MATH 332 — Fall 2026

Euclidean Vector Spaces
Points, directions, and affine solution sets · Anton, Rorres, & Kaul §§3.1–3.5
Assignment 3 due 10/9

Fred J. Hickernell

October 9, 2026

Course Map

Course decks

• Why Linear Algebra?
• Systems and Matrices
• Inverses
• Matrix Transformations
• Determinants
• Euclidean Spaces
• Vector Spaces and Span
• Bases and Coordinates
• Matrix Spaces and Rank
• Eigenvalues
• Inner Products
• Special Topics

In this deck

Separate locations from directions

\[\begin{gather*} Q-P=\text{vector}, \qquad P+\vct{v}=\text{point} \\ \mat{A}\vct{x}=\vct{b} \quad\Longrightarrow\quad \vct{x}=\vct{x}_p+\vct{z}, \qquad \mat{A}\vct{z}=\vct{0} \end{gather*}\]

• Points, Vectors, and Directions
• Solution Sets Have Directions
• Euclidean Tools for Affine Questions
• Big Ideas

Points, Vectors, and Directions

The same coordinate column can play two different geometric roles

• Why distinguish points from vectors?
• The operations reveal the roles
• Choosing an origin hides the distinction
• Lines and planes combine points with directions

Why distinguish points from vectors?

Point

• a location
• written \(P,Q,\ldots\)
• with chosen coordinates \(P=(p_1,p_2,\ldots)\)
• has no intrinsic direction or length

Vector

• a displacement or direction
• written \(\vct{u},\vct{v},\ldots\)
• from origin \(O\), \(\vct{p}=\overrightarrow{OP}=[p_1,p_2,\ldots]^{\mathsf T}\)
• has length and direction; translation leaves it unchanged

\[ \overrightarrow{PQ}=Q-P \]

Points say where; vectors say how to move

Teaching purpose: coordinate notation makes points and vectors look identical, so students may miss why homogeneous solution sets are vector spaces while nonhomogeneous solution sets usually are not. Use the physical language first: a point is where you are, a vector tells you how to move, and the difference between two possible locations is a displacement. Then preview the algebraic payoff: differences between solutions of \(\mat{A}\vct{x}=\vct{b}\) solve the homogeneous system. The coordinate notation is an example after choosing an origin and axes; parentheses versus column brackets are not what makes an object a point or a vector.

The operations reveal the roles

Operation	Result	Meaning
\(Q-P\)	vector	displacement from \(P\) to \(Q\)
\(P+\vct{v}\)	point	move from \(P\) by \(\vct{v}\)
\(\vct{u}+\vct{v}\)	vector	combine displacements
\(c\vct{v}\)	vector	scale or reverse a displacement
\(P+Q\)	—	no origin-independent geometric meaning

\(a\vct{u}+b\vct{v}\) — a linear combination, with scalar weights \(a,b\)

The expression

\[ (1-t)P+tQ=P+t(Q-P) \]

does have geometric meaning because its coefficients sum to \(1\)

The Euclidean vector space \(\reals^n\)

The vectors in \(\reals^n\) have addition and real scalar multiplication that obey the familiar linear rules:

\[ \vct{u}+\vct{v}=(u_i+v_i)_{i=1}^n, \qquad c\vct{u}=(cu_i)_{i=1}^n \]

The standard dot product \(\vct{u}^{\mathsf T}\vct{v}=\sum_{i=1}^n u_iv_i\) lets us define lengths and angles

This familiar example is a Euclidean vector space; Vector Spaces and Span states the vector-space axioms and extends them beyond coordinate vectors

Choosing an origin hides the distinction

After choosing an origin \(O\) and axes in \(\reals^3\):

\[ P=(p_1,p_2,p_3) \quad\longleftrightarrow\quad \vct{p}=\overrightarrow{OP} =\begin{bmatrix}p_1\\p_2\\p_3\end{bmatrix} \]

Here \(P,Q\) are points; \(\vct{p},\vct{q}\) are their position vectors; \(\vct{u},\vct{v}\) are directions. Other symbols can play the same roles

Moving the origin changes \(\vct{p}\) and \(\vct{q}\), but not the displacement

\[ \overrightarrow{PQ}=\vct{q}-\vct{p} \]

Parentheses versus column brackets do not determine the role; the geometry does

The tuple and column forms help display the distinction in this deck; their shapes alone do not determine whether an object is a point or a vector. The role and the chosen origin determine the meaning. The same construction works in \(\reals^n\) with \(n\) coordinates.

