-5-5-4-4-3-3-2-2-1-11122334455Area = 1Area = 1.26p (Probe)p' = Ap
X AxisY Axis

Stress & Area Outputs

Area Scaling Factor (Det)
1.26x original
Stable grid deformation
Stress Eigen-axes
Eigen-Axis 1
Scale (Ī»):1.8
Vector:[0.94, 0.35]
Eigen-Axis 2
Scale (Ī»):0.7
Vector:[0.71, -0.71]
Aligned with Eigenvector! āœ“

Aligned with axis 1. The output vector p' is scaled exactly by Ī» = 1.8 without bending.

Variable Adjuster

Matrix element a1.5
-2.52.5
Matrix element b0.8
-2.52.5
Matrix element c0.3
-2.52.5
Matrix element d1
-2.52.5
Probe Vector X (pā‚“)2
-44
Probe Vector Y (p_y)1
-44

Auto-Sweep Engine

2x

Determinants & Eigenvectors

EIGEN

The determinant measures the area scaling factor of a matrix. Eigenvectors are invariant directions that only stretch or shrink by eigenvalue scale factors (Ī»), without rotating off their original line under transformation.

Avāƒ—=Ī»vāƒ—det⁔(Aāˆ’Ī»I)=0\begin{aligned}A\vec{v} = \lambda\vec{v} \\ \det(A - \lambda I) = 0\end{aligned}

Whiteboard Solver Steps

Step 1

Calculate the Determinant (Area Scaling)

Visual Guide: - The original white dashed unit square has an Area =1= 1. - The transformed parallelogram has an Area =∣det⁔(A)āˆ£ā‰ˆ1.2600= |\det(A)| \approx 1.2600. Physical Significance: - The determinant measures how much areas stretch, shrink, or flip under a transformation. - If det(A) = 0: The entire 2D plane collapses into a single line or point (volume is zero). In structural stress analysis, this represents structural collapse (loss of a dimension).

det⁔(A)=adāˆ’bc=(1.50)(1.00)āˆ’(0.80)(0.30)=1.2600\det(A) = ad - bc = (1.50)(1.00) - (0.80)(0.30) = 1.2600
Step 2

Solve the Characteristic Equation for Eigenvalues

Concept: - Eigenvalues (Ī»\lambda) are scaling ratios where vectors do not get knocked off their structural line under transformation (Avāƒ—=Ī»vāƒ—A\vec{v} = \lambda\vec{v}). - Solving the quadratic characteristic equation gives the scale factors: - Discriminant =1.2100= 1.2100 \ge 0 \implies \text{Real Eigenvalues}.

det⁔(Aāˆ’Ī»I)=0ā€…ā€ŠāŸ¹ā€…ā€ŠĪ»2āˆ’tr(A)Ī»+det⁔(A)=0Ī»2āˆ’2.50Ī»+1.2600=0\begin{aligned}\det(A - \lambda I) = 0 \implies \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \\ \lambda^2 - 2.50\lambda + 1.2600 = 0\end{aligned}
Step 3

Compute Eigenvectors (Invariant Directions)

Visual Guide: - The dashed gold lines represent the Eigenvector axes. - Drag the green probe vector pāƒ—\vec{p} onto one of these gold lines. Notice it stays exactly on the same line, only stretching/shrinking by its eigenvalue Ī»\lambda. Real-World Utility: - In structural physics (buildings, bridges), when forces deform a beam, eigenvectors identify the directions of maximum compression/stretching without bending. - In mechanics, keeping stress aligned with eigenvectors prevents structural failure by eliminating shear strain.

Ī»1ā‰ˆ1.800,vāƒ—1ā‰ˆ[0.9360.351]Ī»2ā‰ˆ0.700,vāƒ—2ā‰ˆ[0.707āˆ’0.707]\begin{aligned}\lambda_{1} \approx 1.800, \quad \vec{v}_{1} \approx \begin{bmatrix} 0.936 \\ 0.351 \end{bmatrix} \\ \lambda_{2} \approx 0.700, \quad \vec{v}_{2} \approx \begin{bmatrix} 0.707 \\ -0.707 \end{bmatrix} \\ \end{aligned}