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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. | 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 ([[User:Jayling|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. | ||
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Revision as of 10:40, 5 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.)
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.
