Exercises – Short Day

Exercises – Short Day#

Exercise 1: Transposed Matrices#

Compute

\[\begin{split}\left[ \begin{array}{cc} 1 & 2\\ 3 & 4\\ 5 & 6\\ \end{array} \right]^T\end{split}\]

both by hand and using SymPy.

Answer

The transposed matrix is

\[\begin{split}\left[ \begin{array}{ccc} 1 & 3 & 5\\ 2 & 4 & 6\\ \end{array} \right].\end{split}\]

In SymPy the transposed of a matrix A can be computed using the command A.transpose().

Exercise 2: Determinants#

We are given the following matrix:

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 3 & 5\\ 2 & 4 & 6\\ 3 & 0 & 0\\ \end{array} \right]. \end{split}\]

Question a#

Determine the matrices \({\mathbf A}(1;1)\), \({\mathbf A}(2;1)\), and \({\mathbf A}(3;1)\) and compute their determinants.

Hint

In Definition 8.1.1 in the textbook the notation \({\mathbf A}(i;j)\) is explained.

Answer

\[\begin{split} {\mathbf A}(1;1)=\left[ \begin{array}{ccc} 4 & 6\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(2;1)=\left[ \begin{array}{ccc} 3 & 5\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(3;1)=\left[ \begin{array}{ccc} 3 & 5\\ 4 & 6\\ \end{array} \right]. \end{split}\]

The determinants are found directly via the same procedure as in Example 8.1.1 in the textbook:

\[\mathrm{det}({\mathbf A(1;1)})=0 \quad \mathrm{det}({\mathbf A(2;1)})=0 \quad \mathrm{det}({\mathbf A(3;1)})=-2.\]

Question b#

Compute \(\mathrm{det}({\mathbf A})\) by hand using Definition 8.1.2 from the textbook. In other words, compute \(\mathrm{det}({\mathbf A})\) by expanding the determinant along the first column.

Answer

By expanding along the first column, Definition 8.1.2 tells us that

\[\mathrm{det}({\mathbf A})=1\cdot \mathrm{det}({\mathbf A}(1;1))-2\cdot \mathrm{det}({\mathbf A}(2;1))+3\cdot \mathrm{det}({\mathbf A}(3;1))=-6.\]

Question c#

Compute the matrices \({\mathbf A}(2;1)\), \({\mathbf A}(2;2)\), and \({\mathbf A}(2;3)\) as well as their determinants.

Answer

\[\begin{split} {\mathbf A}(2;1)=\left[ \begin{array}{ccc} 3 & 5\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(2;2)=\left[ \begin{array}{ccc} 1 & 5\\ 3 & 0\\ \end{array} \right] \quad {\mathbf A}(2;3)=\left[ \begin{array}{ccc} 1 & 3\\ 3 & 0\\ \end{array} \right] \end{split}\]

and

\[\mathrm{det}({\mathbf A(2;1)})=0 \quad \mathrm{det}({\mathbf A(2;2)})=-15 \quad \mathrm{det}({\mathbf A(2;3)})=-9.\]

Question d#

Compute \(\mathrm{det}({\mathbf A})\) by hand by expanding the determinant along the 2nd row.

Answer

By expanding along the 2nd row, we get

\[\mathrm{det}({\mathbf A})=-2\cdot \mathrm{det}({\mathbf A}(2;1))+4\cdot \mathrm{det}({\mathbf A}(2;2))-6\cdot \mathrm{det}({\mathbf A}(2;3))=-6.\]

Question e#

In this case it is convenient to expand the determinant along the 3rd row. Why?

Answer

Due to the many zeroes in the 3rd row of the matrix, we see that:

\[ \mathrm{det}({\mathbf A})=3\cdot \mathrm{det}({\mathbf A}(3;1))=3 \cdot (3 \cdot 6-4 \cdot 5)=-6. \]

Question f#

Lastly, compute \(\mathrm{det}({\mathbf A})\) using SymPy.

Hint

The determinant of a square matrix A can be computed with the command A.det().

