Undergraduate · Matrix conditioning · 486 of 650
Matrix conditioning · Small diagonal entry a=0.15; Final data error b=0.6
Use the current perturbation δ=0.36, so Δb=(0,0.36). Find the solution perturbation and the Euclidean condition number κ₂(A). Givens: Small diagonal entry a=0.15; Final data error b=0.6.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Use A=diag(1,0.15). A right-hand-side perturbation acts only in the second coordinate; its selected final size is 0.6. Use the current perturbation δ=0.36, so Δb=(0,0.36). Find the solution perturbation and the Euclidean condition number κ₂(A).
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A starting point
Apply A⁻¹=diag(1,1/a) to the perturbation. The condition number is the largest singular value divided by the smallest.
Work through the reasoning
Step 1
Identify the model and target
Use A=diag(1,0.15). A right-hand-side perturbation acts only in the second coordinate; its selected final size is 0.6. Use the current perturbation δ=0.36, so Δb=(0,0.36). The governing relation is A=diag(1,a); A⁻¹(1,δ)=(1,δ/a); κ₂(A)=1/a. Apply A⁻¹=diag(1,1/a) to the perturbation. The condition number is the largest singular value divided by the smallest.
Step 2
Substitute and calculate
The inverse second diagonal entry is 1/0.15=6.666667. A data error δ=0.36 therefore becomes δ/a=0.36/0.15=2.4, demonstrating this direction's amplification.
Step 3
Check the mathematical meaning
Multiplying Δx=(0,2.4) by A returns Δb=(0,0.36). Here 0<a≤1, so κ₂=1/a=6.666667; this direction attains the inverse amplification. Data perturbation: 0.36; Solution perturbation: 2.4; Condition number κ₂: 6.667. Decimal values are rounded, so use unrounded intermediate values.
The answer
The inverse second diagonal entry is 1/0.15=6.666667. A data error δ=0.36 therefore becomes δ/a=0.36/0.15=2.4, demonstrating this direction's amplification. Multiplying Δx=(0,2.4) by A returns Δb=(0,0.36). Here 0<a≤1, so κ₂=1/a=6.666667; this direction attains the inverse amplification. Animation check: Data perturbation: 0.36; Solution perturbation: 2.4; Condition number κ₂: 6.667. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
A small residual or small change in data can produce a much larger solution change when an inverse strongly stretches one direction. This diagonal example isolates that mechanism without numerical roundoff. Conditioning is a property of the problem, distinct from the stability of the algorithm used to solve it. Use A=diag(1,0.15). A right-hand-side perturbation acts only in the second coordinate; its selected final size is 0.6. Use the current perturbation δ=0.36, so Δb=(0,0.36). The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
A=diag(1,a); A⁻¹(1,δ)=(1,δ/a); κ₂(A)=1/a
03 · Reflect and transfer
Explain what changes and why.
Would the same perturbation magnitude in the first coordinate be amplified as much? Explain using the inverse matrix.
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.