3.3 Why det(A - λI) = 0 Finds Non-Zero Vectors in the Collapse Nullspace

Almost every student has solved $\det(A - \lambda I) = 0$ to find eigenvalues, yet very few can explain why subtracting $\lambda$ from the diagonal and setting the determinant to zero has anything to do with vectors that don't rotate.

The secret lies in connecting two geometric ideas: first, subtracting $\lambda I$ converts "vectors that get stretched by $\lambda$" into "vectors that get crushed to $\mathbf{0}$"; second, a matrix can only crush a nonzero vector to $\mathbf{0}$ if it squishes space into a lower dimension (zero volume, or zero determinant).

In this Article

• 1. Step 1: Converting $A\mathbf{v} = \lambda\mathbf{v}$ into a Nullspace Problem
• 2. Step 2: Why Nonzero Kernel Vectors Require Zero Volume
• 3. The $2 \times 2$ Trace-Determinant Characteristic Shortcut
• 4. Worked Step-by-Step Eigenpair Derivation
• 5. Active Recall Check and Key Takeaways

1. Step 1: Converting $A\mathbf{v} = \lambda\mathbf{v}$ into a Nullspace Problem

We start with the eigenvector equation $A\mathbf{v} = \lambda\mathbf{v}$. Because we cannot directly factor $\mathbf{v}$ out of $A\mathbf{v} - \lambda\mathbf{v}$ (since $A$ is a $d \times d$ matrix and $\lambda$ is a scalar), we rewrite $\lambda\mathbf{v}$ as $(\lambda I)\mathbf{v}$, where $I$ is the $d \times d$ identity matrix:

$$A\mathbf{v} - \lambda I\mathbf{v} = \mathbf{0} \quad \Longleftrightarrow \quad (A - \lambda I)\mathbf{v} = \mathbf{0}$$

What does this shifted matrix $M_\lambda = A - \lambda I$ do? For any vector $\mathbf{v}$ that $A$ stretches by $\lambda$, subtracting $\lambda\mathbf{v}$ cancels the output completely to $\mathbf{0}$! Thus, the eigenvectors of $A$ with eigenvalue $\lambda$ are literally the nonzero vectors in the nullspace (kernel) of $A - \lambda I$.

2. Step 2: Why Nonzero Kernel Vectors Require Zero Volume

For almost any number $\lambda$ you guess, the matrix $A - \lambda I$ is invertible, meaning the only solution to $(A - \lambda I)\mathbf{v} = \mathbf{0}$ is the useless zero vector $\mathbf{v} = \mathbf{0}$.

   WHEN det(A - lambda I) != 0:                  WHEN det(A - lambda I) == 0 (Tuned to Eigenvalue!):
   Unit square maps to full 2D parallelogram.    Entire 2D plane squishes flat onto a 1D line!
   Only v = [0,0] lands on the origin 0.         A whole line of nonzero vectors v gets crushed to 0!

         ^      +-------+                              ^          /
         |     /       /                               |         /  <-- Squished 1D image (Area = 0)
         |    /   Area != 0                            |        /
   ------+---0-------+-->                        ------+-------0--------->
             ^                                                 ^
       Only 0 lands at 0!                            Whole eigenspace line lands at 0!

3. The $2 \times 2$ Trace-Determinant Characteristic Shortcut

For any $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, expanding $\det(A - \lambda I) = (a-\lambda)(d-\lambda) - bc = 0$ always produces:

$$\lambda^2 - \underbrace{(a+d)}_{\text{tr}(A)}\lambda + \underbrace{(ad-bc)}_{\det(A)} = 0$$

Since this quadratic factors as $(\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2 = 0$, we get two instant invariants:

• Trace Identity: $\lambda_1 + \lambda_2 = a + d = \text{tr}(A)$ (the sum of the eigenvalues equals the sum of the diagonal entries).
• Determinant Identity: $\lambda_1 \lambda_2 = ad - bc = \det(A)$ (the product of the eigenvalues equals the determinant).

4. Worked Step-by-Step Eigenpair Derivation

Let $A = \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$. Let us find its eigenvalues, verify the trace/determinant rules, and watch $A - \lambda I$ collapse to rank 1:

1. Characteristic Polynomial: Here $\text{tr}(A) = 5 + 2 = 7$ and $\det(A) = (5)(2) - (2)(2) = 6$. Thus $\lambda^2 - 7\lambda + 6 = (\lambda - 6)(\lambda - 1) = 0$, giving $\lambda_1 = 6$ and $\lambda_2 = 1$. (Check: $6 + 1 = 7 = \text{tr}(A)$ and $6 \times 1 = 6 = \det(A)$.)
2. Collapse at $\lambda_1 = 6$: Subtract $6$ from the diagonal:

   $$A - 6I = \begin{bmatrix} -1 & 2 \\ 2 & -4 \end{bmatrix}$$

   Look at the two columns: Column 2 is $-2 \times$ Column 1! Because the columns are collinear, any vector $\mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ gets crushed straight to $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

3. Collapse at $\lambda_2 = 1$: Subtract $1$ from the diagonal:

   $$A - 1I = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$$

   Again, the columns are collinear, and $\mathbf{v}_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ lies in the nullspace!

5. Active Recall Check and Key Takeaways

Quick Check: Test Your Understanding

Question: A $2 \times 2$ matrix $A$ has trace $\text{tr}(A) = 10$ and determinant $\det(A) = 21$. What are its two eigenvalues?

Answer: We need two numbers that add to $10$ and multiply to $21$ (or solve $\lambda^2 - 10\lambda + 21 = 0$). Factoring $(\lambda - 7)(\lambda - 3) = 0$ immediately gives $\lambda_1 = 7$ and $\lambda_2 = 3$.

• Nullspace Reformulation: $A\mathbf{v} = \lambda\mathbf{v}$ is equivalent to $(A - \lambda I)\mathbf{v} = \mathbf{0}$.
• Zero Determinant = Volume Collapse: $\det(A - \lambda I) = 0$ tunes $\lambda$ to the exact values where $A - \lambda I$ squishes space and admits nonzero nullspace vectors.
• Canonical References: See Wikipedia Reference on the Characteristic Polynomial and Math Stack Exchange on Determinants as Volume Scaling.

Course Navigation & Chapter Links

This article is part of the complete 24-part Synapticz public course: Intuitive Eigenvalues, Eigenvectors, and Principal Component Analysis (Top-Down & Bottom-Up).

• Chapter 3 Overview: Chapter 3: Visualizing Linear Transformations, Eigenvectors, and Eigenvalues Without Memorization
• Microlearning 3.1: 3.1 Matrices as Geometric Grid Transformations: Stretching, Shearing, and Rotating
• Microlearning 3.2: 3.2 Eigenvectors and Eigenvalues as Invariant Axes of Pure Stretch
• Microlearning 3.3: 3.3 Why det(A - λI) = 0 Finds Non-Zero Vectors in the Collapse Nullspace
• Previous Lesson: 3.2 Eigenvectors and Eigenvalues as Invariant Axes of Pure Stretch
• Next Lesson: Chapter 4: What a Covariance Matrix Really Is — From Scatter Clouds to Orthogonal Stretch Axes