110.202 Linear Algebra
Notes for 2/20/2004
⎡
⎤
1
3.2. #24 Consider the line L spanned by ⎣ 2 ⎦ in R3 . Find a basis of
3
⊥
L .
[Solution]
By definition, we know that
⎧
⎫
⎡ ⎤
1
⎨
⎬
L⊥ = v ∈ R3 | v · ⎣ 2 ⎦ = 0 .
⎩
⎭
3
⎡ ⎤
a
Assume that v = ⎣ b ⎦ ∈ L⊥ where a, b, c ∈ R. Then, we have
c
a + 2b + 3c = 0. By solving this equation, we have
⎡ ⎤ ⎡
⎤ ⎡
⎤
⎡
⎤
a
−2b − 3c
−2
−3
⎣ b ⎦=⎣
⎦ = ⎣ 1 ⎦ b + ⎣ 0 ⎦ c.
b
c
c
0
1
⎤
⎡
⎤
⎤
⎡
⎡
−2
−2
−3
This implies that ⎣ 1 ⎦ and ⎣ 0 ⎦ span L⊥ . Assume α ⎣ 1 ⎦ +
0
0
1
⎤
⎡
−3
⎣
0 ⎦ = 0. By solving this equation, we can easily to get that
β
1
⎡
⎤
⎡
⎤
−2
−3
α = β = 0. Therefore, ⎣ 1 ⎦ and ⎣ 0 ⎦ are linearly independent.
0
1
⎧⎡
⎤ ⎡
⎤⎫
−3 ⎬
⎨ −2
⎣
⎦
⎣
1
0 ⎦ forms a basis of L⊥ .
Hence,
,
⎩ 0
1 ⎭
¥
1
2
3.2. #40 Consider an m × n matrix A and an n × p matrix B. We are
told that the columns of A and the columns of B are linearly
independent. Are the column of the product AB linearly independent as well?
[Solution]
[Method 1]
By Fact 3.2.6, since the columns of Anand
B are
o the columns of
n o
linearly independent, we know ker (A) = 0 and ker (B) = 0 . To
calculate ker (AB), we assume that ABx =n 0owhere x ∈ ker (AB).
Since A (Bx) = 0, we have Bx ∈ ker (A) = 0 . Therefore, Bx = 0.
n o
This implies that x ∈ ker (B) = 0 . Hence, x = 0. This concludes
n o
that ker (AB) = 0 . By Fact 3.2.6, we have the columns of AB are
linearly independent.
[Method 2]
£
¤
Assume that A = v1 v2 · · · vn where v1 , v2 , · · · , vn are the
⎡
⎤
b11 b12 · · · b1p
⎢ b21 b22 · · · b2p ⎥
m × 1 column vectors of A and B = ⎢
.. . .
. ⎥ and u1 =
⎣ ...
. .. ⎦
.
bn1 bn2 · · · bnp
⎡
⎤
⎤
⎡
b1p
b11
⎢ b2p ⎥
⎢ b21 ⎥
⎢ . ⎥ , · · · , up = ⎢ . ⎥ are the n × 1 column vectors of B. We
⎣ .. ⎦
⎣ .. ⎦
bn1
have
bnp
AB =
=
£
£
v1 v2
⎡
⎤
b11 b12 · · · b1p
¤ ⎢ b21 b22 · · · b2p ⎥
· · · vn ⎢
.. . .
. ⎥
⎣ ...
. .. ⎦
.
bn1 bn2 · · · bnp
b11 v1 + b21 v2 + · · · + bn1 vn · · · b1p v1 + b2p v2 + · · · + bnp vn
¤
,
where w1 = b11 v1 +b21 v2 +· · ·+bn1 vn , · · · , wp = b1p v1 +b2p v2 +· · ·+bnp vn
are the column vectors of AB.
