is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Vector Spaces and Subspaces

What is a vector space?

A vector space is just a set V of vectors on which are defined two operations: addition and multiplication. Note: They don't actually have to be addition and multiplication, they just have to hold for these ten axioms, and addition and multiplication hold.

  1. The sum of vectors u and v, denoted u + v is in V (closure under addition)

  2. u + v = v + u

  3. (u + v) + w = u + (v + w)

  4. There is a zero vector 0 (doesn't actually have to be 0) in V such that u + 0 = u

  5. For each u in V, there is a vector -u in V such that u + (-u) = 0

  6. The scalar multiple of u by c, denoted by cu is in V. (closure under scalar multiplication)

  7. c(u + v) = cu + cv

  8. (c + d)u = cu + du

  9. c(du) = cdu

  10. 1u = u

Basically, showing that it's closed under addition and scalar multiplication makes it a vector space given our normal + and *, because of the properties of real numbers. When the operations aren't orthodox, you must show all 10.

What is a subspace?

A subspace is a subset of a vector space that has three properties

  1. It has the zero vector

  2. It's closed under vector addition (Axiom 1 from before)

  3. It's closed under scalar multiplication (Axiom 6 from before)

So, a subspace is a vector space. Conversely, every vector space is a subspace too! (of itself or larger spaces)

Theorem: Span

MathJax TeX Test Page If $v_1 ... v_p$ are in a vector space V, then Span{$v_1...v_p$} is a subspace of V. Proof: $$LS1: \text{ set the coefficients to 0}$$ $$LS2: \text{Let } x_1, x_2 \in Span(v_1 ... v_p)$$ $$x_1 = c_1v_1 + ... + c_nv_n$$ $$x_2 = d_1v_1 + ... + d_nv_n$$ $$x_1 + x_2 = (c_1 + d_1)v_1 + ... + (c_n + d_n)v_n \in Span$$ $$LS2: \text{ Let } x_1 \in Span(v_1 ... v_p)$$ $$x_1 = c_1v_1 + ... + c_nv_n$$ $$cx_1 = cc_1v_1 + ... + cc_nv_n \in Span$$


MathJax TeX Test Page Is $\begin{bmatrix}a-b\\b-c\\c-a\\b\end{bmatrix}$ a subspace of $\mathbb{R}^4$? $$\begin{bmatrix}a-b\\b-c\\c-a\\b\end{bmatrix} = a\begin{bmatrix}1\\0\\-1\\0\end{bmatrix} + b\begin{bmatrix}-1\\1\\0\\1\end{bmatrix} + c\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}$$ $$=\text{Span}\left(\begin{bmatrix}1\\0\\-1\\0\end{bmatrix} ,\begin{bmatrix}-1\\1\\0\\1\end{bmatrix} ,\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}\right)$$ By our theorem, it is a subspace.

Linear Transformations of Linear Subspaces

MathJax TeX Test Page Let V be a linear subspace $\in \mathbb{R}^n$. Show that $T(V) = \{T(x) | x \in V\}$ is a linear subspace as well. $$\text{Prove LS1}: $$ $$\vec{0} \in V \text{ (V satisfies LS1) }$$ $$T\left(\vec{0}\right) = \vec{0}$$ $$\text{Therefore, } \vec{0} \in T$$ $$\text{Prove LS2}: $$ $$\text{Let }x_1, x_2 \in T$$ $$x_1 = T(v_1), x_2 = T(v_2), v_1,v_2 \in V$$ $$x_1 + x_2 = T(v_1) + T(v_2) = T(v_1 + v_2)$$ $$x_1 + x_2 = T(v_1 + v_2) \in T \left(v_1 + v_2 \in V\right)$$ $$\text{Prove LS3}: $$ $$\text{Let }x_1 \in T$$ $$x_1 = T(v_1), v_1 \in V$$ $$cx_1 = cT(v_1) = T(cv_1)$$ $$cx_1 = T(cv_1) \in T \left(cv_1 \in V\right)$$

Affine Subspaces

MathJax TeX Test Page A linear object is one that goes through the origin. A linear line (sounds funny, but is technically correct) is one that goes through the origin. When you add a vector to the line, it becomes affine. So, an affine subspace is defined like this: $$B = \vec{a} + V, V \text{ is a LS}$$ A natural follow-up question we could ask is if T(B) is also an affine subspace. $$T(B) = T(\vec{a} + V) = T(\vec{a}) + T(V)$$ We proved that $T(V)$ is a linear subspace. T(\vec{a}) is another vector. By definition, T(B) is thus an affine subspace.

Null Spaces, Column Spaces, and Linear Transformations

Efficient Way to Calculate Determinants