Determinants

Lecture 20

Minjae Park

Auburn University
MATH 2660 - Spring 2026

February 25, 2026

Attendance

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Let \(A\) be an \(n\times m\) matrix. There are four fundamental subspaces.
Column space \(C(A) \subseteq \mathbb{R}^n\) has dimension \(\mathop{\mathrm{rank}}(A)\).
Row space \(C(A^T) \subseteq \mathbb{R}^m\) has dimension \(\mathop{\mathrm{rank}}(A)\).
Nullspace \(N(A) \subseteq \mathbb{R}^m\) has dimension \(\mathop{\mathrm{null}}(A)\).
Left nullspace \(N(A^T) \subseteq \mathbb{R}^n\) has dimension \(\mathop{\mathrm{null}}(A^T)\).
Rank–nullity theorem:
\[\mathop{\mathrm{rank}}(A) + \mathop{\mathrm{null}}(A) = m,\qquad \mathop{\mathrm{rank}}(A) + \mathop{\mathrm{null}}(A^T) = n.\]
Rank measures how much essential information is encoded in the matrix.

Let
\[A=\begin{bmatrix} 1 & 0 & 1 & -1 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 1 & 2\\ \end{bmatrix}.\]
Compute the RREF of \(A\).
Use the pivot columns to find a basis for \(C(A)\).
Determine \(\mathop{\mathrm{rank}}(A)\).
Use the Rank–Nullity Theorem to compute \(\mathop{\mathrm{null}}(A)\).
Submit your nullity answer in iClicker.

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The system \(A\vec{x}=\vec{b}\) has a solution exactly when \(\vec{b}\in C(A)\).
If \(\vec{b}\notin C(A)\), then there is no exact solution, so we look for \(\vec{x}\) that minimizes the error \(\vec{e}=A\vec{x}-\vec{b}\). The error is minimized precisely when \(\vec{e}\) is orthogonal to the column space \(C(A)\).
Orthogonality to \(C(A)\) is equivalent to \(A^T\vec{e}=\vec{0}\), so \(\vec{e}\in N(A^T)\), the left nullspace.
Since \(\mathop{\mathrm{rank}}(A)+\mathop{\mathrm{null}}(A^T)=n\), the subspaces \(C(A)\) and \(N(A^T)\) are orthogonal complements that together fill \(\mathbb R^n\).
Therefore, every vector \(\vec{b}\in\mathbb R^n\) can be uniquely written as \(\vec{b}=\vec{u}+\vec{e}\) with \(\vec{u}\in C(A)\) and \(\vec{e}\in N(A^T)\), which leads to the normal equations.
In more advanced linear algebra, this is expressed as \(\mathbb R^n=C(A)\oplus N(A^T)\), a direct sum decomposition.

Recall the geometric interpretation of
\[A = [\vec{u}_1 \ \vec{u}_2] = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\]
as a linear transformation on the plane.
This means the standard basis vectors are sent to \(\vec{e}_1 \mapsto \vec{u}_1 = \langle a,c \rangle\) and \(\vec{e}_2 \mapsto \vec{u}_2 = \langle b,d \rangle\).
The new vectors \(\vec{u}_1, \vec{u}_2\) form a new coordinate grid.
The whole plane is “deformed’’ according to how these two vectors move.
Check Linear Transformations in 2D

What happens if two non-zero vectors \(\vec{u}_1\) and \(\vec{u}_2\) are parallel (i.e. linearly dependent)?
Then one vector is a scalar multiple of the other.
Their span is only one-dimensional.
The “grid’’ collapses to a line — all area is flattened.
From RREF:
- If RREF\((A)=I_2\), then \(\mathop{\mathrm{rank}}(A)=2\) and the columns are linearly independent.
- Otherwise \(\mathop{\mathrm{rank}}(A)\le 1\), so the columns are linearly dependent.

