Homework 3 collaboration area
Question from James Down Under (Jayling):
For Page 329 Question 11. Am I meant to calculate all eigenvalues and eigenvectors or just calculate the eigenvector corresponding to the given eigenvalue of 3?
Answer from Steve Bell :
Yes, you are only supposed to find the eigenvector for lambda=3. (The idea here is to spare you from finding the roots of a rather nasty 3rd degree polynomial.)
Jayling: thanks Steve, I did try the hard way first but then started to drown in the algebra.
Question from a student:
Let 3x+4y+2z = 0; 2x+5z= 0 be the system for which I have to find the basis.
When Row Reduced the above system gives [ 1 0 2.5 0 ; 0 1 -1.375 0].
Rank = no of non zero rows = 2 => Dim(rowspace) = 2 ; Nullity = # free variables = 1
Q1: Aren't [ 1 0 2.5] and [0 1 -1.375] called the basis of the system?
A1 from Steve Bell:
Those two vectors form a basis for the ROW SPACE.
The solution space is only 1 dimensional (since the number of free variables is only 1).
Q2: Why is that we get a basis by considering the free variable as some "parameter" and reducing further(and get 1 vector in this case). Isn't that the solution of the system?
A2 from Steve Bell :
If the system row reduces to
[ 1 0 2.5 0 ] [ 0 1 -1.375 0 ]
then z is the free variable. Let it be t. The top equation gives
x = -2.5 t
and the second equation gives
y = 1.375 t
and of course,
z = t.
So the general solution is
[ x ] [ -2.5 ] [ y ] = [ 1.375 ] t [ z ] [ 1 ]
Thus, you can find the solution from the row echelon matrix, but I wouldn't say that you can read it off from there -- not without practice, at least.
Question from a student :
On problem 11, I swapped rows 1 and 2 during row reduction and my final solution has x1 and x2 swapped. Do I need to swap back any row swaps or did I make a mistake along the way? Tlouvar
Answer from Eun Young :
Let's suppose that $ \lambda $ is an eigenvalue of a matrix A. You want to find an eigenvector corresponding to $ \lambda $. To do that you need to solve Ax = $ \lambda $x, which is same as (A- $ \lambda $I) x = 0.
If you solve the 2nd equation (A - $ \lambda $ I) x =0, swapping rows doesn't change your answer.
If you solve the 1 st equation, Ax = $ \lambda $ x, swapping rows changes your answer.
Here's the reason. Let P be a permutation matrix swapping rows 1 and 2.
If you multiply A by P from the left , P will swap the 1 st and 2nd rows of A.
Note that Ax = $ \lambda $x < => PAx = $ \lambda $Px .
This means that if you swap rows of A, rows of x will be swapped too.
However, (A-$ \lambda $I)x = 0 <=> P (A - $ \lambda $I) x = P 0 <=> [P (A - $ \lambda $I)] x = 0 .
This doesn't affect your answer. So, it depends on what equation you use when you swap rows.
