Exercises – Long Day

Exercises – Long Day#

Exercise 1: Span of Vectors in \(\mathbb{R}^3\)#

In this exercise we will investigate the span of the following three vectors from \(\mathbb{R}^3\):

\[\begin{split}{\mathbf w}_1=\left[\begin{array}{r}1\\-1\\2\end{array}\right], {\mathbf w}_2=\left[\begin{array}{r}1\\3\\2\end{array}\right], \ \mathrm{and} \ {\mathbf w}_3=\left[\begin{array}{r}7\\9\\14\end{array}\right]. \end{split}\]

Question a#

Investigate whether \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\) can be spanned by fewer than three vectors. If so, then write one of the vectors as a linear combination of the other two.

Hint

In Example 9.1.5 in the textbook a similar case is worked through.

Answer

The span can indeed be spanned by fewer than three vectors since the three vectors are linearly dependent. For instance, we have \({\mathbf w}_3=3{\mathbf w}_1+4{\mathbf w}_2.\) Hence, according to Theorem 9.1.1, \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2)=\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\).

Question b#

Check that the vectors \({\mathbf w}_1\) and \({\mathbf w}_2\) are linearly independent and conclude that the list \(({\mathbf w}_1,{\mathbf w}_2)\) is an ordered basis for \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\).

Hint

We have already seen that \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2)=\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\). According to Corollary 9.2.2 the last claim follows from the first.

Question c#

Calculate the dimension of \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\).

Hint

In Definition 9.4.1 in the textbook you can read more about how the dimension of a span of vectors is defined.

Exercise 2: Another Span of Vectors in \(\mathbb{R}^3\)#

We are given the following three vectors in \(\mathbb{R}^3\):

\[\begin{split}{\mathbf v}_1=\left[\begin{array}{r}1\\-1\\2\end{array}\right], {\mathbf v}_2=\left[\begin{array}{r}1\\3\\2\end{array}\right], \ \mathrm{and} \ {\mathbf v}_3=\left[\begin{array}{r}1\\0\\0\end{array}\right]. \end{split}\]

Question a#

Show that the three given vectors are linearly independent.

Hint

Different methods can be followed. One approach is to use Theorem 7.1.3 in the textbook. Another is to use Corollary 8.2.5.

Question b#

Determine an ordered basis for \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\). What is the dimension of this span?

Hint

Can Corollary 9.2.2 in the textbook be of use?

Answer

An ordered basis is the list \(({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\). Thus, the dimensionen of \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\) is \(3\).

Exercise 3: Kernel and Column Space of a Matrix#

We are given a matrix \(\mathbf A\) as follows

\[\begin{split} {\mathbf A}= \left[ \begin{array}{rrrr} 1 & 1 & 1 & 1\\ 0 & 2 & 4 & -1\\ 2 & 0& -2 & 3\\ 2 & 4 & 6 & 1\\ \end{array} \right]. \end{split}\]

Question a#

Determine an ordered basis for the kernel of the matrix. What is the nullity of the matrix?

Hint

The nullity is the dimension of the kernel.

Hint

In Example 9.3.1 it is explained for another case how such task is done.

Answer

The reduced row-echelon form of the matrix \(\mathbf A\) turns out to be

\[\begin{split} \left[ \begin{array}{rrrr} 1 & 0& -1 & 3/2\\ 0 & 1 & 2 & -1/2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{array} \right]. \end{split}\]

From this we can see that an ordered basis for the kernel of \(\mathbf A\) is

\[\begin{split}\left( \left[\begin{array}{r}1\\-2\\ 1 \\ 0\end{array}\right], \left[\begin{array}{r}-3/2\\1/2\\0 \\1\end{array}\right] \right).\end{split}\]

As this basis is a list containing two vectors we conclude that the nullity of \(\mathbf A\) is \(2\) (that is, its kernel has a dimension of \(2\)).

Question b#

Determine an ordered basis for the column space of the matrix, and check that the rank-nullity theorem holds (Theorem 9.4.2 in the textbook).

Hint

In Example 9.3.2 it is explained for another case how such task is done.

Answer

An ordered basis for the column space of \(\mathbf A\) is

\[\begin{split}\left( \left[\begin{array}{r}1\\0 \\ 2 \\ 2\end{array}\right], \left[\begin{array}{r}1\\2 \\ 0 \\ 4\end{array}\right] \right).\end{split}\]

Hence, the dimension of the column space is \(2\). This is identical to the rank of the matrix - you can confirm this by counting the pivots in \(\mathbf A\)’s reduced row-echelon form shown in the answer to Question a above. The rank-nullity theorem is thus fulfilled, as it should be, since the sum of the nullity and the rank equals the number of columns/rows in the matrix: \(2+2=4\).

