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.
Question from Dalec
For #2 on page 351, I found my spectrum to be lambda = 2i , and -i. For the case where lambda = 2i , I am trying to find the eigenvectors, and I get a matrix
[ -i 1+i | 0] [ -1+i -2i | 0]
Is there a way to get a 0 in the bottom left, or is this simply overcontrained?
- Chris
I have questions about determinants. For a homogeneous systems, for non-zero determinants we have only the trivial solution while for zero determinant we have infinitely many solutions. For non-homogeneous system, when the determinant is non-zero we have exactly one solution. 1. What will happen if a non-homogeneous system has zero determinant? 2. From the determinant of a non-homogeneous system can we know when the system doesn't have any solution?
- Farhan
Question from Ryan Russon:
About p. 338, #3,6, and 8, are we supposed to be finding eigenvectors here? I noticed that they put them in the back of the book, although it only asks to find the spectrum of each, which was defined as the set of eigenvalues in 8.1? I understand that we are using Thms 1-5 to prove our results and it seems like #3 doesn't require finding eigenvectors to prove that it isn't any of the listed matrices. I hope I am not way off-base here. Thanks!
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