Exercises – Long Day

Exercises – Long Day#

Exercise 1: A few Examples of Vector Spaces#

In this exercise we consider the following examples of vector spaces:

  1. The real vector space \(V_1=\mathbb{R}^{4}\).

  2. The complex vector space \(V_2=\mathbb{C}^{3}\).

Question a#

According to Definition 10.1.1 in the textbook, every vector space has a zero vector \({\mathbf 0}\). State the zero vector in each of the above vector spaces.

Answer

\[\begin{split}{\mathbf 0}= \left[\begin{array}{c} 0\\ 0\\ 0\\ 0 \end{array}\right].\end{split}\]
\[\begin{split}{\mathbf 0}= \left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right].\end{split}\]

Question b#

State an ordered basis as well as a basis for the given vector spaces. What are the dimensions of the spaces?

Hint

For the real vector space \(\mathbb{R}^{4}\) you can find a possible basis in Example 9.2.1.

Answer

We are here giving an answer to the real vector space \(\mathbb{R}^{4}\). A possible ordered basis is the ordered standard basis from Example 9.2.1, which is

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

The dimension of the space is hence \(4\).

A basis for \(\mathbb{R}^{4}\) is achieved by taking the set of vectors that are contained within its ordered standard basis. A possible basis is hence:

\[\begin{split}\left\{ \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right], \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right], \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right] \right\}.\end{split}\]

Exercise 2: The Coordinate Vector of a Vector with respect to an Ordered Basis#

Let \(W= \{a+bZ+cZ^2 \, \mid \, a,b,c\in \mathbb{C}\} \subset \mathbb{C}[Z]\) be the set that contains all polynomials of degree no higher than two.

Question a#

Check that \(W\) is a subspace of the complex vector space \(\mathbb{C}[Z]\).

Hint

Lemma 10.4.2 in the textbook can be of use here.

Question b#

Check that the list \((1,Z,Z^2)\) is an ordered basis for \(W\). Check that the list \((1,1+Z,1+Z+Z^2)\) also is an ordered basis for \(W\).

Hint

According to Definition 10.2.4 we must check two things for each given list: Firstly that the three polynomials in the list are linearly independent and secondly that each polynomial from \(W\) can be written as a linear combination of the three polynomials in the list.

Question c#

We consider the two ordered bases for \(W\) from Question b, those being \(\beta=(1,Z,Z^2)\) and \(\gamma=(1,1+Z,1+Z+Z^2)\). Determine the following coordinate vectors:

\[\left[2+5Z+Z^2\right]_\beta \, , \quad \left[5Z+10Z^2\right]_\beta\]

and

\[\left[2+5Z+Z^2\right]_\gamma \, , \quad \left[5Z+10Z^2\right]_\gamma \, .\]

Hint

If you are in doubt about what a coordinate vector of a vector with respect to an ordered basis is, then see Definition 10.2.5 in the textbook.

Answer

\[\begin{split} \left[2+5Z+Z^2\right]_\beta= \left[\begin{array}{c} 2\\ 5\\ 1\\ \end{array}\right], \quad \left[5Z+10Z^2\right]_\beta= \left[\begin{array}{c} 0\\ 5\\ 10\\ \end{array}\right], \quad \left[2+5Z+Z^2\right]_\gamma= \left[\begin{array}{c} -3\\ 4\\ 1\\ \end{array}\right], \quad \left[5Z+10Z^2\right]_\gamma= \left[\begin{array}{c} -5\\ -5\\ 10\\ \end{array}\right].\end{split}\]

Exercise 3: More Examples of Vector Spaces#

In this exercise we will ask the same questions as in Exercise 1, but this time for these new vector spaces:

  1. The real vector space \(V_3=\mathbb{R}^{4 \times 2}\).

  2. The complex vector space \(V_4=\mathbb{C}[Z]\).

  3. The real vector space \(V_5=\mathbb{C}\).

Question a#

State the zero vector in each of the above vector spaces.

Answer

\[\begin{split}{\mathbf 0}= \left[\begin{array}{cc} 0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ \end{array}\right].\end{split}\]

The zero vector is the zero polynomial \(0\), meaning the polynomial whose coefficients are all zero.

The zero vector is the complex number \(0\).

Question b#

Provide a basis for each of the given vector spaces and determine the dimensions of the spaces.

Hint

You can find a possible basis for each of them in the textbook. See Examples 10.2.5, 10.2.6, and 10.2.7.

Answer

Possible bases for the vector spaces are, respectively,

\[\begin{split}\left\{ \left[\begin{array}{cc} 1 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 1 & 0\\ 0 & 0\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 0 & 0\\ 1 & 0\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 0 & 0\\ 0 & 0\\ 1 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 1\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 0 & 1\\ 0 & 0\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 0 & 0\\ 0 & 1\\ 0 & 0\\ \end{array}\right], \left[\begin{array}{cc} 0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 1\\ \end{array}\right] \right\},\end{split}\]

\(\{1,Z,Z^2,Z^3,\dots\}\), and \(\{1,i\}\). The dimensions are, respectively, \(8, \infty, 2\).

