Exercises – Short Day

Exercise 1: The transpose

Question a

Calculate

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

Answer

The transpose of the matrix is

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

Question b

Calculate

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

Answer

\[\begin{split}\left[ \begin{array}{cc} 35 & 44\\ 44 & 56\\ \end{array} \right].\end{split}\]

The answer is an example of a symmetric matrix, which is a matrix \(\mathbf M\) that satisfies \(\mathbf M^T=\mathbf M\).

Exercise 2: Determinants

We are given the following matrix:

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

The aim of this exercise is to find the determinant of this matrix by expanding it along a row or column.

Question a

Determine the submatrices \(\mathbf A(1;1)\), \(\mathbf A(2;1)\) and \(\mathbf A(3;1)\) as well as their determinants.

Hint

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

Answer

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

You can obtain the determinants directly by using the method from 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

Calculate \(\mathrm{det}(\mathbf A)\) using Definition 8.1.2 in the textbook. In other words, calculate \(\mathrm{det}(\mathbf A)\) by expanding the determinant along the first column.

Answer

By expanding along the first column, we get directly from Definition 8.1.2 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

For the given matrix \(\mathbf A\) it is more convenient to expand the determinant along the third row of the matrix. Why?

Hint

At the beginning of Section 8.2 it is explained what it means to expand a determinant along a row or column.

Answer

Since the third row contains two zeros, only one term of the expansion is non-zero:

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

Question d

To practise the expansion of determinants along a row, we will calculate \(\mathrm{det}(\mathbf A)\) one last time.

Determine the submatrices \(\mathbf A(2;1)\), \(\mathbf A(2;2)\) and \(\mathbf A(2;3)\) as well as their determinants. Now calculate \(\mathrm{det}(\mathbf A)\) by expanding the determinant along the second row.

Answer

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

and

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

Expanding along the second row, we now get

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

Exercise 3: Determinants and linear independence

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

\[\begin{split} \mathbf v_1=\left[ \begin{array}{c} 1\\ 0\\ 4\\ \end{array} \right], \quad \mathbf v_2=\left[ \begin{array}{c} 1\\ 1\\ 7\\ \end{array} \right], \quad \mathbf v_3=\left[ \begin{array}{c} 1\\ -1\\ 1\\ \end{array} \right]. \end{split}\]

Use Corollary 8.2.5 in the textbook to determine whether the vectors \(\mathbf v_1\), \(\mathbf v_2\) and \(\mathbf v_3\) are linearly independent.

Hint

First, form a matrix with columns \(\mathbf v_1\), \(\mathbf v_2\) and \(\mathbf v_3\). Then calculate its determinant by expanding along a row or column of your own choice. A smart choice is the one with the most zeros, as this will reduce your workload.

Hint

In Corollary 8.2.5, the determining factor is whether the determinant of the matrix mentioned in the previous hint is zero or not.

Answer

The three given vectors are linearly dependent.

Exercise 4: Linear dependence

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 all values of \(a\) for which the vectors \(\mathbf u_1\), \(\mathbf u_2\) and \(\mathbf u_3\) are linearly dependent.

Hint

Corollary 8.2.5 in the textbook can again be of use.

Hint

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 intermediate calculations, to be \(a^2+8a+12\). For which values of \(a\) is the determinant equal to zero?

Answer

Since \(a^2+8a+12=0\) if and only if \(a=-2 \, \vee a=-6\), Corollary 8.2.5 implies that the three given vectors are linearly dependent if and only if \(a \in \{-2,-6\}.\)

Exercise 5: Symmetric matrices

A square matrix \(\mathbf M\) is called symmetric if \(\mathbf M^T = \mathbf M\).

Let \(\mathbf A \in \mathbb F^{m \times n}\) be a matrix, where \(m,n\in\Bbb N\). Remember that, in this course, the field symbol \(\mathbb F\) stands for either \(\mathbb R\) or \(\mathbb C\).

Show that \(\mathbf A^T \cdot \mathbf A\) is a symmetric matrix. If you’d like an example, see your answer to Exercise 1b.

Hint

You need to show that \((\mathbf A^T \cdot \mathbf A)^T=\mathbf A^T \cdot \mathbf A\). Can Theorem 7.2.2 be of use?

Exercise 6: Transpose of 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 \(\mathbf B \in \mathbb F^{n \times \ell}\), where \(m,n,\ell\in\Bbb N\).

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

You can choose to let \(m\), \(\ell\) and \(n\) be equal.

Question b

Let \(\mathbf A \in \mathbb F^{n \times n}\) be a matrix with rank \(\rho(\mathbf A)=n\). Use the result from Question a to show that \((\mathbf A^{T})^{-1}=(\mathbf A^{-1})^T.\)

Hint

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

Hint

\(\mathbf A^{-1}\) exists because \(\rho(\mathbf A)=n\). Hence we can define the matrix \(\mathbf B=(\mathbf A^{-1})^T\).

Now use the result from Question a to show that the matrix \(\mathbf B\) has precisely the properties that the inverse of \(\mathbf A^T\) must have according to Definition 7.3.1. If it does, we can conclude that \(\mathbf B=(\mathbf A^T)^{-1}\).

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