Exercises – Long Day

Exercises – Long Day#

Exercise 1: 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 \(V_2=\mathbb{C}^{4}\).

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

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

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

Question a#

According to Definition 9.1.1 in the textbook, each vector space contains a zero vector \({\mathbf 0}\). State the zero vector in each of the above five vector spaces.

Answer

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

The zero vector is the zero polynomial, meaning a polynomial all of whose coefficients are zero.

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

Question b#

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

Hint

For each case you can find a possible basis in the textbook. See Proposition 9.2.4, as well as Examples 9.2.5, 9.2.6, and 9.2.7.

Answer

The dimensions are given as, respectively, \(4, 4, 8, \infty, 2\).

Exercise 2: Coordinates 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 consists of 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 9.3.2 from the textbook can be of use here.

Question b#

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

Hint

According to Definition 9.2.3 one must check two things for each given sequence: 1. that the three given polynomials in the sequence are linearly independent and 2. that each polynomial from \(W\) can be written as a linear combination of the three polynomials in the sequence.

Question c#

We are now given two ordered bases for \(W\), those being \(\beta=(1,Z,Z^2)\) and \(\gamma=(1,1+Z,1+Z+Z^2)\). Determine the coordinate sets of the following:

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

Hint

Definition 9.2.4 from the textbook tells us what coordinates of a vector with respect to an ordered basis are.

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: Linear Dependency vs. Independency#

Investigate whether the vectors are linearly dependent or linearly independent in the following cases. Where you find linear dependency, write one of the vectors as a linear combination of the others.

  1. The vectors \((1,i), (1+i,-1+i)\) 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 \((1,i), (1+i,-1+i)\) 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

In all cases we are to investigate whether a linear combination of the given vectors that gives zero exists without all coefficients being zero.

Case 1. can hence be solved via our theory about systems of linear equations.

Case 2. can be solved directly by investigating whether a linear combination can result in the zero vector. On can also solve the question using Theorem 9.2.3 from the textbook after having chosen an ordered basis for \(\mathbb{R}^{2\times 3}\).

Case 3. can be solved using a method similar to the one for case 2. Note that \(\mathbb{C}^{2}\) is considered as a real vector space in the exercise. Thus the coefficients in a linear combination must be real numbers.

Case 4. takes place in the real, infinite-dimensional vector space \(\mathbb{R}[Z]\), so Theorem 9.2.3 cannot be used directly. But note that

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

if and only if

\[\begin{split}c_1\cdot\left[\begin{array}{c}1\\2\\1\\0\end{array}\right]+c_2\cdot\left[\begin{array}{c}2\\7\\3\\1\end{array}\right]+c_3\cdot\left[\begin{array}{c}3\\12\\5\\2\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right].\end{split}\]

Answer

  1. The two vectors are linearly dependent. For example we have \((1+i,-1+i)=(1+i)\cdot(1,i).\)

  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 \((1+i,-1+i)\) and \((1,i)\) 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 4: Ordered Bases and Coordinates in \(\mathbb{R}^4\)#

In \(\mathbb{R}^4\) we are given five vectors:

\[\begin{split} {\mathbf v}_1=\left[\begin{array}{r}1\\-1\\2\\1\end{array}\right],\, {\mathbf v}_2=\left[\begin{array}{r}0\\1\\1\\3\end{array}\right],\, {\mathbf v}_3=\left[\begin{array}{r}1\\-2\\2\\-1\end{array}\right],\, {\mathbf v}_4=\left[\begin{array}{r}0\\1\\-1\\3\end{array}\right], \, {\mathbf v}=\left[\begin{array}{r}1\\-2\\2\\-3\end{array}\right].\end{split}\]

Justify that \(\beta=({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4)\) is an ordered basis for \(\mathbb{R}^4\,\), and determine the \(\beta\) coordinate vector \([{\mathbf v}]_\beta\). SymPy may be used to compute the reduced row-echelon form of matrices.

Hint

First determine wether the vectors \({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4\) are linearly independent using Theorem 7.1.3. SymPy can be used to determine the reduced row-echelon form of the \(4 \times 4\) matrix that has the vectors \({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4\) as its columns.

Hint

After you have shown that the vectors \({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4\) are linearly independent, then Theorem 9.2.7 from the textbook can be used to conlude that \(\beta\) is an ordered basis.

Answer

The vectors \({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4\) are linearly independent and hence constitute a basis \(\{{\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4\}\) according to Theorem 9.2.7. The sequence \(({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4)\) is thus an ordered basis.

