Math 2270
Quiz 9
1. Find an orthonormal basis for the kernel of the matrix
1 1
1 1
A=
.
1 −1 −1 1
We already know how to find a basis for the kernel of A. First we’ll do this, then we’ll use the Gramm-Schmidt
process to turn that basis into an orthonormal basis. We row-reduce A as
1 1
1 1
1 1
1 1
1 0 0 1
∼
∼
.
1 −1 −1 1
0 −2 −2 0
0 1 1 0
This implies that the homogeneous system has the solution
x1 = −x4
.
x2 = −x3
Let x3 = s and x4 = t. Then the solution vector can be written as

 




x1
−t
0
 x2   −s 
 −1 


 




 x3  =  s  = s  1  + t 
x4
t
0
Let

−1
0 
.
0 
1



−1
0
 0 
 −1 



~v1 = 
 1  and ~v2 =  0  .
1
0
√
√
Then {~v1 , ~v2 } is a basis for ker(A). We have that k~v1 k = 1 + 1 = 2, so we take


0
 − √1 
2 
~u1 = 
 √1  .

2
0
To find ~u2 , we find first ~v2⊥ ⊥ span(~u1 ),


 
0
−1
 − √1  
 0 
 − (0)  1 2  = 
~v2⊥ = ~v2 − (~u1 · ~v2 )~u1 = 
 √  
 0 
2
1
0
√
√
We see that k~v2⊥ k = 1 + 1 = 2, so

  √1 
− 2
−1
  0 
1 
1 ⊥
0
=
.
~u2 = ⊥ ~v2 = √ 
k~v2 k
2 0   0 
√1
1
2

Thus an orthonormal basis for ker(A) is

0


 − √1
 12
 √


2

0

− √12 

  0 
 .
,
  0 


√1
 
2

−1
0 
.
0 
1
2. Suppose that A and B are orthogonal n × n matrices. Is the matrix AB also orthogonal? Justify your
answer.
Since A and B are orthogonal, we have that A−1 = AT and B −1 = B T , so
(AB)(AB)T = (AB)(B T AT ) = ABB T AT = AIn AT = AAT = In ,
which implies that (AB)−1 = (AB)T and that AB is orthogonal.