A.kaw Matrix Algebra Preliminary Study Note: 2. Vectors

"Matrix Algebra Preliminary" (Introduction to Matrix ALGEBRA) course by Prof. A.k.kaw (University of South Florida) is designed and taught.
PDF format Learning note Download (academia.edu)
2nd. Download the course handout (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
  Vectors can added only if they is of the same dimension and the addition was given by $$\vec{a} + \vec{b} = \begin{ bmatrix}a_1\\ \vdots\\ A_n \end{bmatrix} + \begin{bmatrix}b_1\\ \vdots\\ b_n \end{bmatrix} = \begin{bmatrix}a_1+b_1\\ \VD Ots\\ A_n + b_n \end{bmatrix}$$
• 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 $\vec{a}$ are a $n $-dimensional vector, then $ $k \vec{a} = k\begin{bmatrix}a_1\\ \vdots\\ a_n \END{BM Atrix} = \begin{bmatrix}ka_1\\ \vdots\\ ka_n \end{bmatrix}$$
• Linear combination of vectors
  Given $\vec{a}_1$, $\vec{a}_2$, $\cdots$, $\vec{a}_m$ as $m $ vectors of same dimension $n $, and if $k _1$, $k _2$, $\cdots$, $k _m$ was scalars, then $ $k _1\vec{a}_1 + k_2\vec{a}_2 + \cdots + k_m\vec{a}_m$$ is A linear combination of the $m $ vector S.
• linearly independent vectors
  A set of vectors $\vec{a}_1$, $\vec{a}_2$, $\cdots$, $\vec{a}_m$ is considered to be linearly independent if $ $k _1\vec{a} _1 + k_2\vec{a}_2 + \cdots + k_m\vec{a}_m = \vec{0}$$ have only one solution of $k _1 = k_2 = \cdots = K_m =0$.
• Rank
  From a set of $n $-dimension vectors, the maximum number of linearly independent vectors in the set is called the rank of T He set of vectors. Note that the rank of the vectors can never is greater than the vectors dimension.
• Dot Product
  Let $\vec{a} = \begin{bmatrix}a_1, & \cdots, &a_n\end{bmatrix}$ and $\vec{b} = \begin{bmatrix}b_1, & \cdots, & amp;b_n\end{bmatrix}$ be $n $-dimensional vectors. Then the dot product (i.e. inner product) of the the the and the vectors $\vec{a}$ and $\vec{b}$ are defined as $$\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 and then a subset of the $m $ vectors also have to be linearly independent.
  • If a set of vectors is linearly dependent and 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 and then the set of vectors de Pendent.

Selected problems

1. For $$\vec{a} = \begin{bmatrix}2\\9\\-7 \end{bmatrix},\ \vec{b} = \begin{bmatrix}3\\2\\5 \end{bmatrix},\ \vec{C} = \beg In{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$ Find $\vec{a} + \vec{b}$ and $2\vec{a}-3\vec{b} + \vec{c}$.
solution:$$\vec{a} + \vec{b} = \begin{bmatrix}2\\9\\-7 \end{bmatrix} + \begin{bmatrix}3\\2\\5 \end{bmatrix} = \ begin{bmatrix}5\\ 11\\-2 \end{bmatrix}$$ $$2\vec{a}-3\vec{b} + \vec{c} = 2\begin{bmatrix}2\\9\\-7 \end{bmatrix}-3\beg In{bmatrix}3\\2\\5 \end{bmatrix} + \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} -4\\ 13\\ -28 \end{bmatrix}$$

2. is $$\vec{a} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\ \vec{b} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 1\\ 4\\ \end{bmatrix}$$ 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\\ \end{bmatrix} = 0$$ the coefficient matrix is $$\BEGIN{BM Atrix} 1& 1& 1\\ 1& 2& 4\\ 1& 5& + \end{bmatrix} \rightarrow \begin{bmatrix} 1& 1& 1\\ 0& Amp 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}$$ $$\Rightarr ow \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 is linearly independent a nd the rank is 3.

3. is $$\vec{a} = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\ \vec{b} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{C} = \begin{bmatrix} 3\\ 5\\ 7 \end{bmatrix}$$ 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{BMA Trix} 1& 1& 3\\ 1& 2& 5\\ 1& 5& 7 \end{bmatrix} \rightarrow \begin{bmatrix} 1& 1& 1\\ 0&am P 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 A Re linearly independent and the rank is 3.