Exercise 4: Dimension of Kernel and Column Space#

We are given the reduced row-echelon form of a complex matrix \(\mathbf A \in \mathbb{C}^{5 \times 10}\) by

\[\begin{split} \mathrm{rref}(\mathbf A)=\left[ \begin{array}{rrrrrrrrrr} 0 & 1&-1+i& 0 &\sqrt{2}-i&-\pi& 0 & 1 &\pi^2& e^3\\ 0 & 0 & 0 & 1 & 8 &3/2 & 0 & 1 &100 & -1/2\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & i & -2 & 4/5\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array} \right]. \end{split}\]

State, without doing any calculations, the nullity \(\mathrm{null}(\mathbf A)\) of matrix \(\mathbf A\) as well as the dimension of its column space, i.e. its rank \(\rho(\mathbf A)\).

Answer

\(\mathrm{null}(\mathbf A)=7\) and \(\rho(\mathbf A)=3\).

Exercise 5: Spans and Homogeneous Systems of Linear Equations#

Justify that the solution set to the following homogeneous linear equation system over \(\mathbb R\):

\[\begin{split} \left\{ \begin{array}{rcl} x_2 +3x_3 - x_4+2x_5 & = & 0\\ 2x_1+3x_2+x_3+3x_4 & = & 0\\ x_1 + x_2 -x_3 + 2x_4-x_5 & = & 0 \end{array} \right. \end{split}\]

is the span of some vectors in \(\mathbb{R}^5.\) Then choose an ordered basis for the solution set. What is the dimension of the span?

Hint

First, find the general solution to the system just as in Example 6.4.3 in the textbook. Can you see what a possible ordered basis of the solution set could be? You may compare with Corollary 9.2.3.

Answer

The span has a dimension of \(3\). Hence, an ordered basis must consist of three linearly independent vectors. There are many correct ways to choose an ordered basis. If you follow the procedure from Theorem 6.4.4 to find the general solution, you will find the following ordered basis (also see Corollary 9.2.3):

\[\begin{split}\left( \left[\begin{array}{c} 4\\ -3\\ 1\\ 0\\ 0\\ \end{array}\right], \, \left[\begin{array}{c} -3\\ 1\\ 0\\ 1\\ 0\\ \end{array}\right], \, \left[\begin{array}{c} 3\\ -2\\ 0\\ 0\\ 1\\ \end{array}\right] \right).\end{split}\]

Exercise 6: Span of Vectors#

In \(\mathbb{R}^6\) we are given the four vectors:

\[\begin{split}{\mathbf u}_1=\left[\begin{array}{c}1\\0\\1\\0\\1\\0\end{array}\right]\, , \quad {\mathbf u}_2=\left[\begin{array}{c}0\\1\\1\\1\\1\\-1\end{array}\right] \,, \quad {\mathbf v}_1=\left[\begin{array}{c}4\\-5\\-1\\-5\\-1\\5\end{array}\right]\, , \quad {\mathbf v}_2=\left[\begin{array}{c}-3\\2\\-1\\2\\-1\\-2\end{array}\right]\,.\end{split}\]

Question a#

Show that \({\mathbf u}_1\) and \({\mathbf u}_2\) span the same set that \({\mathbf v}_1\) and \({\mathbf v}_2\) span. In other words, show that \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2)=\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2)\).

Hint

First, show that \(({\mathbf u}_1,{\mathbf u}_2)\) is an ordered basis for \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2,{\mathbf v}_1,{\mathbf v}_2)\) by use of Theorem 9.2.1. Then, consider why this implies that \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2) = \mathrm{Span}({\mathbf u}_1,{\mathbf u}_2,{\mathbf v}_1,{\mathbf v}_2).\)

Hint

To show that \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2) = \mathrm{Span}({\mathbf u}_1,{\mathbf u}_2,{\mathbf v}_1,{\mathbf v}_2)\), find the reduced row-echelon form of the \(6 \times 4\) matrix that has the columns \({\mathbf u}_1,{\mathbf u}_2,{\mathbf v}_1,{\mathbf v}_2\).

Question b#

State a vector in \(\mathbb{R}^6\) that is not contained within \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2)\).

Hint

If \({\mathbf u}_3\) is a vector in \(\mathbb{R}^6\) such that the vectors \({\mathbf u}_1,{\mathbf u}_2,\) and \({\mathbf u}_3\) are linearly independent, can \({\mathbf u}_3\) then belong to \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2)\)?