Lines and planes combine points with directions

A line needs one point and one nonzero direction:

\[ L=\{P+t\vct{v}:t\in\reals\} \]

A plane needs one point and two independent directions:

\[ \Pi=\{P+s\vct{u}+t\vct{v}:s,t\in\reals\} \]

\(s\vct{u}+t\vct{v}\) — a linear combination of the direction vectors

Changing the point on the same line or plane does not change its directions

Only when the set passes through the origin can its points also be treated as a collection of vectors closed under addition and scaling

Point \(+\) directions describes a translated, or affine, set

Solution Sets Have Directions

This is why the point–vector distinction matters in linear algebra

• Homogeneous systems produce directions
• Nonhomogeneous systems translate those directions
• Free variables count the directions
• Read the affine structure from elimination

Homogeneous systems produce directions

The solutions of

\[ \mat{A}\vct{z}=\vct{0} \]

include \(\vct{0}\) and are closed under linear combinations: every linear combination of homogeneous solutions is another homogeneous solution

\[ \mat{A}\vct{z}_1=\vct{0}, \quad \mat{A}\vct{z}_2=\vct{0} \quad\Longrightarrow\quad \mat{A}(c_1\vct{z}_1+c_2\vct{z}_2)=\vct{0} \]

These solutions are directions that the matrix cannot detect:

\[ \mat{A}(\vct{x}+\vct{z})=\mat{A}\vct{x} \]

These homogeneous directions form \(\operatorname{null}(\mat{A})\); Vector Spaces and Span develops the null space as a subspace

Vector Spaces and Span will verify the vector-space axioms and connect the null space to the row and column spaces.

Nonhomogeneous systems translate those directions

Suppose \(\vct{x}_p\) is the coordinate column of one solution point of

\[ \mat{A}\vct{x}=\vct{b} \]

Every solution has the form

\[ \vct{x}=\vct{x}_p+\vct{z}, \qquad \mat{A}\vct{z}=\vct{0} \]

Conversely, the difference between any two solutions is homogeneous:

\[ \mat{A}\vct{x}_1=\mat{A}\vct{x}_2=\vct{b} \quad\Longrightarrow\quad \mat{A}(\vct{x}_1-\vct{x}_2)=\vct{0} \]

For a consistent system, the solution set is an affine set:
Solution set \(=\) one solution point \(+\) all homogeneous directions

Free variables count the directions

For a consistent system with \(n\) unknowns

Free variables	Solution geometry
\(0\)	one point
\(1\)	affine line
\(2\)	affine plane
\(k\)	\(k\)-parameter affine set

If there are \(r\) pivot variables, then there are

\[ n-r \]

free variables

The dimension of the affine solution set is \(n-r\), the number of free parameters
It equals the dimension of the homogeneous direction space \(\operatorname{null}(\mat{A})\)

This is a preview. Bases and Coordinates will define basis and dimension and name the relationship rank–nullity.

Read the affine structure from elimination

Consider

\[\begin{align*} x+y+z&=2\\ x-y+z&=0 \end{align*}\]

Elimination gives \(y=1\) and \(x+z=1\), hence

\[ \vct{x} =\underbrace{\begin{bmatrix}1\\1\\0\end{bmatrix}}_{\text{coordinates of one solution point}} +t\underbrace{\begin{bmatrix}-1\\0\\1\end{bmatrix}}_{\text{homogeneous direction}} \]

\(\exstar\) Interpret this solution set Verify that the direction satisfies \(\mat{A}\vct{z}=\vct{0}\) Explain why adding \(t\vct{z}\) leaves both equations unchanged Decide whether the solution line passes through the origin

Euclidean Tools for Affine Questions

Reuse earlier geometry to compare points and construct directions

• Recall what the dot product already gives us
• Projection finds the nearest point on an affine line
• A special 3D operation: the cross product
• Point differences build a plane normal
• The triple product reconnects to determinants
• Dot product and cross product side by side

The companion notebook explores affine solutions, projection, and displacement geometry

Recall what the dot product already gives us

Systems and Matrices introduced the dot product as the geometry on \(\reals^n\)