Exercise 4: Linear Dependence or Independence#

Investigate whether the vectors are linearly dependent or independent in the following cases. In case of linear dependence, write one of the vectors as a linear combination of the others.

  1. The vectors \(\left[ \begin{array}{r} 1 \newline i \end{array}\right], \left[ \begin{array}{r} 1+i \newline -1+i \end{array}\right]\) in the complex vector space \(\mathbb{C}^{2}\).

  2. The matrices \(\left[ \begin{array}{rrr} 1 & 2 & 0 \newline 1 & 1 & 1 \end{array}\right], \left[ \begin{array}{rrr} 1 & 1 & 2 \newline 0 & 0 & 1 \end{array}\right], \left[ \begin{array}{rrr} 2 & 5 & -2 \newline 3 & 3 & 2 \end{array}\right]\) in the real vector space \(\mathbb{R}^{2\times 3}\).

  3. The vectors \(\left[ \begin{array}{r} 1 \newline i \end{array}\right], \left[ \begin{array}{r} 1+i \newline -1+i \end{array}\right]\) in the real vector space \(\mathbb{C}^{2}\).

  4. The polynomials \(1 + 2Z + Z^{2}, \, 2 + 7Z +3Z^{2} + Z^{3}, \, 3 + 12Z + 5Z^{2} +2Z^{3}\) in the real vector space \(\mathbb{R}[Z]\).

Hint

According to Definition 10.2.1 in the textbook we must investigate whether a linear combination of the given vectors can equal zero by use of some non-zero scalars.

Case 1. can be solved directly using our theory on systems of linear equations.

Case 2. can be solved directly by investigating whether a linear combination can produce the zero vector. Alternatively, one can also solve this case using Theorem 10.2.4 in the textbook after having chosen an ordered basis for \(\mathbb{R}^{2\times 3}\).

Case 3. can be solved in a similar manner as Case 2. Although, note that \(\mathbb{C}^{2}\) is considered as a real vector space here. Thus, the scalars in a linear combination must be real numbers.

Case 4. is in principle taking place in the real, infinite-dimensional vector space \(\mathbb{R}[Z]\). Although, note that the equation

\[c_1\cdot(1 + 2Z + Z^{2})+c_2\cdot(2 + 7Z +3Z^{2} + Z^{3})+c_3\cdot(3 + 12Z + 5Z^{2} +2Z^{3})=0\]

immediately gives rise to a system of linear equations in the unknowns \(c_1,c_2,\) and \(c_3\). Which system would that be?

Answer

  1. The two vectors are linearly dependent. For example, we have \(\left[ \begin{array}{r} 1+i \newline -1+i \end{array}\right] = (1+i)\left[ \begin{array}{r} 1 \newline i \end{array}\right]\, . \)

  2. The three matrices are linearly dependent. For example, we have

\[\begin{split} \left[ \begin{array}{rrr} 2 & 5 & -2 \\ 3 & 3 & 2 \end{array}\right] = 3\left[ \begin{array}{rrr} 1 & 2 & 0 \\ 1 & 1 & 1 \end{array}\right] - \left[ \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]\,. \end{split}\]

  1. The vectors \(\left[ \begin{array}{r} 1 \newline i \end{array}\right]\) og \(\left[ \begin{array}{r} 1+i \newline -1+i \end{array}\right]\) are linearly independent if \(\mathbb{C}^2\) is considered as a real vector space.

  2. The three polynomials are linearly dependent. For example, we have

\[ 3 + 12Z + 5Z^{2} +2Z^{3}=-(1 + 2Z + Z^{2})+2\cdot(2 + 7Z +3Z^{2} + Z^{3}). \]

Exercise 5: Subspace of \(\mathbb{R}^{3 \times 3}\)#

Let \(V=\mathbb{R}^{3 \times 3}\) be a real vector space of all \(3 \times 3\) matrices with real coefficients. We are given the following three subsets of \(V\):

  1. \(W_1\) is the set of all upper triangular matrices in \(V\).

  2. \(W_2\) is the set of all diagonal matrices in \(V\).

  3. \(W_3=\left\{ {\mathbf A} \in \mathbb{R}^{3 \times 3} \, \mid \, {\mathbf A} = -{\mathbf A}^T\right\}.\)

Question a#

Check that \(W_1,W_2,\) and \(W_3\) all are subspaces of \(V\).

Hint

Lemma 10.4.2 in the textbook can be of good use for showing that a subset of a vector space is a subspace.

Question b#

Find a basis for the three subspaces \(W_1,W_2,\) and \(W_3\). What are the dimensions of the subspaces?

