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
Eun Young discussed this issue here in a way that is slightly beyond the scope of our course, so I've moved it to here:
Remark from Steve Bell :
Step 1: Find the eigenvalues from det(A - lambda I)=0.
Step 2: Choose an eigenvalue lambda and plug it into the system
(A - lambda I) a = 0
and solve the system for the eigenvector a. Swapping rows does not change the answer, so you are safe here.
Sometimes you might think you are swapping entries of a vector when you are really multiplying by -1. For example , if [1, -1] is an eigenvector, so is [-1, 1].
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
Suggestions from Shawn Whitman
In one step: multiply row 1 by (1+i) and add to row 2.
In two easier steps: Multiply row 1 by i,
[1, (-1+i)]
[(-1+i), -2i]
then multiply row 1 by (1-i) and add to row 2.
[1, (-1+i)]
[0, 0]
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
Suggestion from Ryan Russon
Here is what I understand:
For question 1) If we are thinking of a system of equations, then by looking at the determinant, we are only looking at the left-hand side (LHS) of the system. If the determinant of that system is zero, it means that one or more of those equations are dependent on the others. Said differently, one or more of those expressions can be put together by combining the other expressions from the LHS of the system. This also means that any non-homogenous system formed from the components of the LHS expressions may have more than one way to be combined to get the desired solution (i.e. $ \bar{x} $ is not unique). Now if the expanded system looks like:
[1 4 1 | 4] [0 2 0 | 1] [0 0 0 | 3]
where you have a statement that "0=3" this is obviously a bad system.
For 2) From the determinant alone, it would not be possible to determine if the system has no solutions. If it is zero, it may have infinitely many or it may be an undetermined system.
Please others chime in and correct me if I am flawed in my thinking.
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!
Follow-up question: On p. 338, #6 Are we only to consider $ A \in \mathbb{R}^{n \times n} $ or are we to consider complex matrices as well? Thanks again!
Response from Jake Eppehimer:
I found that #8 is orthogonal, according to theorem 5. It took quite a bit of manipulation with trig identities, but I believe my answer is reasonable. For number 6, I am not exactly sure how to find the eigenvalues. I am considering substituting a couple prime numbers for k and a, but I am unsure if that is the correct way to do it. It doesn't say anything about eigenvectors, and you don't need them to determine what kind of matrix it is.
Response from Ryan Russon
Thanks Jake. I found the eigenvectors for #6 to be: $ \lambda_{1} = a , \lambda_{2} = a+k , \lambda_{3} = a-k $ by using a cofactor expansion which wasn't too bad. And I think I am a little brain dead today as I can answer my own follow-up: We are obviously not considering $ A \in \mathbb{C}^{n \times n} $ because we are talking about 'symmetric, skew-symmetric, and orthogonal' matrices which are only classes of real-valued matrices.
Response from Jayling: Ryan I was confused with the definition of Spectrum also. But Steve did state in the last lecture that it was the set of all eigenvalues of A. Also I found via the index in the text that Spectrum is indeed this (see first paragraph of page 324). In summary no need to calculate eigenvectors if the question is asking for the spectrum.
Also with Question 6 I am getting a very nasty looking characteristic equation, so I am not to sure how to solve for the algebraic roots.
Response from Hzillmer Maybe I'm overthinking things but for Question 3 here I got the e-vals to be 2+8i and 2-8i which fails theorem 5 that the absolute value must be 1. Does anyone have a thought as to what I'm missing here?
Response from Kees
Its not orthogonal so there is nothing to prove in theorem five. It can fail theorem 5 since it has no reason to pass it. Easy question, since it is none of the three, I do not have to prove any theorem's, correct? I only have to prove, for example, the orthogonality theorems when it is orthogonal etc, not every theorem every time?
Response from Jayling: If you just do the calculation AA$ ^T $ you will see that you do not get the Identity Matrix and therefore it is not orthogonal. If your eigenvalues are real then the matrix is symmetric, if your eigenvalues are pure imaginary or zero then the matrix is skew symmetric, and if your eigenvalues are real or have complex conjugate pairs where the absolute value is 1 then you have an orthogonal matrix.
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