We now use only these consequences:

\[\begin{align*} \vct{u}^{\mathsf T}\vct{v}=0 &\quad\Longleftrightarrow\quad \vct{u}\perp\vct{v}\\ \norm{\vct{v}} &=\sqrt{\vct{v}^{\mathsf T}\vct{v}}\\ \operatorname{dist}(P,Q) &=\norm{\vct{q}-\vct{p}} \end{align*}\]

Notice the roles in the last line:

\[ \underbrace{Q-P}_{\text{vector}} \quad\longrightarrow\quad \underbrace{\norm{\vct{q}-\vct{p}}}_{\text{distance between points}} \]

Projection finds the nearest point on an affine line

Let

\[ L=\{P+t\vct{v}:t\in\reals\}, \qquad \vct{v}\ne\vct{0} \]

To compare another point \(Q\) with \(L\), first form the displacement

\[ \overrightarrow{PQ}=\vct{q}-\vct{p} \]

Project that vector onto the line direction:

\[ \vct{p}_* =\vct{p} +\frac{(\vct{q}-\vct{p})^{\mathsf T}\vct{v}} {\vct{v}^{\mathsf T}\vct{v}}\vct{v} \]

The result \(\vct{p}_*\) gives coordinates of the point on \(L\) nearest to \(Q\)

Find the nearest point: a worked example

\[ P=(1,1,0),\qquad \vct{v}=[1,0,1]^{\mathsf T}, \qquad Q=(4,2,1),\qquad L=\{P+t\vct{v}:t\in\reals\} \]

Use the projection formula:

\[\begin{align*} t_*&=\frac{(\vct{q}-\vct{p})^{\mathsf T}\vct{v}}{\vct{v}^{\mathsf T}\vct{v}} =\frac{4}{2}=2\\ \vct{p}_*&=\vct{p}+2\vct{v}=[3,1,2]^{\mathsf T}\\ \vct{q}-\vct{p}_*&=[1,1,-1]^{\mathsf T} \end{align*}\]

Check: \((\vct{q}-\vct{p}_*)^{\mathsf T}\vct{v}=1-1=0\)

Nearest point \(P_*=(3,1,2)\); distance \(\operatorname{dist}(Q,L)=\sqrt{3}\)

The residual is perpendicular to the line direction. For every \(t\), \(\norm{\vct{q}-(\vct{p}+t\vct{v})}^2=3+2(t-2)^2\), so the minimum occurs at \(t=2\). Projecting \(\vct{q}\) without first subtracting \(\vct{p}\) would instead use the line through the origin and give the wrong answer.

Find the nearest point and distance

\[ P=(1,-1,0),\qquad \vct{v}=\begin{bmatrix}1\\2\\0\end{bmatrix}, \qquad Q=(4,0,2),\qquad L=\{P+t\vct{v}:t\in\reals\} \]

\(\exstar\) Project a displacement to find the nearest point Form \(\vct{q}-\vct{p}\) and compute the minimizing parameter \(t_*\) Find the nearest point \(P_*\) and the distance from \(Q\) to \(L\) Verify that \(\vct{q}-\vct{p}_*\) is perpendicular to \(\vct{v}\) Explain why projecting \(\vct{q}\) onto \(\vct{v}\) alone does not solve this problem

The displacement is \((3,1,2)^{\mathsf T}\), so \(t_*=5/5=1\). The nearest point is \(P_*=(2,1,0)\); the residual is \((2,-1,2)^{\mathsf T}\), which has dot product zero with \((1,2,0)^{\mathsf T}\) and norm \(3\). The squared distance at parameter \(t\) is \(9+5(t-1)^2\). Projecting \(\vct{q}\) alone finds the nearest point on the parallel line through the origin, not on the given affine line. Subtract the anchor point first, project the displacement, then add the anchor point back.

A special 3D operation: the cross product

For \(\vct{u},\vct{v}\in\reals^3\), define

\[ \vct{u}\mathbin{\times}\vct{v} =\begin{bmatrix} u_2v_3-u_3v_2\\ u_3v_1-u_1v_3\\ u_1v_2-u_2v_1 \end{bmatrix} \]

The result is perpendicular to both inputs:

\[\begin{gather*} \vct{u}^{\mathsf T}(\vct{u}\mathbin{\times}\vct{v})=0, \\ \vct{v}^{\mathsf T}(\vct{u}\mathbin{\times}\vct{v})=0 \end{gather*}\]

