06-240/Final Exam Preparation Forum

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If you have questions, ask them here and hopefully someone else will know the answer. (Answering questions will probably help you understand it more).

Since many of us (including I) don't really know how to use Wiki's, I suggest that we keep the formatting simple: I will post a template at the top of this page, and if you want to add something just click on the "edit", copy the template, and insert your question. Order the questions according to section (i.e. solved/unsolved; whoever created the question must decide if it is solved, and sort it accordingly), with the newest at the top, except for the template question. In general, I wouldn't retype the question if it's from the book because that's tedious and we all have the book.

(By the way, I think you leave a space between lines in the code to make a new line; that is, simply pressing enter once will not make a new line. Also, you can press a button at the top of the editing textbox that lets you put in simple equations.)

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Solved Questions

Question Template

Q: How many ways are there to get to the nth stair, if at each step you can move either one or two squares up?

A: This question can be easily modeled by the Fibonacci numbers, with the nth number being the ways to get to the nth stair. This is because, to get to the nth stair, you can come only from the n-1th or the n-2th. This is exactly how the Fibonacci numbers are defined; the proof is simple by induction.

Sec 3.2 Ex. 19

Q: "Let A be an m x n matrix with rank m and B be an n x p matrix with rank n. Determine the rank of AB. Justify your answer." I know how to find that the rank can't be more than m (not much of an accomplishment), but I can't finish it.

A: According to Theorem 3.7(a),(c)&(d)(p.159), I would say rank(AB) $\leq$min(m, n).

R: Can we not get any more specific than that?

A: "Let $L_{A},L_{B},L_{AB}$ have their usual meanings. Then $L_{B}:F^{p}->F^{n}$ is onto. Then we get $R(L_{AB})=R(L_{A}L_{B})=L_{A}L_{B}(F^{p})=L_{A}(F^{n})=R(L_{A})$, i.e. $rank(L_{AB})=rank(L_{A})=m$."

Sec. 3.2 Ex. 21

Q: "Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B such that AB=$I_{m}$".

A: "Take any n x m matrix B with rank n. By exercise 19 in the same section rank AB = rank A = m, hence AB is invertible. Let M be the inverse of AB, then (AB)M = A(BM) = I, i.e. BM is the desired matrix."

Sec. 1.3 Thm 1.3 Proof

Q: In the first paragraph of the proof, it says "But also x + 0 = x , and thus 0'=0." How do we know 0 (that is 0 of V) even exists in W? I understand that we know some zero exists (0'), but not why the zero (0) exists.

A: x is in W as well in V. Thus, x + 0 = x (VS 3).

Reply: Oh I see... now it looks so obvious =/. Thanks.

Exam April/May 2006 #3(b)

Q: Let T : M_3x2(C) -> M_2x3(C) be defined as follows. Given A Є M_3x2(C), let B be the matrix obtained from A by adding i times the second row of A to the third row of A. Let T(A) = B^t, where B^t is the transpose of B. (Note: Here, i is a complex number such that i² = -1.) Determine whether the linear transformation T is invertible.

Totally lost on this question :/ Please show some example matrix and how it is transformed as the question asks if possible. I want to see what actually happens to the elements in the matrix rather than the answer (think that would be more important)

A(Matrix Elements): This is my interpretation:

A = ${\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{pmatrix}}$ where $a_{ij}\in C$, then B = ${\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ia_{21}+a_{31}&ia_{22}+a_{32}\end{pmatrix}}$.

Therefore, T(A) = B^t is T${\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{pmatrix}}$=${\begin{pmatrix}a_{11}&a_{21}&ia_{21}+a_{31}\\a_{12}&a_{22}&ia_{22}+a_{32}\end{pmatrix}}$

R: Thx alot, the matricies are really helpful :)

Sec. 2.4 Lemma p. 101

Q: In the proof of the lemma, the second line, we have "T(beta) spans R(T) = W". How do we know that R(T) = W? This would be true if dim V = dim W, because then T would be onto, but we can't assume what we're trying to prove.

