EEL 3105
Fall 2011
Problem Set 2
1.
Consider the vector
[1
2.8
3
3.33]
v
Find vv
T
. Is it symmetric? Find its eigenvalues. Is it invertible?
Solution: We can calculate vv
T
directly. It is a scalar and its value is 28.92. Since it is a scalar, it is
symmetric. Again, since it is a scalar, its (only eigenvalue) is 28.92. Since this eigenvalue is nonzero, it is
invertible.
Although, I did not ask for this, let us consider v
T
v. This is the 4x4 matrix:
1.0000
2.8000
‐
3.0000
3.3300
2.8000
7.8400
‐
8.4000
9.3240
‐
3.0000
‐
8.4000
9.0000
‐
9.9900
3.3300
9.3240
‐
9.9900
11.0889
What if I had asked find eigenvalue of v
T
v? What would you do? Would you go about writing
characteristic polynomial of this 4x4 matrix? If so, that would be a long and tedious process. And it
would certainly not work for Problem2.
On the other hand, we can go back to the last homework where we showed any nonzero eigenvalue of
AB is also a nonzero eigenvalue of BA.
Now, let us set A to be v
T
and B to be v. Then BA has only one
eigenvalue, i. e., 28.92. Therefore, AB which is the above 4x4 matrix must have 28.92 as its eigenvalue.
Can AB= v
T
v have any other non
‐
zero eigenvalue? Well, if it did then that non
‐
zero eigenvalue of v
T
v
would be an eigenvalue of
vv
T
but
vv
T
has only one eigenvalue, 28.92. Thus, all other eigenvalues of v
T
v
must be zero. Thus, the only eigenvalue of v
T
v are 28.92 and zero. In fact, v
T
v has a triple eigenvalue at
zero.
2.
Let v be an nx1 vector. Find eigenvalues of vv
T
. For choices of v is vv
T
invertible?
Solution: First, if v is the zero vector, then vv
T
is the zero matrix and its eigenvalues are all zero
and it is not invertible. Now, suppose v is a non
‐
zero nx1 vector, vv
T
is a nxn matrix. To find its
eigenvalues, we first see that v
T
v is a non
‐
zero scalar and is the one and only eigenvalue of v
T
v.
Now, we know that nonzero eigenvalues of a matrix product AB and BA are the same. With
A=v and B=v
T
, we can use this fact to conclude that the only non
‐
zero eigenvalue of vv
T
is v
T
v

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