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Vectors

Exercises · Vectors and matrices · with worked solutions · Lecture notes (PDF) · Chapter 3: Vectors

Exercise 1

Consider the vectors

\[ \boldsymbol p=\begin{pmatrix}2\\-1\\3\end{pmatrix}, \qquad \boldsymbol w=\begin{pmatrix}-1\\4\\0\end{pmatrix}\in\R^{3}. \]

Compute \(\boldsymbol p+\boldsymbol w\), \(3\boldsymbol p\) and \(2\boldsymbol p-\boldsymbol w\), and verify that \(\boldsymbol p+\boldsymbol w=\boldsymbol w+\boldsymbol p\).

Solution

Sum and scalar multiplication are computed entry by entry:

\[ \boldsymbol p+\boldsymbol w= \begin{pmatrix}2+(-1)\\-1+4\\3+0\end{pmatrix} = \begin{pmatrix}1\\3\\3\end{pmatrix}, \qquad 3\boldsymbol p= \begin{pmatrix}3\cdot 2\\3\cdot(-1)\\3\cdot 3\end{pmatrix} = \begin{pmatrix}6\\-3\\9\end{pmatrix}. \]

For the last vector, \(2\boldsymbol p-\boldsymbol w=2\boldsymbol p+(-1)\boldsymbol w\):

\[ 2\boldsymbol p-\boldsymbol w= \begin{pmatrix}4\\-2\\6\end{pmatrix} - \begin{pmatrix}-1\\4\\0\end{pmatrix} = \begin{pmatrix}4+1\\-2-4\\6-0\end{pmatrix} = \begin{pmatrix}5\\-6\\6\end{pmatrix}. \]

Finally, \( \boldsymbol w+\boldsymbol p= \begin{pmatrix}-1+2 & 4-1 & 0+3\end{pmatrix}' = \begin{pmatrix}1 & 3 & 3\end{pmatrix}' =\boldsymbol p+\boldsymbol w, \) as stated by the commutative property of the sum.

Exercise 2

Consider the vector \( \boldsymbol a=\begin{pmatrix}1 & -2 & 0 & 5\end{pmatrix}'. \)

  1. Write \(\boldsymbol a\) as a column vector and give its dimension.

  2. Write \(\boldsymbol a'\) and \((\boldsymbol a')'\).

  3. Compute \(\boldsymbol a'\boldsymbol a\).

Solution

  1. The vector is the transpose of a row vector with \(4\) entries, hence

    \[ \boldsymbol a=\begin{pmatrix}1\\-2\\0\\5\end{pmatrix}\in\R^{4}, \]

    and its dimension is \(n=4\).

  2. Transposing a column vector gives a row vector, and transposing twice gives back the original vector:

    \[ \boldsymbol a'=\begin{pmatrix}1 & -2 & 0 & 5\end{pmatrix}\in\R^{1\times 4}, \qquad (\boldsymbol a')'=\boldsymbol a=\begin{pmatrix}1\\-2\\0\\5\end{pmatrix}. \]

  3. By definition of scalar product,

    \[ \boldsymbol a'\boldsymbol a=1\cdot 1+(-2)\cdot(-2)+0\cdot 0+5\cdot 5=1+4+0+25=30. \]

    As expected, \(\boldsymbol a'\boldsymbol a\ge 0\), and \(\boldsymbol a'\boldsymbol a\neq 0\) since \(\boldsymbol a\neq\boldsymbol 0\).

Exercise 3

Consider the vectors

\[ \boldsymbol p=\begin{pmatrix}1\\2\\-3\end{pmatrix}, \qquad \boldsymbol w=\begin{pmatrix}4\\-1\\2\end{pmatrix}, \qquad \boldsymbol u=\begin{pmatrix}0\\3\\1\end{pmatrix}\in\R^{3}. \]

  1. Compute \(\boldsymbol p'\boldsymbol w\) and \(\boldsymbol w'\boldsymbol p\).

  2. Compute \(\boldsymbol p'(\boldsymbol w+\boldsymbol u)\) and verify that it is equal to \(\boldsymbol p'\boldsymbol w+\boldsymbol p'\boldsymbol u\).

