INGENIA

LIN-04

2×2 eigenvalues

Trace and det.

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Lineareigen 2

Governing equation

lambda=mathrmtr/2pmsqrt(mathrmtr/2)2−det\\lambda=\\mathrm{tr}/2\\pm\\sqrt{(\\mathrm{tr}/2)^2-\\det}

a∈[−10,10]a \in \left[-10, 10\right]

where

a
a (—)
b
b (—)
c
c (—)
d
d (—)
\lambda_+
λ+ (—)
\lambda_-
λ- (—)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

LIN-04 · gauge
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①

Governing law — red

lambda=mathrmtr/2pmsqrt(mathrmtr/2)2−det\\lambda=\\mathrm{tr}/2\\pm\\sqrt{(\\mathrm{tr}/2)^2-\\det}

a∈[−10,10]a \in \left[-10, 10\right]

a=−10a = -10λ+=3\lambda_+ = 3

②

Domain of the variable — blue

amin⁡=−10,amax⁡=10a_{\min} = -10,\quad a_{\max} = 10

a∈[−10,10]a \in \left[-10, 10\right]

a=−10a = -10a=−10a = -10

③

Response curve — green

lambda=mathrmtr/2pmsqrt(mathrmtr/2)2−det\\lambda=\\mathrm{tr}/2\\pm\\sqrt{(\\mathrm{tr}/2)^2-\\det}

a∈[−10,10]a \in \left[-10, 10\right]

a=−10a = -10λ+=3\lambda_+ = 3

④

Rate of change — amber

dλ+da\frac{\mathrm{d}\lambda_+}{\mathrm{d}a}

a∈[−10,10]a \in \left[-10, 10\right]

a=−9.874a = -9.874λ+′=0\lambda_+^{\prime} = 0

Lecture brief

Historical brief

Determinants, eigenvalues, least squares and Cayley–Hamilton are the nineteenth-century matrix craft behind every coupled system. The sheets invert, project and diagonalise small n. This sheet (LIN-04 — 2×2 eigenvalues) is the form associated with eigen 2. Working symbols: aa, bb, cc, dd →\rightarrow λ+\lambda_+, λ−\lambda_-. Trace and det.

Purpose

Purpose: compute λ+\lambda_+, λ−\lambda_- from aa, bb, cc, dd in Linear algebra via lambda=mathrmtr/2pmsqrt(mathrmtr/2)2−det\\lambda=\\mathrm{tr}/2\\pm\\sqrt{(\\mathrm{tr}/2)^2-\\det} Trace and det. Use it when a real linear algebra question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given a=2.000 —a = 2.000\,\mathrm{—}, b=1.000 —b = 1.000\,\mathrm{—}, c=0.000 —c = 0.000\,\mathrm{—}, d=3.000 —d = 3.000\,\mathrm{—}, the governing relation lambda=mathrmtr/2pmsqrt(mathrmtr/2)2−det\\lambda=\\mathrm{tr}/2\\pm\\sqrt{(\\mathrm{tr}/2)^2-\\det} yields λ+=3.0000 —\lambda_+ = 3.0000\,\mathrm{—}, λ−=2.0000 —\lambda_- = 2.0000\,\mathrm{—}. Trace and det. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • λ+ \lambda_+3.0000 —
  • λ- \lambda_-2.0000 —
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Narration of this film

Trace and det.

Trace and det.

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Watch on YouTube