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Linear Algebra I 1016-
Test 2 Solution
- (6p) We know that the following matrix A has the reduced row echelon form R:
A =
= R.
(a) Find a basis for row(A).
The nonzero rows from R:
[
]
[
]
(b) Find a basis for col(A).
The pivot columns from A:
(c) Find a basis for null(A).
We solve the system given by Ax = 0: if the variables are x, y, z and u, then from
R we see that y and u are free and z = −u and x = u − 2 y. This means that the
solutions are (^)
x
y
z
u
u − 2 y
y
−u
u
This shows that the answer is
- (6p) Find the dimension of
span{
This span is just the row space of the previous A matrix. The dimension is 2.
- (6p) Find the rank of A
T A, if
A =
[
]
In class, we proved that the rank of A
T A is the same as the rank of A. So the answer
is 2 , as we can see from here:
A =
[
]
R 2 − 2 R 1 ;
[
]
- (6p) Find the matrix of the linear transformation which projects a vector v in R
3
onto the vector u =
As we know, this is
P =
uu
T
uT^ u
[
]
[
]
- (6p) Find the inverse of the matrix
a b c d
We compute:
A =
t 1 1 1
1 t 1 1
1 1 t 1
1 1 1 t
R 1 ↔R 4 ;
1 1 1 t
1 t 1 1
1 1 t 1
t 1 1 1
R 2 −R 1 ,R 3 −R 1 ,R 4 −tR 1 ;
1 1 1 t
0 t − 1 0 1 − t
0 0 t − 1 1 − t
0 1 − t 1 − t 1 − t
2
R4+R 2 ;
1 1 1 t
0 t − 1 0 1 − t
0 0 t − 1 1 − t
0 0 1 − t 2 − t − t
2
R4+R 3 ;
1 1 1 t
0 t − 1 0 1 − t
0 0 t − 1 1 − t
0 0 0 3 − 2 t − t
2
Now 3 − 2 t − t
2 = −(t − 1)(t + 3), so when t = −3 , then the rank is 3. (For the
other possible value, t = 1, the rank is 1.)
- (6p) Let A be an n × m matrix. Show that for any choice of b ∈ R
n , the linear
system of equations
A
T Ax = A
T b
is consistent (x ∈ R
m ).
We know that a system Ax = b is consistent if rank(A) = rank(A|b). We also proved
in class that rank(AB) ≤ rank(A) and that rank(A
T A) = rank(A) = rank(A
T ). Now
rank(A
T ) = rank(A
T A) ≤ rank(A
T A|A
T b) = rank(A
T (A|b)) ≤ rank(A
T ).
Looking at the two ends shows that all these ranks are equal, and we are done.
- (6p) Suppose that A is an n × n matrix and A
3 = O (the zero matrix). Show
that I − A is invertible.
We compute:
(I − A)(I + A + A
2 ) = I + A + A
2 − A − A
2 − A
3 = I.
We found the inverse.
- (6p) Suppose that A and B are n × n matrices such that AB = O (the zero
matrix). Show that
rank(A) + rank(B) ≤ n.
As we know, elements in the column space of B are given by the form Bx. Now
A(Bx) = ABx = Ox = 0, which shows that col(B) ⊆ null(A), and then the dimen-
sion of the column space of B (which is just rank(B)) is less than or equal to the
dimension of the null space of A, which is the nullity of A. So we obtain that
rank(A) + rank(B) ≤ rank(A) + nullity(A) = n
and we are done.
