Difference between revisions of "HWK3MA527Fall2013" - Rhea

Revision as of 06:22, 6 September 2013

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 the Linear Algebra Noobee (Jayling) regarding the Lesson 7 material

An observation from me with eigenvalues and eigenvectors on the 2x2 matrices examples that you presented is that the column space of A-λI when you solve for the first eigenvalue corresponds to the column vector of the second eigenvalue? For the example on page 2 when you REF the A-λI the pivot yields the column space vector in the original matrix of [-2 1]T which exactly matches the eigenvector for when λ=-4, and similarly vice versa when λ=-1 and you REF the A-λI matrix it yields a column space vector of [1 1] T. Is this true for all matrix eigenvalue problems in general, I noticed that it is also true for the complex example that follows? Is this an obvious fact or just a coincidence? Your insight on this would be appreciated.

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 :

To find a eigenvector corresponding to a eigenvalue, you basically need to solve a system of linear equations. When you solve a system of linear equations, swapping rows doesn't change your answer. For example, consider the following system of equations:

2x+3y =1

4x+y= 3.

If you swap rows, you have

4x+y=3

2x + 3y = 1.

Two systems are same and have the same solution.

Back to MA527, Fall 2013

Back to MA527, Fall 2013