To check if the column vectors of AB are linearly independent,
we assume that α1 w1 + α2 w2 + · · · + αp wp = 0. Then, we have
α1 (b11 v1 + b21 v2 + · · · + bn1 vn ) + α2 (b12 v1 + b22 v2 + · · · + bn2 vn ) + · · · +
αp (b1p v1 + b2p v2 + · · · + bnp vn ) = 0, that is, (α1 b11 + α2 b12 + · · · + αp b1p ) v1 +
(α1 b21 + α2 b22 + · · · + αp b2p ) v2 +· · ·+(α1 bn1 + α2 bn2 + · · · + αp bnp ) vn =
3
0. Since v1 , v2 , · · · , vn are linearly independent, we have
⎧
α1 b11 + α2 b12 + · · · + αp b1p = 0
⎪
⎪
⎨ α1 b21 + α2 b22 + · · · + αp b2p = 0
.
..
⎪
.
⎪
⎩
α1 bn1 + α2 bn2 + · · · + αp bnp = 0
Rewrite it into the matrix form, we have
⎡
⎤
⎡
α1
α1
⎢ α2 ⎥ £
¤ ⎢ α2
⎢
⎥
0=B⎢
⎣ ... ⎦ = u1 u2 · · · up ⎣ ...
αp
⎤
⎥
⎥ = α1 u1 +α2 u2 +· · ·+αp up .
⎦
αp
Again, since u1 , u2 , · · · , up are linearly independent, we have α1 = α2 =
· · · = αp = 0. This implies that w1 , w2 , · · · , wp are linearly independent
¥
3.3. #26 Consider the matrix
A=
∙
0 1 2 0 3
0 0 0 1 4
¸
.
Find a matrix B such that ker (A) = im (B).
[Solution]
To find ker (A), we assume
∙
Then, we have
0 1 2 0 3
0 0 0 1 4
½
Therefore,
⎡ ⎤ ⎡
a
a
⎢ b ⎥ ⎢ −2c − 3e
⎢ ⎥ ⎢
⎢ c ⎥=⎢
c
⎢ ⎥ ⎢
⎣ d ⎦ ⎣ −4e
e
e
⎡
¸⎢
⎢
⎢
⎢
⎣
a
b
c
d
e
⎤
⎥ ∙ ¸
⎥
⎥= 0 .
⎥
0
⎦
b + 2c + 3e = 0
.
d + 4e = 0
⎤
⎡
⎥ ⎢
⎥ ⎢
⎥=⎢
⎥ ⎢
⎦ ⎣
1
0
0
0
0
⎤
⎡
⎢
⎥
⎢
⎥
⎥a + ⎢
⎢
⎥
⎣
⎦
0
−2
1
0
0
⎤
⎡
⎢
⎥
⎢
⎥
⎥c +⎢
⎢
⎥
⎣
⎦
0
−3
0
−4
1
⎤
⎥
⎥
⎥ e.
⎥
⎦
4
⎧⎡
⎪
⎪
⎪
⎪
⎨⎢
⎢
It is very easy to check that ⎢
⎢
⎪
⎪
⎣
⎪
⎪
⎩
of ker (A). Write
1
0
0
0
0
⎤ ⎡
⎥ ⎢
⎥ ⎢
⎥,⎢
⎥ ⎢
⎦ ⎣
0
−2
1
0
0
⎤ ⎡
⎥ ⎢
⎥ ⎢
⎥,⎢
⎥ ⎢
⎦ ⎣
0
−3
0
−4
1
⎤⎫
⎪
⎪
⎪
⎥⎪
⎥⎬
⎥ forms a basis
⎥⎪
⎦⎪
⎪
⎪
⎭
⎤
1 0
0
⎢ 0 −2 −3 ⎥
⎥
⎢
⎥.
