A.Kaw矩阵代数初步学习笔记 2. Vectors

“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载（Academia.edu）
第2章课程讲义下载（PDF）

Summary

• Vector
  A vector is a collection of numbers in a definite order. If it is a collection of $n$ numbers, it is called a $n$-dimensional vector. For example, $$\vec{A} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix},\ \vec{B} = \begin{bmatrix}4 & 5 & 6 \end{bmatrix}.$$
• Addition of vectors
  Two vectors can be added only if they are of the same dimension and the addition is given by $$\vec{A} + \vec{B} = \begin{bmatrix}a_1\\ \vdots\\ a_n \end{bmatrix} + \begin{bmatrix}b_1\\ \vdots\\ b_n \end{bmatrix} = \begin{bmatrix}a_1+b_1\\ \vdots\\ a_n + b_n \end{bmatrix}$$
• Null vector
  A null vector (i.e. zero vector) is where all the components of the vector are zero. For example, $$\begin{bmatrix}0\\ 0\\ 0\\ 0 \end{bmatrix}$$
• Unit vector
  A unit vector $\vec{U}$ is defined as $$\vec{U} = \begin{bmatrix}u_1\\ \vdots\\ u_n \end{bmatrix}$$ where $$\sqrt{u_1^2 + \cdots + u_{n}^2 = 1}$$
• Scalar multiplication of vectors
  If $k$ is a scalar and $\vec{A}$ is a $n$-dimensional vector, then $$k\vec{A} = k\begin{bmatrix}a_1\\ \vdots\\ a_n \end{bmatrix} = \begin{bmatrix}ka_1\\ \vdots\\ ka_n \end{bmatrix}$$
• Linear combination of vectors
  Given $\vec{A}_1$, $\vec{A}_2$, $\cdots$, $\vec{A}_m$ as $m$ vectors of same dimension $n$, and if $k_1$, $k_2$, $\cdots$, $k_m$ are scalars, then $$k_1\vec{A}_1 + k_2\vec{A}_2 + \cdots + k_m\vec{A}_m$$ is a linear combination of the $m$ vectors.
• Linearly independent vectors
  A set of vectors $\vec{A}_1$, $\vec{A}_2$, $\cdots$, $\vec{A}_m$ are considered to be linearly independent if $$k_1\vec{A}_1 + k_2\vec{A}_2 + \cdots + k_m\vec{A}_m = \vec{0}$$ has only one solution of $k_1 = k_2 = \cdots = k_m =0$.
• Rank
  From a set of $n$-dimension vectors, the maximum number of linearly independent vectors in the set is called the rank of the set of vectors. Note that the rank of the vectors can never be greater than the vectors dimension.
• Dot product
  Let $\vec{A} = \begin{bmatrix}a_1, & \cdots, &a_n\end{bmatrix}$ and $\vec{B} = \begin{bmatrix}b_1, & \cdots, &b_n\end{bmatrix}$ be two $n$-dimensional vectors. Then the dot product (i.e. inner product) of the two vectors $\vec{A}$ and $\vec{B}$ is defined as $$\vec{A}\cdot\vec{B} = a_1b_1+\cdots+a_nb_n = \sum_{i=1}^{n}a_ib_i$$
• Some useful results
  • If a set of vectors contains the null vector, the set of vectors is linearly dependent.
  • If a set of $m$ vectors is linearly independent, then a subset of the $m$ vectors also has to be linearly independent.
  • If a set of vectors is linearly dependent, then at least one vector can be written as a linear combination of others.
  • If the dimension of a set of vectors is less than the number of vectors in the set, then the set of vectors is linearly dependent.

Selected Problems

1. For $$\vec{A} = \begin{bmatrix}2\\9\\-7 \end{bmatrix},\ \vec{B} = \begin{bmatrix}3\\2\\5 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$ find $\vec{A} + \vec{B}$ and $2\vec{A} - 3\vec{B} + \vec{C}$.

Solution:
$$\vec{A} + \vec{B} = \begin{bmatrix}2\\9\\-7 \end{bmatrix} + \begin{bmatrix}3\\2\\5 \end{bmatrix} = \begin{bmatrix}5\\ 11\\ -2 \end{bmatrix}$$ $$2\vec{A} - 3\vec{B} + \vec{C} = 2\begin{bmatrix}2\\9\\-7 \end{bmatrix} - 3\begin{bmatrix}3\\2\\5 \end{bmatrix} + \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} -4\\ 13\\ -28 \end{bmatrix}$$

2. Are $$\vec{A} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\ \vec{B} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 1\\ 4\\ 25 \end{bmatrix}$$ linearly independent? What is the rank of the above set of vectors?