Answer

  1. A possible basis for \(W_1\) is

\[\begin{split}\left\{ \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\\ \end{array}\right] \right\}.\end{split}\]

Hence, \(\dim_{\mathbb R}(W_1)=6.\)

  1. A possible basis for \(W_2\) is

\[\begin{split}\left\{ \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\\ \end{array}\right] \right\}.\end{split}\]

Hence, \(\dim_{\mathbb R}(W_2)=3.\)

  1. A possible basis for \(W_3\) is

\[\begin{split}\left\{ \left[ \begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\\ \end{array}\right], \left[ \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\\ \end{array}\right] \right\}.\end{split}\]

Hence, \(\dim_{\mathbb R}(W_3)=3.\)

Exercise 6: Subspaces of \(\mathbb{C}[Z]\)#

Which of the following three sets are subspaces of the complex vector space \(\mathbb{C}[Z]\)?

  1. \(W_1=\{p(Z) \in \mathbb{C}[Z] \, \mid \, p(0)=0\}.\)

  2. \(W_2=\{p(Z) \in \mathbb{C}[Z] \, \mid \, 0 \text{ is a root of } p(Z) \text{ with a multiplicity of } 1\}.\)

  3. \(W_3=\{p(Z) \in \mathbb{C}[Z] \, \mid \, Z\cdot p'(Z)=p(Z)\},\) where \(p'(Z)\) denotes the derivative of \(p(Z)\). Meaning, if \(p(Z)=a_0+a_1Z+a_2Z^2+\cdots+a_nZ^n\), then \(p'(Z)=a_1+2a_2Z+\cdots+na_nZ^{n-1}\).

Answer

  1. Lemma 10.4.2 in the textbook can be used to show that \(W_1\) is a subspace of \(\mathbb{C}[Z]\).

  2. \(W_2\) is not a subspace of \(\mathbb{C}[Z]\). The problem is that a linear combination of polynomials that each have the root \(0\) with a multiplicity of \(1\) can have the root \(0\) with a multiplicity strictly greater than \(1\). For example: \(Z\cdot(Z-1)=Z^2-Z\) and \(Z\) is in \(W_2\), but \(Z\cdot(Z-1)+Z=Z^2\) is not in \(W_2\), since \(Z^2\) has the root \(0\) with a multiplicity of \(2\).

  3. \(p(Z)=p'(Z)\) precisely if \(p(Z)=a_1Z\) with \(a_1 \in \mathbb{C}\). Therefore we have \(W_3=\{a_1Z \, \mid \, a_1 \in \mathbb{C}\}.\) But then it applies that \(W_3=\mathrm{Span}(Z)\), and Proposition 10.4.3 in the textbook then implies that \(W_3\) is a subspace of \(\mathbb{C}[Z]\).

Exercise 7: Subspaces and Systems of Linear Equations#

We consider the following system of linear equations over \(\mathbb R\):

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

Note that we investigated this system in Exercise 5 from Long Day in week 8. The solution set of the system is denoted by \(W\).

Question a#

Justify that \(W\) is a subspace in \(\mathbb{R}^5\,\), determine \(\dim(W)\), and choose a basis for \(W\).

Hint

In Exercise 5 from Long Day in week 8 it has already been shown that the solution set is a span of some vectors. This implies that the solution set is a subspace in \(\mathbb{R}^5\,\)?

Answer

Proposition 10.4.3 along with our observation from the hint imply that \(W\) is a subspace in \(\mathbb{R}^5\,\). In Exercise 5 from Long Day in week 8 we already found an ordered basis for \(W\) as well as its dimension, which is \(3\). If we use the ordered basis given in the answer to Exercise 5 from Long Day in week 8, then we have the following possible basis for \(W\):

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

Question b#

Let \(\beta\) be an ordered basis for \(W\). We are given three vectors \(\mathbf{v}_1,\mathbf{v}_2,\) and \(\mathbf{v}_3\) in \(W\) that fulfill that

\[\begin{split}[\mathbf{v}_1]_\beta=\left[\begin{array}{c} 1\\ 0 \\ -2\end{array}\right]\, , \quad [\mathbf{v}_2]_\beta=\left[\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right] \quad \text{og} \quad [\mathbf{v}_3]_\beta=\left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right].\end{split}\]

Is \(\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}\) a basis for \(W\)?

Hint

First, try showing that the set \(\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}\) is linearly independent. Start by reading Theorem 10.2.4 in the textbook.

Hint

Can Theorem 10.2.7 in the textbook be of use?

Answer

Yes, \(\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}\) is a possible basis for \(W\).

Exercise 8: Subspaces#

Let \(V\) be a vector space over a field \(\mathbb F\).

Question a#

Given two subspaces \(W_1\), \(W_2\) in \(V\), show that \(W_1 \cap W_2\) is also a subspace in \(V\).

Question b#

Find an example of a real vector space \(V\) along with two subspaces \(W_1\), \(W_2\) in \(V\) such that \(W_1 \cup W_2\) is not a subspace in \(V\).

Remark#

The Python test on the notebook from Python module 2 released last week will open in Möbius at 15:30 and close at 17:00.

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