0
1
0
B=⎢
⎥
⎢
⎣ 0 0 −4 ⎦
0 0
1
⎧⎡ ⎤ ⎡
⎤⎫
⎤ ⎡
⎪
0
0
1
⎪
⎪
⎪
⎪
⎪
⎪⎢ 0 ⎥ ⎢ −2 ⎥ ⎢ −3 ⎥⎪
⎨
⎥⎬
⎥ ⎢
⎢ ⎥ ⎢
⎥
⎢
⎥
⎢
⎥
⎢
We have im (B) = span ⎢ 0 ⎥ , ⎢ 1 ⎥ , ⎢ 0 ⎥ = ker (A).
⎪
⎪
⎪
⎪
⎪⎣ 0 ⎦ ⎣ 0 ⎦ ⎣ −4 ⎦⎪
⎪
⎪
⎩ 0
1 ⎭
0
¥
3.4. #8 Consider the plane x1 + 2x2 + x3 = 0. ⎡Find a⎤ basis B of this
∙
¸
1
2
for x = ⎣ −1 ⎦.
plane such that [x]B =
−1
1
[Solution]
Since [x]B has two coordinates, we can assume B has two vectors, v1
and v2 . Therefore, x = 2v1 −v2 . Choose v1 on the
⎡ plane
⎤ x1 +2x2 +x3 = 0
1
which is not parallel to x. We can pick v1 = ⎣ 0 ⎦. Then we have
−1
⎡
⎤ ⎡
⎤ ⎡
⎤
1
1
1
v2 = 2v1 − x = 2 ⎣ 0 ⎦ − ⎣ −1 ⎦ = ⎣ 1 ⎦ .
−1
1
−3
⎡
To show that v1 and v2 are linearly independent, we assume c1 v1 +
c2 v2 = 0, that is,
⎡
⎤
⎡
⎤ ⎡
⎤
1
1
c1 + c2
⎦.
c2
0 = c1 ⎣ 0 ⎦ + c2 ⎣ 1 ⎦ = ⎣
−1
−3
−c1 − 3c2
Therefore, we have c2 = 0 and c1 + c2 = 0 which inplies c1 = c2 = 0.
This tells us that v1 and v2 are linearly independent. To show that
v1 and v2 span the plane x1 + 2x2 + x3 = 0. We have to show that
every vector y on the plane can be written as a linear combination
5
of v1 and v2 , that is, there exist
⎡ two⎤coefficients a1 and a2 such that
y1
⎣
y = a1 v1 + a2 v2 . For every y = y2 ⎦ on the plane x1 + 2x2 + x3 = 0,
y3
we have y1 + 2y2 + y3 = 0. Hence,
⎡
⎤ ⎡
⎤
⎡
⎤
⎡
⎤
y1
−2
−1
−2y2 − y3
⎦ = y2 ⎣ 1 ⎦ + y3 ⎣ 0 ⎦ .
y2
y = ⎣ y2 ⎦ = ⎣
0
1
y3
y3
⎤
⎤
⎡
⎡
−2
−1
To write ⎣ 1 ⎦ and ⎣ 0 ⎦ into linear combinations of v1 and v2 ,
0
1
⎤
⎤
⎡
⎡
−2
−1
we assume ⎣ 1 ⎦ = t1 v1 + t2 v2 and ⎣ 0 ⎦ = s1 v1 + s2 v2 . Solve
0
1
these two systems of equations, we have t1 = −3, t2 = 1, s1 = −1 and
s2 = 0. Then we have
⎡
⎤
⎡
⎤
−2
−1
y = y2 ⎣ 1 ⎦ + y3 ⎣ 0 ⎦
0
1
= y2 (−3v1 + 1v2 ) + y3 (−1v1 + 0v2 )
= (−3y2 − y3 ) v1 + y2 v2 .
Set a1 = −3y2 − y3 and a2 = y2 . We have y = a1 v1 + a2 v2 . Therefore,
{v1 , v2 } spans the plane x1 + 2x2 + x3 = 0. And, we show that {v1 , v2 }
is a basis of the plane x1 + 2x2 + x3 = 0.
¥