  3. Compute \((2\boldsymbol p)'\boldsymbol w\) and compare it with \(2\,(\boldsymbol p'\boldsymbol w)\).

Solution

1.

$$
\boldsymbol p'\boldsymbol w=1\cdot 4+2\cdot(-1)+(-3)\cdot 2=4-2-6=-4,
\qquad
\boldsymbol w'\boldsymbol p=4\cdot 1+(-1)\cdot 2+2\cdot(-3)=-4.
$$

The two values coincide (symmetry of the scalar product).

  1. We have \(\boldsymbol w+\boldsymbol u=\begin{pmatrix}4 & 2 & 3\end{pmatrix}'\), hence

    \[ \boldsymbol p'(\boldsymbol w+\boldsymbol u)=1\cdot 4+2\cdot 2+(-3)\cdot 3=4+4-9=-1. \]

    On the other hand, \( \boldsymbol p'\boldsymbol u=1\cdot 0+2\cdot 3+(-3)\cdot 1=3 \), so \(\boldsymbol p'\boldsymbol w+\boldsymbol p'\boldsymbol u=-4+3=-1\), as expected.

  2. We have \(2\boldsymbol p=\begin{pmatrix}2 & 4 & -6\end{pmatrix}'\), hence

    \[ (2\boldsymbol p)'\boldsymbol w=2\cdot 4+4\cdot(-1)+(-6)\cdot 2=8-4-12=-8=2\cdot(-4)=2\,(\boldsymbol p'\boldsymbol w). \]

Exercise 4

Two vectors \(\boldsymbol p,\boldsymbol w\in\R^{n}\) are called orthogonal if their scalar product is zero, i.e., \(\boldsymbol p'\boldsymbol w=0\).

  1. Verify that \(\begin{pmatrix}1 & 2 & 2\end{pmatrix}'\) and \(\begin{pmatrix}2 & -2 & 1\end{pmatrix}'\) are orthogonal.

  2. Find the value of \(\alpha\in\R\) such that \( \boldsymbol p=\begin{pmatrix}1 & \alpha & 2\end{pmatrix}' \) and \( \boldsymbol w=\begin{pmatrix}3 & -1 & \alpha\end{pmatrix}' \) are orthogonal.

Solution

  1. \( \begin{pmatrix}1 & 2 & 2\end{pmatrix}\begin{pmatrix}2\\-2\\1\end{pmatrix}=1\cdot 2+2\cdot(-2)+2\cdot 1=2-4+2=0, \) hence the two vectors are orthogonal.

  2. We compute the scalar product as a function of \(\alpha\):

    \[ \boldsymbol p'\boldsymbol w=1\cdot 3+\alpha\cdot(-1)+2\cdot\alpha=3+\alpha. \]

    Imposing \(\boldsymbol p'\boldsymbol w=0\) we get \(3+\alpha=0\), i.e., \(\alpha=-3\). Check: with \(\alpha=-3\), \(\boldsymbol p=\begin{pmatrix}1 & -3 & 2\end{pmatrix}'\), \(\boldsymbol w=\begin{pmatrix}3 & -1 & -3\end{pmatrix}'\) and \(\boldsymbol p'\boldsymbol w=3+3-6=0\).

Exercise 5

Consider the vectors \( \boldsymbol v_1=\begin{pmatrix}1 & 0 & 2\end{pmatrix}' \) and \( \boldsymbol v_2=\begin{pmatrix}-1 & 3 & 1\end{pmatrix}'. \) Compute the linear combinations \(3\boldsymbol v_1-2\boldsymbol v_2\) and \(-\boldsymbol v_1+\boldsymbol v_2\).