Exercise 7: Ordered Bases#

As in Exercise 2 we will consider the span of the following three vectors from \(\mathbb{R}^3\):

\[\begin{split}{\mathbf v}_1=\left[\begin{array}{r}1\\-1\\2\end{array}\right], {\mathbf v}_2=\left[\begin{array}{r}1\\3\\2\end{array}\right], \ \mathrm{and} \ {\mathbf v}_3=\left[\begin{array}{r}1\\0\\0\end{array}\right]. \end{split}\]

Question a#

As we saw in Exercise 2, \(({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\) is a ordered basis for \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\) and thus the span has a dimension of \(3\). Show that every vector \({\mathbf v}\) in \(\mathbb{R}^3\) can be written as a linear combination of \({\mathbf v}_1,{\mathbf v}_2,\) and \({\mathbf v}_3\), and conclude that \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)=\mathbb{R}^3\).

Note: The intention behind this Exercise is to illustrate how any 3-dimensional span contained within \(\mathbb{R}^3\) is identical to \(\mathbb{R}^3\).

Hint

Investigate whether there exist scalars \(c_1,c_2,c_3 \in \mathbb{R}\) such that \(c_1{\mathbf v}_1+c_2{\mathbf v}_2+c_3{\mathbf v}_3={\mathbf v}\). In other words, investigate whether the equation

\[\begin{split}{\mathbf A} \left[ \begin{array}{r}c_1\\c_2\\c_3\end{array} \right]={\mathbf v}\end{split}\]

has a solution, where \({\mathbf A}\) is the \(3\times 3\) matrix whose columns are \({\mathbf v}_1, {\mathbf v}_2,\) and \({\mathbf v}_3\).

Answer

We know from Exercise 2, Question a, that the vectors \({\mathbf v}_1,{\mathbf v}_2,\) and \({\mathbf v}_3\) are linearly independent. Theorem 7.1.3 in the textbook tells that this implies that the matrix \({\mathbf A}\) has a rank of \(3\). Hence, \({\mathbf A}\) is invertible according to Equation (7.10) in the textbook. This reveals that the equation

\[\begin{split}{\mathbf A} \left[ \begin{array}{r}c_1\\c_2\\c_3\end{array} \right]={\mathbf v}\end{split}\]

has a solution, which is

\[\begin{split}\left[ \begin{array}{r}c_1\\c_2\\c_3\end{array} \right]={\mathbf A}^{-1}{\mathbf v}.\end{split}\]

Question b#

Show that \(({\mathbf e}_1,{\mathbf e}_2,{\mathbf e}_3)\), where

\[\begin{split}{\mathbf e}_1=\left[ \begin{array}{r}1\\0\\0\end{array} \right], {\mathbf e}_2=\left[ \begin{array}{r}0\\1\\0\end{array} \right], \ \text{and} \ {\mathbf e}_3=\left[ \begin{array}{r}0\\0\\1\end{array} \right]\,,\end{split}\]

also is an ordered basis for \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3)\).

Note: This Question illustrates how the ordered basis described in Theorem 9.2.1 is not always the “nicest” choice of ordered basis that is possible.

Answer

This follows from the result in Question a. In fact, \(({\mathbf e}_1,{\mathbf e}_2,{\mathbf e}_3)\) is known as the ordered standard basis for \(\mathbb{R}^3\) and is often considered the simplest possible choice of an ordered basis. See Example 9.2.1 in the textbook to see more about the ordered standard basis for \(\mathbb{F}^m\).

Exercise 8: Dimension and Span#

As in Exercise 1 we consider the span of the following three vectors from \(\mathbb{R}^3\):

\[\begin{split}{\mathbf w}_1=\left[\begin{array}{r}1\\-1\\2\end{array}\right], {\mathbf w}_2=\left[\begin{array}{r}1\\3\\2\end{array}\right], \ \mathrm{and} \ {\mathbf w}_3=\left[\begin{array}{r}7\\9\\14\end{array}\right]. \end{split}\]

Can \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\) be spanned by just on vector? Note that we saw in Exercise 1 that this span has a dimension of 2 - thus we would intuitively expect the answer to be no.

Hint

If \(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\) can be spanned by just one vector, say \({\mathbf u}\), then \({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3\) are all contained within \(\mathrm{Span({\mathbf u})}\). Is this possible?

Hint

If both \({\mathbf w}_1\) and \({\mathbf w}_2\) are scalar multiples of some vector \({\mathbf u}\), can the vectors \({\mathbf w}_1\) and \({\mathbf w}_2\) then be linearly independent?

Answer

\(\mathrm{Span}({\mathbf w}_1,{\mathbf w}_2,{\mathbf w}_3)\) cannot be spanned by just one vector.

“Exercise 9”: Thematic Python Module#

The Jupyter Notebook for module 2 will be released today at 15:30 on the course’s DTU Learn module.

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