4. Is $$\vec{a} = \begin{bmatrix} 1\\ 2\\ 5 \end{bmatrix},\ \vec{b} = \begin{bmatrix} 2\\ 4\\ \end{bmatrix},\ \vec{ C} = \begin{bmatrix} 1.1\\ 2.2\\ 5.5 \end{bmatrix}$$ 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\\ \end{bmatrix} + X_3\BEGIN{BMA Trix} 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 \en D{bmatrix} \rightarrow x_1 = -2x_2-1.1x_3$$ which exists non-trivial solutions. Thus They is 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 $\vec{b} = \BEGIN{BMA Trix}-3 & 2 & 1 & 2.5 \end{bmatrix}$.
solution: $$\vec{a}\cdot\vec{b} = 2\times ( -3) + 1\times2 + 2.5 \times1 + 3\times2.5 = 6$$

6. If $\vec{u}$, $\vec{v}$, $\vec{w}$ is three non-zero vector of 2-dimensions, then is they independent?
solution:
Suppose the three 2-dimensional Non-zero Vectors is $\vec{u}=\begin{bmatrix}u_1\\ u_2\end{bmatrix}$, $\vec{v}=\begin{bmatrix}v_1\\ v_2\end{bmatrix}$, 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}$$, 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 is linearly dependent.
In general cases, if the dimension of a set of vectors are less than the number of vectors in the set and then the set of Vectors is linearly dependent.

7. $\vec{u}$ and $\vec{v}$ are and non-zero vectors of dimension $n $. Prove that if $\vec{u}$ and $\vec{v}$ be 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$$ Note that neither $x _1$ nor $x _2$ are zero, otherwise for instance, $x _1 = 0$ and $x _2\neq0$. We have $x _2\vec{v} = 0\rightarrow x_2 = 0$ or $\vec{v} = 0$. Either of these is contradiction (both of the vectors are Non-zero). Thus $x _1\neq0$ and $x _2\neq0$, and we have $$\vec{v} =-{x_1\over x_2}\vec{u}$$ (IS), $\vec{v} = q\vec{u}$, where $q =- {X_1\over x_2}$.

8. $\vec{u}$ and $\vec{v}$ are both non-zero vectors of dimension $n $. Prove that if there is a scalar $q $ such this $\vec{v} = q\vec{u}$, 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 contradic tion.
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: $$|\vec{v}| = \sqrt{5^2+ ( -3) ^2+2^2} = \sqrt{38}$$

What is the rank of the set of the vectors $$\begin{bmatrix}2\\3\\7 \end{bmatrix},\ \begin{bmatrix}6\\9\\21 \END{BM atrix},\ \begin{bmatrix}3\\2\\7 \end{bmatrix}.$$ solution: $$\ begin{bmatrix}2& 6& 3\\ 3& 9& 2\\ 7& 21& 7 \end{bmatrix} \rightarrow\begin{cases}2r_2-3r_1\\ {1\o ver7}r_3\end{cases}\begin{bmatrix}2& 6& 3\\ 0& 0& -5\\ 1& 3& 1 \end{bmatrix}$$ $$\Rightarrow\ begin{cases}r_1-2r_3\\-{1\over5}r_2\end{cases}\begin{bmatrix}0& 0& 1\\ 0& 0& 1\\ 1& 3& 1 \end{ Bmatrix} \rightarrow\begin{cases}r_1-r_2\\ r_3-r_2 \end{cases}\begin{bmatrix}0& 0& 0\\ 0& 0& 1\\ 1& 3& 0 \end{bmatrix}$$ Thus The rank of this set of vectors is 2.

If $\vec{a} = \begin{bmatrix}5 & 2 & 3\end{bmatrix}$ and $\vec{b} = \begin{bmatrix}6 &-7 & 3\end{bmat rix}$, then what's $4\vec{a} + 5\vec{b}$?
solution: $$4\vec{a} + 5\vec{b} = 4\begin{bmatrix}5 & 2 & 3\end{bmatrix} + 5\begin{bmatrix}6 &-7 &am P 3\end{bmatrix}$$ $$=\begin{bmatrix}20+30 & 8-35 & 12+15\end{bmatrix} = \begin{bmatrix}50 & -27 & 27\end{b matrix}$$

What is the dot product of vectors $$\begin{cases}\vec{a} = 3i+5j+7k\\ \vec{b}=11i+13j+17k\end{cases}$$ Soluti On: $$\vec{a}\cdot\vec{b} = 3\TIMES11+5\TIMES13+7\TIMES17 = 217$$

What is the angle between-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 both vectors is $\arccos0.98 98774$.

A.kaw Matrix Algebra Preliminary Study Note: 2. Vectors