Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 8/21/2025
Problem 1. Solved Homogeneous System of Linear Equations
Find the dimension and a basis for the solution space W⊂R4 of⎩⎨⎧2x1+x2−x3+3x4=0,x1−x2+2x3−x4=0,3x1+x2+x3+2x4=0.Gaussian elimination to echelon form (stop before RREF):2131−10−1213−12R1↔R2123−1102−11−132R2←R2−2R1,R3←R3−3R1100−1332−5−5−155R3←R3−R2100−1302−50−150=echelon form.This corresponds to the system {x1−x2+2x3−x4=0,3x2−5x3+5x4=0,so x3,x4 are free and x2=35x3−35x4.Back-substitute into the first equation: x1=x2−2x3+x4=(35x3−35x4)−2x3+x4=−31x3−32x4.Basis via the “free-var = 1/0” method:(i) Let x3=1,x4=0⇒x2=35,x1=−31.u1=(−31,35,1,0).(ii) Let x3=0,x4=1⇒x2=−35,x1=−32.u2=(−32,−35,0,1).(Optional integer scaling by 3: 3u1=(−1,5,3,0),3u2=(−2,−5,0,3)).Answer: dimW=2,basis {(−1,5,3,0),(−2,−5,0,3)}.General solution: (x1,x2,x3,x4)=a(−1,5,3,0)+b(−2,−5,0,3),a,b∈R.
Problem 2. Finding the dimension and a basis for the solution space.
Find the dimension and a basis for the solution space W⊂R5 of⎩⎨⎧x1+2x2−x3+0x4+3x5=0,x1+x2+2x3+x4−x5=0,2x1+5x2−x3+x4+5x5=0.Answer: dimW=3,basis {(5,−2,1,0,0),(2,−1,0,1,0),(−5,1,0,0,1)}.
Problem 3. Solve Homogeneous System
Solve the homogeneous system depending on a∈R:⎩⎨⎧x1+2x2+ax3=0,x2+ax3=0,ax3=0.For which a is the trivial solution the only one? If nontrivial solutions exist, give dimW and a basis.Answer: {a=0:x3=0⇒x2=0⇒x1=0⇒dimW=0(trivial only);a=0:free x3=t,x2=0,x1=0⇒dimW=1,basis {(0,0,1)}.
Problem 4. Proof - W subspace
Let A∈Rm×n and W={x∈Rn:Ax=0}.Prove that W is a subspace of Rn and dimW=n−rank(A).Answer (outline): ∙Zero vector∙Closed under addition∙Closed under scalarsHence W is a subspace.∙Row-reduce A to RREF; the number of pivot columns is rank(A).∙Number of free variables is n−rank(A).Each free variable yields one basis vector of W.⇒dimW=n−rank(A)(Rank–Nullity Theorem).
Problem 5. Echelon Form
The homogeneous system in variables (x1,x2,x3,x4) has echelon form100−2103−40010000.Find dimW and a basis for the solution space W.Answer: Free variables x3=s,x4=t.x2=4s−t,x1=5s−2t.⇒(x1,x2,x3,x4)=s(5,4,1,0)+t(−2,−1,0,1).dimW=2,basis {(5,4,1,0),(−2,−1,0,1)}.
Problem 6. Solve HLE - #2
Find the dimension and a basis for the solution space W⊂R5 of⎩⎨⎧x1−x2+x3+2x4−x5=0,2x1−2x2+x3+3x4−2x5=0,x1−x2+2x3+x4−x5=0.Answer: dimW=2,basis {(1,1,0,0,0),(1,0,0,0,1)}.
Problem 7. Solve HLE - #3
Find the dimension and a basis for the solution space W⊂R5 of⎩⎨⎧x1+x2+x3+x4+x5=0,x1−x2+2x3−x4+0x5=0,2x1+0x2+3x3+0x4+x5=0.Answer: dimW=3,basis {(−3,1,2,0,0),(0,−1,0,1,0),(−1,−1,0,0,2)}.
Problem 8. Solve HLE - #4
Find the dimension and a basis for the solution space W⊂R5 of⎩⎨⎧x1−2x2+3x3−x4+4x5=0,2x1−4x2+6x3−2x4+8x5=0,−3x1+6x2−9x3+3x4−12x5=0.Answer: dimW=4,basis {(2,1,0,0,0),(−3,0,1,0,0),(1,0,0,1,0),(−4,0,0,0,1)}.