A normal whose length is parallelogram area; the right-hand rule chooses its direction

\[ \norm{\vct{u}\mathbin{\times}\vct{v}} =\norm{\vct{u}}\norm{\vct{v}}\sin\theta, \qquad \vct{v}\mathbin{\times}\vct{u}=-(\vct{u}\mathbin{\times}\vct{v}) \]

Reversing the inputs reverses the normal direction; this vector-valued operation is special to 3D

The natural generalization in other dimensions is the exterior, or wedge, product. It records the oriented parallelogram determined by two vectors, but its output is a bivector rather than another ordinary vector. In three dimensions, perpendicularity identifies that oriented area with the vector \(\vct{u}\mathbin{\times}\vct{v}\). There is also an exceptional vector cross product in dimension seven, but it is outside the scope of this course. The wedge product belongs to exterior algebra, which generalizes oriented areas, volumes, determinants, and the three-dimensional cross product.

A determinant pattern remembers the cross product

Using the standard basis vectors, remember

\[ \vct{u}\mathbin{\times}\vct{v} =\begin{vmatrix} \vct{e}_1&\vct{e}_2&\vct{e}_3\\ u_1&u_2&u_3\\ v_1&v_2&v_3 \end{vmatrix} \]

Expand across the first row with signs \(+,-,+\):

\[\begin{align*} \vct{u}\mathbin{\times}\vct{v} &=\vct{e}_1\begin{vmatrix}u_2&u_3\\v_2&v_3\end{vmatrix} -\vct{e}_2\begin{vmatrix}u_1&u_3\\v_1&v_3\end{vmatrix} +\vct{e}_3\begin{vmatrix}u_1&u_2\\v_1&v_2\end{vmatrix} \end{align*}\]

Each \(2\times2\) determinant uses \(ad-bc\)

A mnemonic, or memory aid, not a real determinant: the first row contains vectors

Why the cross-product norm is area

Squaring the component formula and collecting terms gives

\[\begin{align*} \norm{\vct{u}\mathbin{\times}\vct{v}}^2 &=(u_2v_3-u_3v_2)^2+(u_3v_1-u_1v_3)^2+(u_1v_2-u_2v_1)^2\\ &=(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2) -(u_1v_1+u_2v_2+u_3v_3)^2\\ &=\norm{\vct{u}}^2\norm{\vct{v}}^2-(\vct{u}^{\mathsf T}\vct{v})^2 \end{align*}\]

For nonzero inputs, use the dot-product cosine formula:

\[ \norm{\vct{u}\mathbin{\times}\vct{v}}^2 =\norm{\vct{u}}^2\norm{\vct{v}}^2(1-\cos^2\theta) =\norm{\vct{u}}^2\norm{\vct{v}}^2\sin^2\theta \]

Parallelogram area \(=\) base \(\times\) perpendicular height \(=\norm{\vct{u}}\bigl(\norm{\vct{v}}\sin\theta\bigr) =\norm{\vct{u}\mathbin{\times}\vct{v}}\)

If either vector is zero, both the area and the cross product are zero

The angle between nonzero vectors lies in \([0,\pi]\), so \(\sin\theta\ge0\); taking the nonnegative square root gives the stated norm. Parallel and antiparallel inputs have zero height and area. In the algebraic identity, the terms \(u_i^2v_i^2\) cancel between the product of sums and the squared dot product; the remaining square and mixed terms are exactly those obtained by expanding the three cross-product components.

Compute a normal and an area

\[ \vct{u}=\begin{bmatrix}1\\2\\0\end{bmatrix}, \qquad \vct{v}=\begin{bmatrix}0\\1\\3\end{bmatrix} \]

\[ \vct{u}\mathbin{\times}\vct{v} =\begin{bmatrix}2\cdot3-0\cdot1\\0\cdot0-1\cdot3\\1\cdot1-2\cdot0\end{bmatrix} =\begin{bmatrix}6\\-3\\1\end{bmatrix} =\vct{n} \]

Check perpendicularity:

\[ \vct{u}^{\mathsf T}\vct{n}=6-6=0, \qquad \vct{v}^{\mathsf T}\vct{n}=-3+3=0 \]

Parallelogram area \(=\norm{\vct{n}}=\sqrt{46}\); triangle area \(=\sqrt{46}/2\)

Point differences build a plane normal

For three noncollinear points \(P,Q,R\in\reals^3\), first form two directions:

\[ \vct{u}=\overrightarrow{PQ}=\vct{q}-\vct{p}, \qquad \vct{v}=\overrightarrow{PR}=\vct{r}-\vct{p} \]

