Comprehensive Exercises on Vectors

Approach: Use the Euclidean norm, which is the square root of the sum of the squared components, and then divide the vector by that norm to obtain the vector of length 1 pointing in the same direction.

Steps:

1. Norm: $\|v\| = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
2. Unit vector: $\hat{v} = \frac{v}{\|v\|} = \frac{1}{5}(3, 4) = \left(\frac{3}{5}, \frac{4}{5}\right)$
3. Check: $\|\hat{v}\| = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1$

Answer:$\|v\| = 5$ and $\hat{v} = \left(\frac{3}{5}, \frac{4}{5}\right)$

Approach: Addition and scalar multiplication act component by component, and the parallelogram rule says that the area of the parallelogram spanned by two vectors is the absolute value of their two-dimensional determinant.

Steps:

1. $v + w = (2 + 1,\; 1 + 3) = (3, 4)$
2. $2v - w = (4, 2) - (1, 3) = (3, -1)$
3. Area: $S = |v_x w_y - v_y w_x| = |2 \cdot 3 - 1 \cdot 1| = |6 - 1| = 5$

Answer:(1) $v + w = (3, 4)$；(2) $2v - w = (3, -1)$；(3) the area is $5$

Approach: Set up the component equations for the coefficients and see how many free parameters they leave. A free parameter means the representation is not unique, which happens precisely because the three vectors are linearly dependent.

Steps:

1. Write $a v_1 + b v_2 + c v_3 = (a + c,\; b + c) = (3, 2)$
2. This gives $a + c = 3$ and $b + c = 2$, so $a = 3 - c$ and $b = 2 - c$
3. The parameter $c$ is free, so there are infinitely many representations. For instance:
   • $c = 0$: $(3, 2) = 3v_1 + 2v_2$
   • $c = 1$: $(3, 2) = 2v_1 + 1v_2 + 1v_3$

Answer:$(3, 2) = 3v_1 + 2v_2 = 2v_1 + v_2 + v_3$; the representation is not unique

Approach: Look for scalars that are not all zero and that make the linear combination vanish. Since one vector is a multiple of the other, such a relation must exist.

Steps:

1. Let $c_1(1, 2) + c_2(2, 4) = (0, 0)$
2. Both component equations give the same condition: $c_1 + 2c_2 = 0$
3. Take $c_1 = 2$, $c_2 = -1$: $2(1, 2) - (2, 4) = (2, 4) - (2, 4) = (0, 0)$
4. The coefficients are not all zero, so the set is linearly dependent

Answer:The set is linearly dependent, with the relation $2v_1 - v_2 = \mathbf{0}$

Approach: Compare the third vector with the first two. If one of them can be written as a combination of the others, the set is dependent and the span is already determined by the remaining vectors.

Steps:

1. Notice that $(1, 1) = (1, 0) + (0, 1)$, that is, $v_3 = v_1 + v_2$
2. This is a nontrivial relation, so the three vectors are linearly dependent
3. The vectors $(1, 0)$ and $(0, 1)$ are not collinear, so they already span the whole plane
4. Adding a dependent vector does not enlarge the span, so $\operatorname{span}\{(1,0),(0,1),(1,1)\} = \mathbb{R}^2$

Answer:The set is linearly dependent and its span is the whole plane $\mathbb{R}^2$

Approach: Look for a vector that is a combination of the others, discard it, and check that the remaining vectors are not collinear. The number of surviving vectors is the dimension, which is also the rank of the original set.

Steps:

1. Compare the third vector with the first two: $(1, 2, 0) + (0, 1, 1) = (1, 3, 1)$, so $v_3 = v_1 + v_2$
2. The set is therefore dependent, and $v_3$ may be removed without changing the span
3. The remaining vectors $(1, 2, 0)$ and $(0, 1, 1)$ are not multiples of each other, so they are linearly independent and span the same space
4. Hence a basis is $\{(1, 2, 0), (0, 1, 1)\}$, and the dimension (the rank of the original set) is 2

Answer:A basis is $\{(1, 2, 0), (0, 1, 1)\}$ and the dimension is $2$

Approach: A set of three vectors in $\mathbb{R}^3$ is a basis when its column matrix has a nonzero determinant. Once that is settled, write down the coordinate system and solve it.

Steps:

1. Form the matrix whose columns are the three vectors: $\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$
2. Its determinant is $1(1 \cdot 1 - 0 \cdot 1) - 0(1 \cdot 1 - 0 \cdot 0) + 1(1 \cdot 1 - 1 \cdot 0) = 1 + 0 + 1 = 2 \neq 0$, so the vectors are linearly independent and form a basis of $\mathbb{R}^3$
3. Solve for the coordinates: $\begin{cases} x_1 + x_3 = 2 \\ x_1 + x_2 = 3 \\ x_2 + x_3 = 4 \end{cases}$
4. Adding the three equations gives $2(x_1 + x_2 + x_3) = 9$, so $x_1 + x_2 + x_3 = \frac{9}{2}$
5. Hence $x_3 = \frac{9}{2} - 3 = \frac{3}{2}$, $x_2 = \frac{9}{2} - 2 = \frac{5}{2}$ and $x_1 = \frac{9}{2} - 4 = \frac{1}{2}$

Answer:The set is a basis of $\mathbb{R}^3$, and the coordinates of $\beta$ are $\left(\frac{1}{2}, \frac{5}{2}, \frac{3}{2}\right)$

Approach: Check the closure requirements and the eight axioms, noting that they are inherited from the real numbers through the usual arithmetic of polynomials. Then count the free coefficients.