There is a geometric way to detect linear independence.
Consider the area of the parallelogram formed by \(\vec{u}_1\) and \(\vec{u}_2\).
If the area is zero, what does that mean?
It means the vectors lie on the same line → they are linearly dependent.
If the area is nonzero → they are linearly independent.

Let \(A = [\vec{u}_1 \ \vec{u}_2] = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\).
The parallelogram formed by \(\vec{u}_1=\langle a,c \rangle\) and \(\vec{u}_2=\langle b,d \rangle\) has area \[\text{Area} = |ad - bc|.\]
Check Geometric Proof.

Let \(A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\).
We define the determinant of \(A\) by \(\det(A)=\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc\).
The condition \(\det(A)=0\) holds if and only if the column vectors \(\vec{u}_1\) and \(\vec{u}_2\) are linearly dependent.
The determinant measures signed area of the parallelogram formed by the columns of \(A\).
The absolute value \(|\det(A)|\) gives the area, while the sign records the orientation (explained next).

The geometric area of a parallelogram is always positive, but \(\det(A)\) may be positive or negative.
For \(A=[\vec{u}_1\ \vec{u}_2]\), the sign of \(\det(A)\) records the orientation of the ordered pair \((\vec{u}_1,\vec{u}_2)\).
If the smaller angle from \(\vec{u}_1\) to \(\vec{u}_2\) is counterclockwise, then \(\det(A)>0\) and the pair has positive orientation.
If the smaller angle from \(\vec{u}_1\) to \(\vec{u}_2\) is clockwise, then \(\det(A)<0\) and the pair has negative orientation.
The standard basis \(\{\vec{e}_1,\vec{e}_2\}\) has positive orientation, while \(\{\vec{e}_2,\vec{e}_1\}\) has negative orientation.

Right-hand rule: curl the fingers of your right hand from \(\vec{u}_1\) toward \(\vec{u}_2\) through the smaller angle; if your thumb points toward you, the orientation is positive, and if it points away from you, the orientation is negative.
If two coordinate systems have the same orientation, one can be continuously deformed into the other without collapsing area.
If they have opposite orientation, a “flip’’ is required to pass from one to the other.
Explore this visually at Orientation and Sign of the Determinant.
Swapping the two columns of \(A\) reverses the orientation and changes the sign of the determinant.

A linear transformation scales area uniformly across the plane.
Every unit square in the original grid is transformed into the same parallelogram determined by the column vectors of \(A\).
Since total area is computed by assembling many tiny squares (the idea behind integration), the entire area scales by a single constant factor.
That constant scaling factor is \(|\det(A)|\).
The sign of \(\det(A)\) does not affect area size, but it determines whether the transformation preserves or reverses orientation.
Explore this visually at Linear Transformations in 2D.

The area of a unit circle is \(\pi\).
For \(a, b>0\), consider the linear transformation \[A=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}.\]
This stretches by factor \(a\) in the \(x\)-direction and \(b\) in the \(y\)-direction.
The determinant is \[\det(A)=ab.\]
The unit circle transforms into an ellipse with semi-axes \(a\) and \(b\).
Since area scales by \(\det(A)\), the new area is \(\pi ab\).

The determinant idea extends naturally to higher dimensions.
Let \(A=[\vec{u}_1\ \vec{u}_2\ \vec{u}_3]\) be a \(3\times3\) matrix with column vectors \(\vec{u}_i\in\mathbb R^3\).
The determinant \(\det(A)\) measures the signed volume of the parallelepiped formed by \(\vec{u}_1,\vec{u}_2,\vec{u}_3\).
If this volume is nonzero, the three vectors form a genuine three-dimensional grid and are linearly independent.
If the volume is zero, the grid collapses into a plane or even a line, so the vectors are linearly dependent.
You can visualize this at Linear Transformations in 3D.
Therefore, \(\det(A)\) provides a test for linear independence in \(\mathbb R^3\).
The explicit formula is more complicated than in the \(2\times2\) case; we will develop it using cofactor expansion next class.