A: The first line of the Lemma states, "Let T be an invertible linear trans..." So, T is onto(and 1-1), thus "T(beta) spans R(T) = W".

R: Yes. My trouble was with the fact that invertibility implies onto-ness. I thought that if we had $T:P_{2}(R)->P_{6}(R)$, and T(f) = xf, then T would still be invertible since you can 'recover' the f if you were given xf. I guess it makes more sense to not call T invertible in this case, because $T^{-1}$ is technically only one-to-one over the range of T.

R: T has to be both Onto and 1-1 so that it's invertible. In your example, some of the $f\in P_{6}(R)$ will not be 'recovered' because they weren't mapped from $P_{2}(R)$. Furthermore, T^-1 has to map the whole vector space W back to V(as defined on p.99) but not the range only. In other words, if T is 1-1 only, $T^{-1}\circ T(v)=v,\forall v\in V$ but $T\circ T^{-1}(w)\neq w,\exists w\in W$, because some T^-1(w) are not defined.

R: That nicely rigorizes what I was thinking, and I'm convinced. Thanks.

Exam April/May 2006 #7

Q: Let T : V -> V and U : V -> V be linear operators on a finite-dimensional vector space V. Assume that U is invertible and T is diagonalizable. Prove that the linear operator UTU^-1 = U o T o U^-1.

I dont know where or how to start this question ><.

A: I think we need to prove that UTU^-1 is diagonalizable instead of proving UTU^-1 = U o T o U^-1.

I started by letting A = UTU^-1, then multiplying U^-1 and U to the both sides, we get U^-1AU = U^-1UTU^-1U iff U^-1AU = T. Since T is diagonalizable, therefore there exists an invertible matrix Q s.t Q^-1TQ = D, where D is a diagonal matrix. Therefore, Q^-1(U^-1AU)Q = D iff (UQ)^-1A(UQ) = D (because U and Q invertible, Q^-1U^-1 = (UQ)^-1), it follows that A is diagonalizable.

Exam April 2004 #6(a)

Q: Suppose A is an invertible matrix for which the sum of entries of each row is a scalar $\lambda$. Show that the sum of entries of each row of A^-1 is $1/\lambda$. (Hint: find an eigenvector for A with eigenvalue $\lambda$.) If A is a diagonal matrix, then it's obvious that the sum of entries of each row is $\lambda$ and the sum of entries of each row of A^-1 is $1/\lambda$. I was stuck with a more general invertible matrix.

A: Following the hint, you can see that an eigenvector corresponding to $\lambda$ is (1, 1, 1, ...)*. Therefore $Av=\lambda v$, and rearranging you get $A^{-1}v=1/\lambda v$. Using the same logic as before, you can show that since this $\lambda$ corresponds to a homogeneous system of equations with the same eigenvector v = (1, 1, 1, ...), the sum of each row is equal to $1/\lambda$.

• Just to elaborate on the first part, you are looking for a vector $v=(x_{1},x_{2},x_{3},...)$ so that $A-\lambda I=0$. This corresponds to the system:

${\begin{pmatrix}(a_{11}-\lambda )x_{1}&a_{12}x_{2}&a_{13}x_{3}&...\\a_{21}x_{1}&(a_{22}-\lambda )x_{2}&a_{23}x_{3}&...\\a_{11}x_{1}&a_{12}x_{1}&(a_{13}-\lambda )x_{3}&...\end{pmatrix}}$, and so in each row you can see that $x_{1}=1,x_{2}=1,x_{3}=1$ works because then all the a's in each row add up to $\lambda$.

• Also, does anyone know how to do part (b) of that question? My guess is to make one subspace {0}, the second (t,0,0) and the third (0,r,s) for all t,r,s,. Does that look okay?

R: Thanks. I think the subspaces are {0}, {(t,0,0)} and {(0,s,0)} so that $R^{3}\neq W_{1}\oplus W_{2}\oplus W_{3}$.

R: We need them to add up to $R_{3}$ though. Anyway, hopefully we won't need to know about direct sums.