Solution
\[ 3\boldsymbol v_1-2\boldsymbol v_2 = \begin{pmatrix}3\\0\\6\end{pmatrix} + \begin{pmatrix}2\\-6\\-2\end{pmatrix} = \begin{pmatrix}5\\-6\\4\end{pmatrix}, \qquad -\boldsymbol v_1+\boldsymbol v_2 = \begin{pmatrix}-1\\0\\-2\end{pmatrix} + \begin{pmatrix}-1\\3\\1\end{pmatrix} = \begin{pmatrix}-2\\3\\-1\end{pmatrix}. \]

Exercise 6

Consider the vectors \( \boldsymbol v_1=\begin{pmatrix}1 & 1 & 0\end{pmatrix}' \) and \( \boldsymbol v_2=\begin{pmatrix}0 & 1 & 2\end{pmatrix}'. \) Determine whether the following vectors are linear combinations of \(\boldsymbol v_1\) and \(\boldsymbol v_2\) and, if so, find the coefficients:

\[ \boldsymbol w=\begin{pmatrix}2\\5\\6\end{pmatrix}, \qquad \boldsymbol u=\begin{pmatrix}1\\2\\3\end{pmatrix}. \]
Solution

We look for \(\lambda_1,\lambda_2\in\R\) such that \(\lambda_1\boldsymbol v_1+\lambda_2\boldsymbol v_2\) equals the given vector. Since \( \lambda_1\boldsymbol v_1+\lambda_2\boldsymbol v_2=\begin{pmatrix}\lambda_1 & \lambda_1+\lambda_2 & 2\lambda_2\end{pmatrix}', \) we compare the entries.

  • For \(\boldsymbol w\):

    \[ \begin{cases} \lambda_1=2\\ \lambda_1+\lambda_2=5\\ 2\lambda_2=6 \end{cases} \]

    The first equation gives \(\lambda_1=2\), the second one \(\lambda_2=3\), and the third one is satisfied (\(2\cdot 3=6\)). Hence \(\boldsymbol w=2\boldsymbol v_1+3\boldsymbol v_2\).

  • For \(\boldsymbol u\):

    \[ \begin{cases} \lambda_1=1\\ \lambda_1+\lambda_2=2\\ 2\lambda_2=3 \end{cases} \]

    The first two equations give \(\lambda_1=1\) and \(\lambda_2=1\), but then \(2\lambda_2=2\neq 3\). The system has no solution, hence \(\boldsymbol u\) is not a linear combination of \(\boldsymbol v_1\) and \(\boldsymbol v_2\).

Exercise 7

Determine whether the following pairs of vectors of \(\R^2\) are linearly independent:

\[ \text{(a)}\quad \begin{pmatrix}2\\1\end{pmatrix},\ \begin{pmatrix}4\\2\end{pmatrix}; \qquad\qquad \text{(b)}\quad \begin{pmatrix}2\\1\end{pmatrix},\ \begin{pmatrix}1\\3\end{pmatrix}. \]
Solution

  • (a) The second vector is twice the first one: \(\begin{pmatrix}4 & 2\end{pmatrix}'=2\begin{pmatrix}2 & 1\end{pmatrix}'\). Hence, with \(\lambda_1=2\) and \(\lambda_2=-1\),

    \[ 2\begin{pmatrix}2\\1\end{pmatrix}-1\begin{pmatrix}4\\2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}, \]

    a linear combination equal to \(\boldsymbol 0\) with coefficients not all zero: the vectors are linearly dependent.

  • (b) The condition \(\lambda_1\begin{pmatrix}2 & 1\end{pmatrix}'+\lambda_2\begin{pmatrix}1 & 3\end{pmatrix}'=\boldsymbol 0\) reads

    \[ \begin{cases} 2\lambda_1+\lambda_2=0\\ \lambda_1+3\lambda_2=0 \end{cases} \]

    From the second equation \(\lambda_1=-3\lambda_2\); substituting in the first one, \(-6\lambda_2+\lambda_2=-5\lambda_2=0\), hence \(\lambda_2=0\) and \(\lambda_1=0\). The vectors are linearly independent.

