Definition of a Vector
A vector is an ordered collection of numbers, which are called the components or entries of the vector. A vector can be represented as:
where each $v_i$ is a scalar, and $n$ is the dimension of the vector. A vector in $n$-dimensional space is denoted as a column vector.
Example:
For a vector in 3-dimensional space:
The vector $\mathbf{v}$ has 3 components, so it is a 3-dimensional vector.
Size of a Vector
The size of a vector refers to the number of components or entries in the vector. A vector with $n$ components is said to have size $n$ or be a $n$-dimensional vector.
- If , then the size of $\mathbf{v}$ is $n$.
Operations on Vectors
- Addition: Given two vectors and of the same size, their sum $\mathbf{v} + \mathbf{w}$ is:
- Scalar Multiplication: If is a vector and $c$ is a scalar, then the scalar multiplication $c \cdot \mathbf{v}$ is:
Norm of a Vector
The norm of a vector $\mathbf{v}$, denoted as $||\mathbf{v}||$, is a measure of the vector’s length or magnitude. There are different types of norms:
- L1 Norm (Manhattan Norm): The L1 norm of a vector is defined as: $$
||\mathbf{v}||_1 = |v_1| + |v_2| + \dots + |v_n|$$
Example: , then: $$
||\mathbf{v}||_1 = |2| + |-3| + |4| = 2 + 3 + 4 = 9$$ - L2 Norm (Euclidean Norm): The L2 norm of a vector is defined as: $$
||\mathbf{v}||_2 = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}
$$ Example: If , then: $$
||\mathbf{v}||_2 = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \approx 5.385
$$
Unit Vector
A unit vector is a vector with a magnitude of 1. To convert any given vector to a unit vector, we divide the vector by its L2 norm.
The L2 norm (Euclidean norm) of a vector is defined as:
$$
||\mathbf{v}||_2 = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}
$$
To convert a vector $\mathbf{v}$ to a unit vector $\hat{\mathbf{v}}$, we divide each component of $\mathbf{v}$ by its L2 norm:
$$
\hat{\mathbf{v}} = \frac{\mathbf{v}}{||\mathbf{v}||_2}
$$
Example: Converting a Vector to a Unit Vector
Consider the vector:
- Compute the L2 norm of $\mathbf{v}$:
$$
||\mathbf{v}||_2 = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
$$
- Divide the vector by its L2 norm to obtain the unit vector:
Thus, the unit vector $\hat{\mathbf{v}}$ is:
Now, the vector $\hat{\mathbf{v}}$ has a magnitude of 1, making it a unit vector.
Orthogonal Vectors
Two vectors and are said to be orthogonal if their dot product is zero:
$$
\mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + \dots + v_n w_n = 0
$$
In other words, if $\mathbf{v}$ and $\mathbf{w}$ are orthogonal, they are at a right angle to each other.
Orthonormal Vectors
Two vectors $\mathbf{v}$ and $\mathbf{w}$ are said to be orthonormal if they are both orthogonal and normalized (i.e., their L2 norm is 1):
$$
\mathbf{v} \cdot \mathbf{w} = 0 \quad \text{and} \quad ||\mathbf{v}||_2 = ||\mathbf{w}||_2 = 1
$$
Orthonormal vectors are often used in orthogonal bases where each vector is both orthogonal to the others and has a unit length.
Orthonormal Matrix
An orthonormal matrix is a square matrix $Q$ whose columns (or rows) are orthonormal vectors. This means that:
$$
Q^T \cdot Q = I_n
$$
where $Q^T$ is the transpose of the matrix $Q$ and $I_n$ is the identity matrix of size $n \times n$. This property implies that the matrix $Q$ is invertible, and its inverse is its transpose:
$$
Q^{-1} = Q^T
$$
Linearly Independent Vectors
A set of vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are said to be linearly independent if the only solution to the equation:
$$
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_k \mathbf{v}_k = 0
$$
is $c_1 = c_2 = \dots = c_k = 0$. In other words, no vector in the set can be written as a linear combination of the others.
If there exists a non-trivial solution (where not all $c_i$ are zero), the vectors are linearly dependent.
Example of Linearly Independent Vectors (LI)
Consider the following vectors in $\mathbb{R}^2$:
To check if these vectors are linearly independent, we set up the equation:
$$
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = 0
$$
Substituting the values of the vectors:
This leads to the system of linear equations:
$$
c_1 + 3c_2 = 0
$$
$$
2c_1 + 4c_2 = 0
$$
From the first equation, we have $c_1 = -3c_2$. Substituting into the second equation:
$$
2(-3c_2) + 4c_2 = 0
$$
$$
-6c_2 + 4c_2 = 0
$$
$$
-2c_2 = 0 \quad \Rightarrow \quad c_2 = 0
$$
Since $c_2 = 0$, we substitute into $c_1 = -3c_2$ to get $c_1 = 0$. Therefore, the only solution is $c_1 = c_2 = 0$, which means that the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly independent.
Example of Linearly Dependent Vectors (LD)
Consider the following vectors in $\mathbb{R}^2$:
These vectors are linearly dependent because $\mathbf{v}_1$ is a scalar multiple of $\mathbf{v}_2$. Specifically:
$$
\mathbf{v}_1 = \frac{1}{2} \mathbf{v}_2
$$
Thus, the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly dependent.