Solution:
Suppose $$x_1\vec{A} + x_2\vec{B} + x_3\vec{C} = 0$$ $$\Rightarrow x_1\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} + x_2\begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix} + x_3\begin{bmatrix} 1\\ 4\\ 25 \end{bmatrix} = 0$$ The coefficient matrix is $$\begin{bmatrix} 1& 1& 1\\ 1& 2& 4\\ 1& 5& 25 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 1& 1\\ 0& 1& 3\\ 0& 4& 24 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 1& 1\\ 0& 1& 3\\ 0& 1& 6 \end{bmatrix}\Rightarrow \begin{bmatrix} 1& 0& -5\\ 0& 0& -3\\ 0& 1& 6 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} 1& 0& -5\\ 0& 0& 1\\ 0& 1& 6 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 0& 0\\ 0& 0& 1\\ 0& 1& 0 \end{bmatrix} \Rightarrow x_1=x_2=x_3=0$$ Thus they are linearly independent and the rank is 3.

3. Are $$\vec{A} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\ \vec{B} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 3\\ 5\\ 7 \end{bmatrix}$$ linearly independent? What is the rank of the above set of vectors?

Solution:
Suppose $$x_1\vec{A} + x_2\vec{B} + x_3\vec{C} = 0$$ $$\Rightarrow x_1\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} + x_2\begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix} + x_3\begin{bmatrix} 3\\ 5\\ 7 \end{bmatrix} = 0$$ The coefficient matrix is $$\begin{bmatrix} 1& 1& 3\\ 1& 2& 5\\ 1& 5& 7 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 1& 1\\ 0& 1& 2\\ 0& 4& 4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 1& 1\\ 0& 1& 2\\ 0& 1& 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 0& 0\\ 0& 0& 1\\ 0& 1& 1 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} 1& 0& 0\\ 0& 0& 1\\ 0& 1& 0 \end{bmatrix} \Rightarrow x_1=x_2=x_3=0$$ Thus they are linearly independent and the rank is 3.

4. Are $$\vec{A} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{B} = \begin{bmatrix} 2\\ 4\\ 10 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 1.1\\ 2.2\\ 5.5 \end{bmatrix}$$ linearly independent? What is the rank of the above set of vectors?

Solution:
Suppose $$x_1\vec{A} + x_2\vec{B} + x_3\vec{C} = 0$$ $$\Rightarrow x_1\begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix} + x_2\begin{bmatrix} 2\\ 4\\ 10 \end{bmatrix} + x_3\begin{bmatrix} 1.1\\ 2.2\\ 5.5 \end{bmatrix} = 0$$ The coefficient matrix is $$\begin{bmatrix} 1& 2& 1.1\\ 2& 4& 2.2\\ 5& 10& 5.5 \end{bmatrix} \Rightarrow \begin{bmatrix} 1& 2& 1.1\\ 0& 0& 0\\ 0& 0& 0 \end{bmatrix} \Rightarrow x_1 = -2x_2-1.1x_3$$ which exists non-trivial solutions. Thus they are linearly dependent and the rank is 1.

5. Find the dot product of $\vec{A} = \begin{bmatrix}2& 1 & 2.5 &3 \end{bmatrix}$ and $\vec{B} = \begin{bmatrix}-3 & 2 & 1 & 2.5 \end{bmatrix}$.

Solution:
$$\vec{A}\cdot\vec{B} = 2\times(-3) + 1\times2 + 2.5\times1 + 3\times2.5 = 6$$

6. If $\vec{u}$, $\vec{v}$, $\vec{w}$ are three non-zero vector of 2-dimensions, then are they independent?

Solution:
Suppose the three 2-dimensional non-zero vectors are $\vec{u}=\begin{bmatrix}u_1\\ u_2\end{bmatrix}$, $\vec{v}=\begin{bmatrix}v_1\\ v_2\end{bmatrix}$, and $\vec{w}=\begin{bmatrix}w_1\\ w_2\end{bmatrix}$. We have $$x_1\vec{u} + x_2\vec{v} + x_3\vec{w} = 0$$ $$\Rightarrow \begin{cases} x_1u_1+x_2v_1+x_3w_1 = 0 \\ x_1u_2+ x_2v_2 + x_3 w_3 = 0\end{cases}$$ That is, the number of unknown is greater than the number of equations. Thus it has non-trivial solutions for $x_1$, $x_2$, $x_3$, which means they are linearly dependent.
In general cases, if the dimension of a set of vectors is less than the number of vectors in the set, then the set of vectors is linearly dependent.

