Math 1300: Tips for Assignment 4 (includes a copy of the assignment)

Published: Sat, 12/10/16

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Did you read my tips on how to study and learn this course?  If not, here is a link to those important suggestions:
Did you miss my Tips for Assignment 1? Click here.
Did you miss my Tips for Assignment 2? Click here.
Did you miss my Tips for Assignment 3? Click here.
Tips for Assignment 4
Here is a link to the actual assignment, in case you don't have it handy:
Study Lesson 14 (Markov Analysis), Lesson 15 (Linear Transformations) and Lesson 16 (Eigenvalues and Eigenvectors) from my Linear Algebra & Vector Geometry book to prepare for this assignment.
Question 1
I do not discuss "onto" in my book.  Onto means that there is at least one vector in the original vector space that can be transformed into any vector you can name in the target vector space (the space you are mapping onto).  So, essentially, a linear transformation is onto if its range is the same as the vector space you are mapping onto.  You can add this statement to the Equivalent Statements Theorem I list on page 514:
The matrix is onto.

This question basically combines every part of what I discuss in Lesson 15

Also note that the Equivalent Statements Theorem for Linear Transformations on page 514 of my book applies here since all of these transformations are from R^3 to R^3 the same vector space.  Thus, they will all be onto, provided they are a linear transformation in the first place!  To be onto is to mean that the range is the entire R^3 vector space in this case.

Read page 515 of my book to learn how to prove a transformation is linear or not, and then take a look at Lesson 15, question 8 for examples.  Then, if you have proved it is linear, compute the standard matrix as I teach you and perhaps use Lesson 15, question 5 for examples.  Then, follow my advice from the Equivalent Statements Theorem on page 514.

Essentially, you should proceed like so in this question:
  1. Using my tips at the bottom of page 507 (the first page of Lesson 15), determine by inspection if the transformation is linear or not.  Then, actually prove it is, or is not, a linear transformation as I discuss on page 515, and illustrate in my question 8.
  2. Of course, if you have shown that T is not linear, you are done.  If you have shown it is linear, proceed to find the standard matrix as I discuss on pages 508 and 509, and illustrate in my questions 2-4. 
  3. Use the Equivalent Statements Theorem to determine if T is one-to-one and/or onto.  The easiest way is to use determinants.  Compute the determinant of the standard matrix you found in Step 2.
    1. If the determinant is ZERO the transformation is not one-to-one, and therefore it is also not onto.
    2. If the determinant is not ZERO the transformation is one-to-one, and therefore it is also onto.
  4. In the unlikely event you are doing a question where your standard matrix [T] is not a square matrix, the never fail procedure is to row-reduce [T] to RREF.  Then:
    1. If every COLUMN in the RREF matrix has a leading one, T is one-to-one.
    2. If every ROW in the RREF matrix has a leading one, T is onto.
Additional Hints:
For some reason, in some of the parts they write v but don't bother to then express it as (v1, v2, v3).  In any part where they give you v, make sure you replace v with (v1, v2, v3) and then do the math.  For example, in part (a), compute the cross product of (v1, v2, v3) and (1,2,3).  I recommend that you write the answer for that cross product in column form, like they are doing with their vectors in the other parts.  That should make it easier for you to identify if the transformation is linear or not.

You should also use (v1, v2, v3) to compute the given dot products in parts (d) and (e).
Question 2
Similar to my Lesson 15, questions 2-4.

Part (b)
This link gives you the formula you need.  Ridiculous! As is much of this assignment:
Question 3
Similar to my Lesson 16, questions 3-4.  Those general vectors with parameter t that I come up with for each eigenvalue in my question 4 are what they mean by the associated eigenspace.
Question 4
Part (a)
This way is slow, but never fails.  When given T(v1) = w1, T(v2) = w2, and T(v3) = w3 like we have been given here, the never fail way to find the standard matrix A is:
  1. We can create a matrix V where the first column of V is v1, the second column of V is v2 and the third column of V is v3.  We can also create a matrix W where the first column of W is w1, the second column of W is w2 and the third column of W is w3.
  2. Then, we know the standard matrix A is such that:    AV = W
  3. Therefore, we know A = W*V-inverse.  Thus we must find V-inverse.
  4. Proceed to find V-inverse, the inverse of matrix V, and then multiply W times V-inverse to find the standard matrix A.  Make sure you are multiplying in the correct order!  Must be W in front of V-inverse.  Order matters in matrix equations!
Faster method for the clever students:
We can always very quickly establish the standard matrix A if we know the transformations of the elementary vectors e1, e2, and e3.  So, we need to know T(e1), T(e2), and T(e3).  If you look closely at the given transformations, you can use the properties of linear transformations to derive these required transformations.
  • For example, note how similar v3 is to e1.  Fact about linear transformations is that 1/2 times T(v3) = 1/2 times w3, which is to say, you can multiply both sides of the transformation equation by 1/2 (or any scalar for that matter).  But, 1/2 T(v3) = T(1/2 v3), by properties of a linear transformation.
  • Note how v2 = e1 + e2.  Something similar is going on for v1.
  • If you see what I am getting at, try it; if you don't, there is the never fail way above.
Part (b)
Find the eigenvalues as I teach in Lesson 16, and as you just did in question 4 above, using the standard matrix you found in part (a).

Part (c)
I don't discuss diagonalization in my book, unfortunately.  The key requirement is, in this case you should have a 3by3 matrix A (which they are calling [T] in this part) found in part (a) above.  Since A is 3by3, you must have come up with 3 separate eigenvalues in part (b).  I assume that is what will happen here.  If you get less than 3 eigenvalues, it may still be diagonalizable, but then you would have to find the eigenvectors to be sure.  I have not done the question myself, so I will leave it in your hands.  If you do get three eigenvalues, then you can safely state it is diagonalizable.  If you get less than three, you will have to compute the eigenvectors.  You must be able to come up with three independent eigenvectors.

Part (d)
Use the determinant of A to see if it is invertible.
Question 5
There is a way to do this using eigenvalues, but I don't see how it helps if you don't know anything about Transition matrices, so really, the best way to do this is the method I teach in Lesson 14 (Markov Analysis).

Set up the 2by2 transition matrix, T, labelling the columns B and G, respectively, and also the rows B and G, respectively.  B stands for B wins, and G stands for G wins.  Note that, if B loses, that must mean G wins, so you have been given the entries to put in the B,B and G,G positions of your matrix.  You can then fill in the rest knowing that each column must add up to 1.  Make sure you express those percents in decimal form.

Now, proceed to find the steady-state vector as I instruct in Lesson 14.  My questions 1 and 2 in Lesson 14 are similar.  The fact that Green won the first day is irrelevant.  Do note that your steady-state vector will be in the same order as T, so B first, then G, so the first entry will tell you the probability B wins in the long run.
Question 6
We have to go to the initial definition of an eigenvalue, λ.  We know, to find the eigenvalues for a matrix, A, we are looking for the value, λ, such that:
Ax = λx for any vector x.

Now, realize that A^3 is really A*A*A or AAA.  Thus, AAA = A.  We can then post-multiply, or right multiply both sides by x, to say
AAAx = Ax
Just add a set of brackets.
AA(Ax) = Ax
But, Ax can be replaced by λx.
AA(λx) = λx
But, you can factor that λ out to the front, since it is just a scalar multiple.
λAAx = λx
But that gives you another Ax that you can replace with λx!

If you keep at this, you will eventually get an equation very similar to the given A equation but with λ in place of A.  Then you just need to solve that cubic equation for λ.  Pull everything to one side and factor.