MA 1112: Linear Algebra II Tutorial problems, January 29, 2019
1.We begin with computing “convenient” bases of these subspaces. First, we form the transpose matrix of the matrix made of columns of the first system of vectors, and compute its reduced row echelon form:
0 3 −2 2
−9 8 2 −3
4 1 1 1
(1)↔(3),14(1),(2)+9(1)
7→
1 14 14 14 0 414 174 −34 0 3 −2 2
4
41(2),(1)−14(2),(3)−3(2)
7→
1 0 416 1141 0 1 1741 −413 0 0 −13341 9141
−13341(3),(1)−416(3),(2)−1741(3)
7→
1 0 0 197 0 1 0 194 0 0 1 −1319
,
next we do the same to the second system of vectors:
6 0 −3 1
3 3 0 5
9 −1 −5 0
1
6(1),(2)−3(1),(3)−9(1)
7→
1 0 −12 16 0 3 32 92 0 −1 −12 −32
1
3(2),(3)+(2)
7→
1 0 −12 16 0 1 12 32
0 0 0 0
This means thatU1 has a basis consisting of the vectors
g1=
1 0 0
7 19
, g2 =
0 1 0
4 19
, g3=
0 0 1
−1319
,
andU2 has a basis consisting of the vectors
h1=
1 0
−12
1 6
, h2 =
0 1
1 23 2
.
2.The intersection is described by the system of equations a1g1+a2g2+a3g3−b1h1−b2h2=0.
The matrix of this system is
1 0 0 −1 0
0 1 0 0 −1
0 0 1 12 −12
7 19
4
19 −1319 −16 −32
.
Let us compute its reduced row echelon form:
1 0 0 −1 0
0 1 0 0 −1
0 0 1 12 −12
7 19
4
19 −1319 −16 −32
(4)−197(1),(4)−194(2),(4)+1319(3)
7→
1 0 0 −1 0 0 1 0 0 −1 0 0 1 12 −12 0 0 0 3157 −3119
57
31(4),(3)−12(4),(1)+(4)
7→
1 0 0 0 −3 0 1 0 0 −1 0 0 1 0 1 0 0 0 1 −3
so b2 is the only free variable, and once we set b2 = t, we have b1 = 3t, a3= −t,a2 =t, and a1=3t. Therefore, the general vector of the intersec- tion is of the form
3t
1 0 0
7 19
+t
0 1 0
4 19
−t
0 0 1
−1319
=t
3 1
−1 2
,
and the intersection is spanned by
3 1
−1 2
.
3.Let us reduce the basis vectors ofU1 using the basis vector ofU1∩U2 that we found:
3 1 −1 2 1 0 0 197 0 1 0 194 0 0 1 −1319
1
3(1),(2)−(1)
7→
1 13 −13 23 0 −13 13 −1757 0 1 0 194 0 0 1 −1319
Now, let us find the reduced row echelon form of the resulting matrix:
0 −13 13 −1757 0 1 0 194 0 0 1 −1319
−3(1),(2)−(1)
7→
0 1 −1 1719 0 0 1 −1319 0 0 1 −1319
(3)−(2),(1)+(2)
7→
0 1 −1 194 0 0 1 −1319
0 0 0 0
We conclude that the vectors
0 1
−1
4 19
,
0 0 1
−1319
can be chosen for a relative basis.
4. ForU= span(v1, v2) to be invariant, it is necessary and sufficient to haveϕ(v1), ϕ(v2)∈U. Indeed, this condition is necessary because we must have ϕ(U) ⊂ U, and it is sufficient because each vector of U is a linear combination ofv1 and v2.
We have ϕ(v1) = Av1 =
−3 8
−8
and ϕ(v2) = Av2 =
−1 8
−8
. It just
remains to see if there are scalars x, y such that ϕ(v1) = xv1+ yv2 and scalars z, tsuch that ϕ(v2) =zv1+tv2. Solving the corresponding systems of linear equations, we see that there are solutions: ϕ(v1) = −3v1+5v2 and ϕ(v2) = −v1+7v2. Therefore, this subspace is invariant.