Their cross product is a normal vector:

\[ \vct{n}=\vct{u}\mathbin{\times}\vct{v}, \qquad \vct{n}^{\mathsf T}\vct{u} =\vct{n}^{\mathsf T}\vct{v}=0 \]

The plane is therefore

\[ \vct{n}^{\mathsf T}(\vct{x}-\vct{p})=0 \]

Example: a normal from three points

\[ P=(1,0,1),\qquad Q=(2,1,1),\qquad R=(1,2,3) \]

First form directions from the same point:

\[ \vct{u}=\overrightarrow{PQ}=(1,1,0)^{\mathsf T}, \qquad \vct{v}=\overrightarrow{PR}=(0,2,2)^{\mathsf T} \]

Their cross product is

\[ \vct{n}=\vct{u}\mathbin{\times}\vct{v}=(2,-2,2)^{\mathsf T} \]

Check: \(\vct{n}^{\mathsf T}\vct{u}=2-2=0\) and \(\vct{n}^{\mathsf T}\vct{v}=-4+4=0\)

Example: the plane and its area

Keep \(P=(1,0,1)\) and \(\vct{n}=(2,-2,2)^{\mathsf T}\)

The plane through \(P,Q,R\) is

\[ \vct{n}^{\mathsf T}(\vct{x}-\vct{p})=0 \quad\Longleftrightarrow\quad 2(x-1)-2y+2(z-1)=0 \]

Equivalently, \(x-y+z=2\)

The triangle has half the parallelogram area:

\[ \operatorname{area}(PQR)=\tfrac12\norm{\vct{n}} =\tfrac12\sqrt{2^2+(-2)^2+2^2}=\sqrt3 \]

Exchanging \(Q\) and \(R\) reverses \(\vct{n}\) but leaves the plane and area unchanged

Find a plane and its triangle area

\[ P=(0,1,0),\qquad Q=(2,1,0),\qquad R=(0,2,2) \]

\(\exstar\) Use point differences and a cross product Find \(\vct{u}=\overrightarrow{PQ}\) and \(\vct{v}=\overrightarrow{PR}\) Compute \(\vct{n}=\vct{u}\mathbin{\times}\vct{v}\) and verify that it is perpendicular to both directions Find an equation for the affine plane through \(P,Q,R\) Find the area of triangle \(PQR\) using the original cross product

The directions are \((2,0,0)^{\mathsf T}\) and \((0,1,2)^{\mathsf T}\); the cross product is \((0,-4,2)^{\mathsf T}\). Both dot products are zero. The plane equation is \(-4(y-1)+2z=0\), or \(2y-z=2\). The triangle area is \(\sqrt5\). A scaled normal gives the same plane but its norm does not give the original parallelogram area.

Another plane and triangle area

\[ P=(1,1,0),\qquad Q=(2,1,1),\qquad R=(1,3,0) \]

\(\exstar\) Find a normal, the plane equation, and the triangle area Check your normal against both point-difference vectors Explain what changes if \(Q\) and \(R\) are exchanged

The directions are \((1,0,1)^{\mathsf T}\) and \((0,2,0)^{\mathsf T}\). Their cross product is \((-2,0,2)^{\mathsf T}\), perpendicular to both. The plane is \(-2(x-1)+2z=0\), or \(x-z=1\), and the triangle area is \(\sqrt2\). Exchanging \(Q\) and \(R\) negates the normal and leaves the plane and area unchanged.

The triple product reconnects to determinants

For vectors \(\vct{u},\vct{v},\vct{w}\in\reals^3\),

\[ \vct{u}^{\mathsf T}(\vct{v}\mathbin{\times}\vct{w}) =\det\begin{bmatrix}\vct{u}&\vct{v}&\vct{w}\end{bmatrix} \]

• absolute value — parallelepiped volume
• sign — orientation of the ordered directions
• zero — the directions lie in one plane

Point coordinates enter only through differences such as

\[ \overrightarrow{PQ},\quad \overrightarrow{PR},\quad \overrightarrow{PS} \]

Determinants measure the geometry of displacement vectors, independent of where the points are located

Why can the triple product vanish?

\[ \vct{u}^{\mathsf T}(\vct{v}\mathbin{\times}\vct{w}) =\det\begin{bmatrix}\vct{u}&\vct{v}&\vct{w}\end{bmatrix} \]

