Difference between revisions of "HWK3MA527Fall2013" - Rhea

Revision as of 05:14, 8 September 2013

Homework 3 collaboration area

MA527 Fall 2013

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

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