๐ What is a Basis and Dimension?
In linear algebra, a basis for a vector space is a set of linearly independent vectors that span the entire space. This means any vector in the space can be written as a linear combination of the basis vectors. The dimension of a vector space is the number of vectors in any basis for that space.
๐ History and Background
The concepts of basis and dimension evolved from the study of linear equations and vector spaces in the 19th century. Key figures like Hermann Grassmann and Giuseppe Peano contributed significantly to the formalization of these ideas. These concepts are fundamental to modern linear algebra and its applications across various fields.
๐ Key Principles
- ๐ Linear Independence: A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. Formally, if $c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_n\vec{v}_n = \vec{0}$ implies $c_1 = c_2 = ... = c_n = 0$, then the vectors are linearly independent.
- ๐ฏ Spanning Set: A set of vectors spans a vector space if every vector in the space can be written as a linear combination of the vectors in the set.
- ๐ข Dimension: The dimension of a vector space is the number of vectors in a basis. All bases for a given vector space have the same number of vectors.
- โ Basis Uniqueness: While the basis itself isn't unique, the number of vectors in any basis *is* unique for a given vector space.
๐ Real-World Examples
Basis and dimension are crucial in many areas:
- ๐ป Computer Graphics: Representing 3D objects requires a basis for 3D space. Transformations like rotations and scaling can be easily expressed using matrix operations on basis vectors.
- ๐ก Signal Processing: Signals can be decomposed into a linear combination of basis functions (e.g., Fourier basis) for efficient analysis and compression.
- ๐ Data Analysis: Principal Component Analysis (PCA) uses eigenvectors as a basis to reduce the dimensionality of data while preserving important information.
๐ Practice Problems
Let's put your knowledge to the test!
Question 1
Determine if the following vectors form a basis for $\mathbb{R}^2$: $\vec{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$, $\vec{v}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.
Question 2
Determine if the following vectors form a basis for $\mathbb{R}^3$: $\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, $\vec{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, $\vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. What is the dimension of $\mathbb{R}^3$?
Question 3
Find a basis for the subspace of $\mathbb{R}^4$ spanned by the vectors $\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $\vec{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\vec{v}_3 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$, $\vec{v}_4 = \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}$. What is the dimension of this subspace?
Question 4
Determine if the following vectors are linearly independent: $\vec{v}_1 = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$, $\vec{v}_2 = \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$, $\vec{v}_3 = \begin{bmatrix} 4 \\ -1 \\ 3 \end{bmatrix}$.
Question 5
Find a basis for the null space of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix}$.
Question 6
What is the dimension of the vector space of all $2 \times 2$ matrices?
Question 7
Let $V$ be a vector space with dimension 5. If $S$ is a linearly independent subset of $V$ with 3 vectors, can $S$ be a basis for $V$? Explain.
๐ก Solutions
Solution 1
No, these vectors do not form a basis because they are linearly dependent. Notice that $\vec{v}_2 = 2\vec{v}_1$.
Solution 2
Yes, these vectors form a basis for $\mathbb{R}^3$ (the standard basis). They are linearly independent and span $\mathbb{R}^3$. The dimension of $\mathbb{R}^3$ is 3.
Solution 3
A basis for the subspace is {$\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$}. The dimension of this subspace is 3. Note that $\vec{v}_4 = \vec{v}_1 + \vec{v}_2$.
Solution 4
The vectors are linearly dependent. Notice that $2\vec{v}_1 + \vec{v}_2 = \vec{v}_3$.
Solution 5
The null space of $A$ is the set of all vectors $\vec{x}$ such that $A\vec{x} = \vec{0}$. A basis for the null space is {$\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}$}.
Solution 6
The dimension is 4. A basis for the space of all $2 \times 2$ matrices is {$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$}.
Solution 7
No, $S$ cannot be a basis for $V$ because a basis for $V$ must have 5 vectors, which is the dimension of $V$. Since $S$ only has 3 vectors, it cannot span the entire space $V$.
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Conclusion
Understanding basis and dimension is essential for mastering linear algebra. By practicing with these problems and reviewing the key principles, you can strengthen your understanding and excel in your studies. Keep practicing!