Exercise 8

Determine whether the following sets of vectors of \(\R^3\) are linearly independent. If they are dependent, find a linear combination equal to \(\boldsymbol 0\) with coefficients not all zero.

  1. \(\boldsymbol v_1=\begin{pmatrix}1 & 2 & 1\end{pmatrix}'\), \(\boldsymbol v_2=\begin{pmatrix}0 & 1 & 1\end{pmatrix}'\), \(\boldsymbol v_3=\begin{pmatrix}1 & 3 & 2\end{pmatrix}'\).

  2. \(\boldsymbol w_1=\begin{pmatrix}1 & 0 & 0\end{pmatrix}'\), \(\boldsymbol w_2=\begin{pmatrix}1 & 1 & 0\end{pmatrix}'\), \(\boldsymbol w_3=\begin{pmatrix}1 & 1 & 1\end{pmatrix}'\).

Solution

  1. The condition \(\lambda_1\boldsymbol v_1+\lambda_2\boldsymbol v_2+\lambda_3\boldsymbol v_3=\boldsymbol 0\) reads

    \[ \begin{cases} \lambda_1+\lambda_3=0\\ 2\lambda_1+\lambda_2+3\lambda_3=0\\ \lambda_1+\lambda_2+2\lambda_3=0 \end{cases} \]

    From the first equation \(\lambda_1=-\lambda_3\). Substituting in the second one: \(-2\lambda_3+\lambda_2+3\lambda_3=0\), i.e., \(\lambda_2=-\lambda_3\). The third equation becomes \(-\lambda_3-\lambda_3+2\lambda_3=0\), which holds for every \(\lambda_3\). Choosing \(\lambda_3=1\) we get \(\lambda_1=\lambda_2=-1\):

    \[ -\boldsymbol v_1-\boldsymbol v_2+\boldsymbol v_3 = \begin{pmatrix}-1-0+1\\-2-1+3\\-1-1+2\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}. \]

    The vectors are linearly dependent (indeed \(\boldsymbol v_3=\boldsymbol v_1+\boldsymbol v_2\)).

  2. The condition \(\lambda_1\boldsymbol w_1+\lambda_2\boldsymbol w_2+\lambda_3\boldsymbol w_3=\boldsymbol 0\) reads

    \[ \begin{cases} \lambda_1+\lambda_2+\lambda_3=0\\ \lambda_2+\lambda_3=0\\ \lambda_3=0 \end{cases} \]

    Solving from the last equation upwards: \(\lambda_3=0\), then \(\lambda_2=0\), then \(\lambda_1=0\). The vectors are linearly independent.

Exercise 9

Without solving any system, explain why the following sets of vectors are linearly dependent, and exhibit a linear combination equal to \(\boldsymbol 0\) with coefficients not all zero.

  1. \(\begin{pmatrix}1 & 0 & 0\end{pmatrix}'\), \(\begin{pmatrix}0 & 1 & 0\end{pmatrix}'\), \(\begin{pmatrix}0 & 0 & 1\end{pmatrix}'\), \(\begin{pmatrix}1 & 2 & 3\end{pmatrix}'\in\R^{3}\).

  2. \(\begin{pmatrix}1 & 2\end{pmatrix}'\), \(\begin{pmatrix}0 & 0\end{pmatrix}'\in\R^{2}\).

  3. \(\begin{pmatrix}1 & 2\end{pmatrix}'\), \(\begin{pmatrix}-1 & -2\end{pmatrix}'\in\R^{2}\).

