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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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Show that, for any \(\mathbf{u}\) and \(\mathbf{v}\) in a normed vector space, $$\|\mathbf{u}+\mathbf{v}\| \geq |\|\mathbf{u}\|-\|\mathbf{v}\|$$

In \(C[0,1],\) with inner product defined by \((3),\) compute (a) \(\left\langle e^{x}, e^{-x}\right\rangle\) (b) \(\langle x, \sin \pi x\rangle\) (c) \(\left\langle x^{2}, x^{3}\right\rangle\)

Prove that the transpose of an orthogonal matrix is an orthogonal matrix.

Let $$A=\left(\begin{array}{ll} 2 & 1 \\ 1 & 1 \\ 2 & 1 \end{array}\right) \quad \text { and } \quad \mathbf{b}=\left(\begin{array}{r} 12 \\ 6 \\ 18 \end{array}\right)$$ (a) Use the Gram-Schmidt process to find an orthonormal basis for the column space of \(A\) (b) Factor \(A\) into a product \(Q R,\) where \(Q\) has an orthonormal set of column vectors and \(R\) is upper triangular. (c) Solve the least squares problem \(A \mathbf{x}=\mathbf{b}\)

Prove: If \(A\) is an \(m \times n\) matrix and \(\mathbf{x} \in \mathbb{R}^{n},\) then cither \(A \mathbf{x}=0\) or there exists \(\mathbf{y} \in R\left(A^{T}\right)\) such that \(\mathbf{x}^{T} \mathbf{y} \neq 0 .\) Draw a picture similar to Figure 5.2 .2 to illustrate this result geometrically for the case where \(N(A)\) is a two-dimensional subspace of \(\mathbb{R}^{3}\)
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