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Chapter 5: Problem 6

Is it possible for a matrix to have the vector (3,1,2) in its row space and \((2,1,1)^{T}\) in its null space? Explain.

Short Answer

Expert verified
No, it is not possible for a matrix to have the vector (3,1,2) in its row space and \((2,1,1)^{T}\) in its null space simultaneously. This is because the two vectors have a nonzero dot product, which implies that they are not orthogonal. In order for a vector to be in the null space, it must be orthogonal to the row space.

Step by step solution

01

Understand the basics of row space and null space

The row space of a matrix consists of all linear combinations of its rows. It describes the space spanned by the rows of that matrix. The null space of a matrix is the set of all vectors that satisfy the equation Ax=0, with A being the matrix, and x being any vector satisfying the equation.
02

Explain the dimension of the row space and null space

A matrix has a specific dimension called the rank. The rank of a matrix is equal to the dimension of its row space (row rank) and column space (column rank) since they are equal for any matrix. On the other hand, the dimension of the null space can be found using the formula n - rank, where n is the number of columns.
03

Analyze the relationship between row and null spaces

If a vector is in the row space, then any orthogonal vector to that vector will be in the null space (Ax=0). In this case, we have the vector (3,1,2) in the row space, and we need to check if the orthogonal vector is \((2,1,1)^{T}\).
04

Determine if \((2,1,1)^{T}\) is orthogonal

To check if the two vectors are orthogonal, we need to take the dot product of the two vectors. If it is zero, they are orthogonal. Let's take the dot product: Dot product = \((3)(2) + (1)(1) + (2)(1) =\) \(6 + 1 + 2 = 9\) The dot product is not zero, meaning the two vectors are not orthogonal.
05

Conclude the possibility of having both vectors in respective spaces

Since the given vectors (3,1,2) and \((2,1,1)^{T}\) are not orthogonal and have a nonzero dot product, it is not possible for a matrix to have (3,1,2) in its row space and \((2,1,1)^{T}\) in its null space simultaneously.

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Most popular questions from this chapter

Let \(A\) be an \(m \times n\) matrix of rank \(n\) and let \(P=A\left(A^{T} A\right)^{-1} A^{T}\) (a) Show that \(P \mathbf{b}=\mathbf{b}\) for every \(\mathbf{b} \in R(A) .\) Ex plain this property in terms of projections. (b) If \(\mathbf{b} \in R(A)^{\perp},\) show that \(P \mathbf{b}=\mathbf{0}\) (c) Give a geometric illustration of parts and (b) if \(R(A)\) is a plane through the origin in \(\mathbb{R}^{3}\)

Show that $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|$$ defines a norm on \(\mathbb{R}^{n}\)

Dimension Theorem Let \(U\) and \(V\) be subspaces of \(\mathbb{R}^{n} .\) In the case that \(U \cap V=\\{\mathbf{0}\\},\) we have the following dimension relation \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V \\] (See Exercise 18 in Section 4 of Chapter 3 .) Make use of the result from Exercise 14 to prove the more general theorem \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V-\operatorname{dim}(U \cap V) \\]

Show that, in any vector space with a norm, $$\|-\mathbf{v}\|=\|\mathbf{v}\|$$

Let \(S\) be the subspace of \(\mathbb{R}^{3}\) spanned by \(\mathbf{x}=\) \((1,-1,1)^{T}\) (a) Find a basis for \(S^{\perp}\) (b) Give a geometrical description of \(S\) and \(S^{\perp}\)
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