\(\exstar\) Explain each zero case without calculating coordinates If \(\vct{v}\) and \(\vct{w}\) are nonparallel and \(\vct{u}\) lies in the plane they span, why is the triple product zero? If \(\vct{v}\) and \(\vct{w}\) are parallel, why is the triple product zero for every \(\vct{u}\)? Connect both explanations to zero parallelepiped volume and dependent determinant columns

In the first case, \(\vct{u}=a\vct{v}+b\vct{w}\), while \(\vct{v}\mathbin{\times}\vct{w}\) is perpendicular to both spanning directions; their dot product is zero. In the second case, \(\vct{v}\mathbin{\times}\vct{w}=\vct{0}\). In either case the three columns are linearly dependent and the parallelepiped has zero volume.

Dot product and cross product side by side

For nonzero vectors \(\vct{u}\) and \(\vct{v}\):

	Dot product \(\vct{u}^{\mathsf T}\vct{v}\)	Cross product \(\vct{u}\mathbin{\times}\vct{v}\)
Inputs	Vectors in \(\reals^n\)	Vectors in \(\reals^3\) for the standard cross product
Output	A scalar	A vector perpendicular to both inputs
Angle and size	\(\norm{\vct{u}}\norm{\vct{v}}\cos\theta\) measures alignment	\(\norm{\vct{u}\mathbin{\times}\vct{v}}=\norm{\vct{u}}\norm{\vct{v}}\sin\theta\) gives parallelogram area
Zero when	The inputs are perpendicular	The inputs are parallel or antiparallel
Reverse order	\(\vct{u}^{\mathsf T}\vct{v}=\vct{v}^{\mathsf T}\vct{u}\)	\(\vct{u}\mathbin{\times}\vct{v}=-(\vct{v}\mathbin{\times}\vct{u})\)

When the cross product is nonzero, the right-hand rule gives its direction; its length is area

Both products give zero if either input is zero; the angle and the zero tests in the table assume nonzero vectors. The standard vector-valued cross product used in this course is defined on \(\reals^3\). The right-hand rule distinguishes the two possible normal directions: curl the fingers of the right hand from \(\vct{u}\) toward \(\vct{v}\); the thumb points along \(\vct{u}\mathbin{\times}\vct{v}\). The dot product extends to every \(\reals^n\) and gives a signed scalar, while the cross-product norm is nonnegative.

Big Ideas

• Points are locations; vectors are displacements or directions
• Vectors in \(\reals^n\), with ordinary addition and real scalar multiplication, form a Euclidean vector space
• Choosing an origin lets both be represented by coordinate columns, but does not erase their different roles
• Homogeneous solutions form a vector space of directions through the origin
• A consistent nonhomogeneous solution set is one solution point plus all homogeneous directions
• The dot product tests orthogonality and measures point differences
• Cross products provide a special three-dimensional connection among point differences, normals, and determinants

How Far We Have Come

From matrix equations to the geometry of their solutions

• Systems and Matrices — \([\,\mat{A}\mid\vct{b}\,]\) tests consistency and identifies pivot variables and free parameters
• Matrix Transformations — PLU reuses solves; products compose and inverses undo actions; determinants give signed scale and detect collapse
• Points and directions — point differences are vectors; \(\vct{p}+t\vct{d}\) is a line through \(\vct{p}\) when \(\vct{d}\ne\vct{0}\)
• Homogeneous directions — \(\operatorname{null}(\mat{A})=\{\vct{z}:\mat{A}\vct{z}=\vct{0}\}\) supplies the differences between solutions
• Dot products and projection — dot products measure lengths and angles; projection finds nearest points
• Cross products in 3D — \(\vct{u}\times\vct{v}\) is perpendicular to both; \(\|\vct{u}\times\vct{v}\|\) is the parallelogram area

For a consistent system, all solutions are \(\vct{x}_p+\operatorname{null}(\mat{A})\); free variables count independent directions

What Comes Next

Concrete picture

• vectors in \(\reals^n\)
• homogeneous directions
• point plus directions
• number of free parameters

Next names

• formal vector-space axioms and examples beyond \(\reals^n\)
• subspace and null space
• span and linear independence
• basis and dimension

Next: Vector Spaces, Subspaces, and Span formalizes the structure used here and extends it beyond Euclidean vectors