The solution to the inhomogeneous system of linear equations

\[c_1\cdot{\mathbf v}_1+c_2\cdot{\mathbf v}_2+c_3\cdot{\mathbf v}_3+c_4\cdot{\mathbf v}_4={\mathbf v}\]

provides the wanted \(\beta\) coordinate vector. More precisely we have that

\[\begin{split} [{\mathbf v}]_\beta=\left[\begin{array}{r}2\\-1\\-1\\-1\end{array}\right]\, . \end{split}\]

The solution is found using SymPy by computing the reduced row-echelon form of the \(4 \times 5\) matrix that has the vectors \({\mathbf v}_1,{\mathbf v}_2,{\mathbf v}_3,{\mathbf v}_4,{\mathbf v}\) as its columns.

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

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

  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 9.3.2 from the textbook can be used to show 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 we have \(\dim_{\mathbb R} W_1=6.\)

  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 we have \(\dim_{\mathbb R} W_2=3.\)

  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 we have \(\dim_{\mathbb R} W_3=3.\)

Exercise 6: Subspace 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. \(W_2=\{p(Z) \in \mathbb{C}[Z] \, \mid \, 0 \text{ is a root in } p(Z) \text{ with a multiplicity of } 1\}.\)

  3. \(W_3=\{p(Z) \in \mathbb{C}[Z] \, \mid \, Z\cdot p'(Z)=p(Z)\},\) where \(p'(Z)\) denotes the derivative of \(p(Z)\).

Answer

  1. Lemma 9.3.2 from 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 has \(0\) as a root with a multiplicity of \(1\) can have \(0\) as a root with a multiplicity strictly greater than \(1\). For example: \(Z\cdot(Z-1)=Z^2-Z\) and \(Z\) are in \(W_2\), but \(Z\cdot(Z-1)+Z=Z^2\) is not in \(W_2\), because \(Z^2\) has \(0\) as a root with a multiplicity of \(2\).

  3. If \(p(Z)=a_0+a_1Z+a_2Z^2+\cdots+a_nZ^n\), then it applies that \(p'(Z)=a_1+2a_2Z+\cdots+na_nZ^{n-1}\). This can be used to show that \(W_3=\{a_1Z \, \mid \, a_1 \in \mathbb{C}\}.\) Lemma 9.3.2 from the textbook can now be used to show that \(W_3\) is a subspace of \(\mathbb{C}[Z]\).

Exercise 7: Subspaces and Systems of Linear Equations#

Justify that the solution set to the 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 a subspace in \(\mathbb{R}^5\,\). State the dimension of the subspace, and determine a basis for this subspace.

Hint

First find the general solution to the system as in Example 6.4.3 in the textbook, e.g. using SymPy. Can you see a possible basis for the solution set? You can compare with Corollary 9.3.4.

Answer

The dimension of the subspace is \(3\). The basis must contain three linearly independent vectors in its solution space. Hence there are many possibilities of a correct answer. Following the procedure from Theorem 6.4.4 to find the general solution will provide the following basis:

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

Exercise 8: Span of Vectors#

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

\[{\mathbf u}_1=(1,0,1,0,1,0)\, , \quad {\mathbf u}_2=(0,1,1,1,1,-1)\,, \quad {\mathbf v}_1=(4,-5,-1,-5,-1,5)\, , \quad {\mathbf v}_2=(-3,2,-1,2,-1,-2)\,.\]

Question a#

Show that \({\mathbf u}_1\) and \({\mathbf u}_2\) span the same subspace in \(\mathbb{R}^6\) as \({\mathbf v}_1\) and \({\mathbf v}_2\). 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 \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2) \subseteq \mathrm{Span}({\mathbf v}_1,{\mathbf v}_2)\) and afterwards that \(\mathrm{Span}({\mathbf v}_1,{\mathbf v}_2) \subseteq \mathrm{Span}({\mathbf u}_1,{\mathbf u}_2)\).

Hint

To show that \(\mathrm{Span}({\mathbf u}_1,{\mathbf u}_2) \subseteq \mathrm{Span}({\mathbf v}_1,{\mathbf v}_2)\) it is sufficient to show that \({\mathbf u}_1 \in \mathrm{Span}({\mathbf v}_1,{\mathbf v}_2)\) and \({\mathbf u}_2 \in \mathrm{Span}({\mathbf v}_1,{\mathbf v}_2)\).

Question b#

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

Exercise 9: Subspaces#

Let \(V\) be a vector space over a field \(\mathbb F\). Given two subspaces \(W_1\), \(W_2\) of \(V\), show that \(W_1 \cap W_2\) is also a subspace of \(V\).