbf{x}_{3}, \quad \mathbf{y}_{3}=\mathbf{x}_{3}+\mathbf{x}_{1}$$ Are \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) linearly independent? Prove your answer.
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