Solution

  1. These are \(4>3\) vectors of \(\R^3\), hence they are linearly dependent. Explicitly, the first three vectors are the canonical basis \(\boldsymbol e_1,\boldsymbol e_2,\boldsymbol e_3\) and \(\begin{pmatrix}1 & 2 & 3\end{pmatrix}'=1\boldsymbol e_1+2\boldsymbol e_2+3\boldsymbol e_3\), so

    \[ 1\boldsymbol e_1+2\boldsymbol e_2+3\boldsymbol e_3-1\begin{pmatrix}1\\2\\3\end{pmatrix}=\boldsymbol 0. \]

  2. The set contains the zero vector, hence it is linearly dependent: \( 0\begin{pmatrix}1 & 2\end{pmatrix}'+1\begin{pmatrix}0 & 0\end{pmatrix}'=\boldsymbol 0. \)

  3. The second vector is a linear combination of the first one, \(\begin{pmatrix}-1 & -2\end{pmatrix}'=-1\begin{pmatrix}1 & 2\end{pmatrix}'\), hence \( 1\begin{pmatrix}1 & 2\end{pmatrix}'+1\begin{pmatrix}-1 & -2\end{pmatrix}'=\boldsymbol 0. \)

Exercise 10

Consider the vectors

\[ \boldsymbol b_1=\begin{pmatrix}1\\1\\0\end{pmatrix}, \qquad \boldsymbol b_2=\begin{pmatrix}0\\1\\1\end{pmatrix}, \qquad \boldsymbol b_3=\begin{pmatrix}1\\0\\1\end{pmatrix}\in\R^{3}. \]

  1. Prove that \(\boldsymbol b_1,\boldsymbol b_2,\boldsymbol b_3\) form a basis of \(\R^3\).

  2. Find the coordinates of \(\boldsymbol x=\begin{pmatrix}4 & 3 & 5\end{pmatrix}'\) with respect to this basis and with respect to the canonical basis.

Solution

  1. We must show that the vectors are linearly independent and that their span is \(\R^3\). Given \(\boldsymbol y=\begin{pmatrix}a & b & c\end{pmatrix}'\in\R^3\), the equation \(\lambda_1\boldsymbol b_1+\lambda_2\boldsymbol b_2+\lambda_3\boldsymbol b_3=\boldsymbol y\) reads

    \[ \begin{cases} \lambda_1+\lambda_3=a\\ \lambda_1+\lambda_2=b\\ \lambda_2+\lambda_3=c \end{cases} \]

    Summing the three equations: \(2(\lambda_1+\lambda_2+\lambda_3)=a+b+c\), i.e., \(\lambda_1+\lambda_2+\lambda_3=s\) with \(s=\frac{a+b+c}{2}\). Subtracting from this each equation we get the unique solution

    \[ \lambda_1=s-c,\qquad \lambda_2=s-a,\qquad \lambda_3=s-b. \]

    • Since a solution exists for every \(\boldsymbol y\), the span is \(\R^3\).

    • For \(\boldsymbol y=\boldsymbol 0\) (\(a=b=c=0\)) we get \(s=0\) and \(\lambda_1=\lambda_2=\lambda_3=0\): the vectors are linearly independent.

    Hence \(\boldsymbol b_1,\boldsymbol b_2,\boldsymbol b_3\) form a basis of \(\R^3\).

  2. For \(\boldsymbol x\) we have \(a=4\), \(b=3\), \(c=5\), so \(s=\frac{4+3+5}{2}=6\) and \(\lambda_1=6-5=1\), \(\lambda_2=6-4=2\), \(\lambda_3=6-3=3\). Check:

    \[ 1\begin{pmatrix}1\\1\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\1\end{pmatrix}+3\begin{pmatrix}1\\0\\1\end{pmatrix} = \begin{pmatrix}1+0+3\\1+2+0\\0+2+3\end{pmatrix} = \begin{pmatrix}4\\3\\5\end{pmatrix}. \]

    With respect to the canonical basis the coordinates are simply the entries of \(\boldsymbol x\): \(\boldsymbol x=4\boldsymbol e_1+3\boldsymbol e_2+5\boldsymbol e_3\).