7. $\vec{u}$ and $\vec{v}$ are two non-zero vectors of dimension $n$. Prove that if $\vec{u}$ and $\vec{v}$ are linearly dependent, there is a scalar $q$ such that $\vec{v} = q\vec{u}$.

Solution:
Suppose we have $$x_1\vec{u} + x_2\vec{v} = 0$$ Note that neither $x_1$ nor $x_2$ is zero, otherwise for instance, $x_1 = 0$ and $x_2\neq0$. Then we have $x_2\vec{v} = 0\Rightarrow x_2 = 0$ or $\vec{v} = 0$. Either of these is contradiction (both of the vectors are non-zero). Thus $x_1\neq0$ and $x_2\neq0$, and we have $$\vec{v} = -{x_1\over x_2}\vec{u}$$ that is, $\vec{v} = q\vec{u}$, where $q=-{x_1\over x_2}$.

8. $\vec{u}$ and $\vec{v}$ are two non-zero vectors of dimension $n$. Prove that if there is a scalar $q$ such that $\vec{v} = q\vec{u}$, then $\vec{u}$ and $\vec{v}$ are linearly dependent.

Solution:
Since $$\vec{v} = q\vec{u} \Rightarrow q\vec{u}-\vec{v} = 0$$ Note that $q\neq0$, otherwise $\vec{v}=0$ which is contradiction.
Thus $\vec{u}$ and $\vec{v}$ are linearly dependent.

9. What is the magnitude of the vector $\vec{V}=\begin{bmatrix}5 & -3 & 2 \end{bmatrix}$?

Solution:
$$|\vec{V}| = \sqrt{5^2+(-3)^2+2^2} = \sqrt{38}$$

10. What is the rank of the set of the vectors $$\begin{bmatrix}2\\3\\7 \end{bmatrix},\ \begin{bmatrix}6\\9\\21 \end{bmatrix},\ \begin{bmatrix}3\\2\\7 \end{bmatrix}.$$
Solution:
$$\begin{bmatrix}2& 6& 3\\ 3& 9& 2\\ 7& 21& 7 \end{bmatrix} \Rightarrow\begin{cases}2R_2-3R_1\\ {1\over7}R_3\end{cases}\begin{bmatrix}2& 6& 3\\ 0& 0& -5\\ 1& 3& 1 \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_1-2R_3\\ -{1\over5}R_2\end{cases}\begin{bmatrix}0& 0& 1\\ 0& 0& 1\\ 1& 3& 1 \end{bmatrix} \Rightarrow\begin{cases}R_1-R_2\\ R_3-R_2 \end{cases}\begin{bmatrix}0& 0& 0\\ 0& 0& 1\\ 1& 3& 0 \end{bmatrix}$$ Thus the rank of this set of vectors is 2.

11. If $\vec{A} = \begin{bmatrix}5 & 2 & 3\end{bmatrix}$ and $\vec{B} = \begin{bmatrix}6 & -7 & 3\end{bmatrix}$, then what is $4\vec{A} + 5\vec{B}$?

Solution:
$$4\vec{A} + 5\vec{B} = 4\begin{bmatrix}5 & 2 & 3\end{bmatrix} + 5\begin{bmatrix}6 & -7 & 3\end{bmatrix}$$ $$=\begin{bmatrix}20+30 & 8-35 & 12+15\end{bmatrix} = \begin{bmatrix}50 & -27 & 27\end{bmatrix}$$

12. What is the dot product of two vectors $$\begin{cases}\vec{A} = 3i+5j+7k\\ \vec{B}=11i+13j+17k\end{cases}$$
Solution:
$$\vec{A}\cdot\vec{B} = 3\times11+5\times13+7\times17 = 217$$

13. What is the angle between two vectors $$\begin{cases}\vec{A} = 3i+5j+7k\\ \vec{B}=11i+13j+17k\end{cases}$$

Solution:
$$\cos < \vec{A}, \vec{B} > = {\vec{A}\cdot\vec{B}\over |\vec{A}|\cdot|\vec{B}|}$$ $$={217\over\sqrt{9+25+49}\cdot\sqrt{121+169+289}} = 0.9898774$$ Thus the angle between the two vectors is $\arccos0.9898774$.