Steps:

1. Closure under addition: the sum of two polynomials of degree $\leq 2$ still has degree $\leq 2$
2. Closure under scalar multiplication: $c \cdot p(x)$ still has degree $\leq 2$
3. Associativity, commutativity, the zero element, additive inverses, distributivity and the identity element all hold because they hold for real coefficients
4. Hence the set is a vector space
5. Every polynomial $a_0 + a_1x + a_2x^2$ is uniquely $a_0 \cdot 1 + a_1 \cdot x + a_2 \cdot x^2$, so a basis is $\{1, x, x^2\}$ and the dimension is 3

Answer:It is a vector space with basis $\{1, x, x^2\}$ and dimension $3$

Approach: A $2\times2$ matrix has four free entries, and each entry can be isolated in a matrix unit. Those four matrix units span the space and are independent.

Steps:

1. Every $2\times2$ matrix can be written as $A = a_{11}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + a_{12}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + a_{21}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + a_{22}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$
2. The four coefficients $a_{11}, a_{12}, a_{21}, a_{22}$ are recovered from the entries of $A$, so the representation is unique and the four matrix units are linearly independent
3. The dimension equals the number of basis vectors, that is, the number of free entries

Answer:A basis is $\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\}$ and the dimension is $4$

Approach: The columns of $B$ are the new basis vectors written in standard coordinates, and the new coordinates are obtained by multiplying by $B^{-1}$.

Steps:

1. The change-of-basis matrix is $B = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
2. Its inverse is $B^{-1} = \frac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{pmatrix}$
3. The new coordinates are $[v]_\mathcal{B} = B^{-1}v = \begin{pmatrix} 0.5 \cdot 3 + 0.5 \cdot 2 \\ 0.5 \cdot 3 - 0.5 \cdot 2 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0.5 \end{pmatrix}$
4. Check: $2.5\mathbf{b}_1 + 0.5\mathbf{b}_2 = 2.5(1,1) + 0.5(1,-1) = (3, 2)$

Answer:$B = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, $B^{-1} = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{pmatrix}$, and $[v]_\mathcal{B} = (2.5,\; 0.5)^T$

Approach: The inner product multiplies the components pairwise and adds the results; the norms are the square roots of the self inner products; orthogonality means the inner product vanishes.

Steps:

1. Inner product: $v \cdot w = 1 \cdot 4 + 2 \cdot (-1) + 3 \cdot 2 = 4 - 2 + 6 = 8$
2. Norms: $\|v\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$, $\|w\| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{21}$
3. Orthogonality: since $v \cdot w = 8 \neq 0$, the vectors are not orthogonal
4. Angle: $\cos\theta = \frac{v \cdot w}{\|v\|\|w\|} = \frac{8}{\sqrt{14}\sqrt{21}} = \frac{8}{\sqrt{294}}$

Answer:$v \cdot w = 8$, $\|v\| = \sqrt{14}$, $\|w\| = \sqrt{21}$; the vectors are not orthogonal, and $\cos\theta = \frac{8}{\sqrt{294}}$

Approach: Keep the first vector and subtract the projection of the second one onto it. Once the orthogonal set is known, divide each vector by its own length to make it orthonormal.

Steps:

1. $\vec{b}_1 = \vec{a}_1 = (1,1,0)$
2. Compute $\vec{b}_2$: $(\vec{a}_2, \vec{b}_1) = (1,0,1) \cdot (1,1,0) = 1$, $(\vec{b}_1, \vec{b}_1) = (1,1,0) \cdot (1,1,0) = 2$ $\vec{b}_2 = (1,0,1) - \frac{1}{2}(1,1,0) = (0.5, -0.5, 1)$
3. Check orthogonality: $(1,1,0) \cdot (0.5,-0.5,1) = 0.5 - 0.5 + 0 = 0$
4. Lengths: $\|\vec{b}_1\| = \sqrt{2}$ and $\|\vec{b}_2\| = \sqrt{0.25 + 0.25 + 1} = \frac{\sqrt{6}}{2}$
5. Unit vectors: $\vec{e}_1 = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$ and $\vec{e}_2 = \left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)$

Answer:The orthogonal set is $\{(1,1,0), (0.5,-0.5,1)\}$ and the orthonormal set is $\left\{\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right), \left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)\right\}$

Approach: An orthogonal matrix is characterised by $Q^T Q = I$. Compute the product and the determinant, which must be $\pm 1$ for an orthogonal matrix.

Steps:

1. $Q^T = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
2. $Q^T Q = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$
3. Determinant: $|Q| = 0 \cdot 0 - (-1) \cdot 1 = 1$

Answer:$Q$ is an orthogonal matrix, and $|Q| = 1$