Exercise 3: Determinants and Linear Independency#

We are given the following four vectors in \(\mathbb{C}^4\):

\[\begin{split} {\mathbf v}_1=\left[ \begin{array}{c} 1\\ 1\\ 0\\ i\\ \end{array} \right] \quad {\mathbf v}_2=\left[ \begin{array}{c} 4+i\\ 1\\ -i\\ 0\\ \end{array} \right] \quad {\mathbf v}_3=\left[ \begin{array}{c} 1+i\\ 1-i\\ 5i\\ -7\\ \end{array} \right] \quad {\mathbf v}_4=\left[ \begin{array}{c} -6\\ 0\\ 7i\\ -8+i\\ \end{array} \right] . \end{split}\]

Use Corollary 8.3.5 from the textbook to determine whether the vectors \({\mathbf v}_1, {\mathbf v}_2, {\mathbf v}_3,\) and \({\mathbf v}_4\) are linearly independent or not. SymPy can be used to compute determinants.

Hint

Complex numbers such as \(7i\) and \(-8+i\) are written in SymPy as 7*I and -8+I.

Answer

The four given vectors are linearly dependent.

Exercise 4: Linear Dependency#

We are given the vectors

\[\begin{split} {\mathbf u}_1=\left[ \begin{array}{c} 1\\ 1\\ -6\\ \end{array} \right] \quad {\mathbf u}_2=\left[ \begin{array}{c} 1\\ 1\\ a\\ \end{array} \right] \quad {\mathbf u}_3=\left[ \begin{array}{c} a\\ -2\\ 0\\ \end{array} \right], \end{split}\]

where \(a \in \mathbb{R}\) is a constant. Determine by hand all values of the constant \(a\) such that the vectors \({\mathbf u}_1\), \({\mathbf u}_2\), and \({\mathbf u}_3\) are linearly independent.

Hint

Corrolary 8.3.5 from the textbook can again be of use.

Answer

The determinant of the matrix

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 1 & a\\ 1 & 1 & -2\\ -6 & a & 0\\ \end{array} \right] \end{split}\]

turns out, after a few calculation steps, to be equal to \(a^2+8a+12\). Since \(a^2+8a+12=0\) if and only if \(a=-2 \, \vee a=-6\), we can thus via Corollary 8.3.5 conclude that the three given vectors are linearly dependent if and only if \(a \in \{-2,-6\}.\)

Exercise 5: Transposed Square Matrices#

We know from Theorem 7.2.2 in the textbook that \(({\mathbf A} \cdot {\mathbf B})^T={\mathbf B}^T \cdot {\mathbf A}^T\) for all \({\mathbf A} \in \mathbb{F}^{m \times n}\) and \({\mathbf B} \in \mathbb{F}^{n \times \ell}\). The symbol \(\mathbb{F}\) represents \(\mathbb{R}\) or \(\mathbb{C}.\)

Question a#

Conclude that \(({\mathbf A} \cdot {\mathbf B})^T={\mathbf B}^T \cdot {\mathbf A}^T\) for all \({\mathbf A},{\mathbf B} \in \mathbb{F}^{n \times n}\).

Hint

\(m\) and \(\ell\) can be set equal to \(n\).

Question b#

Assume that \(\mathbf A \in \mathbb{F}^{n \times n}\) and that \(\rho({\mathbf A})=n\). Use the result from Question a to realise that \(({\mathbf A}^{T})^{-1}=({\mathbf A}^{-1})^T.\)

Hint

Begin by figuring out which proporties the inverse matrix of \({\mathbf A}^{T}\) should have according to Definition 7.3.1.

Hint

\({\mathbf A}^{-1}\) exists since \(\rho({\mathbf A})=n\). Hence we can define the matrix \(\mathbf B=({\mathbf A}^{-1})^T\). Now use the result from Question a to realise that matrix \(\mathbf B\) has exactly those proporties that the inverse of \({\mathbf A}^T\) should have according to Definition 7.3.1. If these proporties are achieved, then one can conclude that \({\mathbf B}=({\mathbf A}^T)^{-1}\).