maths 1B last minute questions (2 Viewers)

Drsoccerball · Oct 30, 2016

$1) T is a linear transformation in $ L(U,V) $ and $ u_1, u_2,..u_n \in U. $ Given that $ T(u_1), T(u_2)...T(u_n) $ are linear independent elements of V, prove that $u_1,u_2...u_n $ are also linear independent. Also prove the converse is false.$

$2) Prove that the Eigenvalues of an n x n matrix is equal to the Eigenvalues of its transpose.$

$3) A is an n x n matrix with real elements. Prove that if $ \lambda $ is an Eigenvalue of A, then $ \lambda^k $ is an Eigenvalue of $ A^k $ where k is a positive integer.$

$$ 4) $ V_1 , V_2, V_3 $ are vector spaces over the field F. $ T_1 \in L(V_1,V_2) $ and $ T_2 \in L(v_2,v_3), $ where $ T_2(V_2) = V_3 $ and $ Ker(T_2) = T_1(V_1) $ Prove that: $ \\ Dim(V_1) - Dim(V_2) + Dim(V_3) = Dim(Ker(T_1))$

He-Mann · Oct 30, 2016

Using definition of linear independence:

$1) $There exists $\lambda_1, \dots, \lambda_n,$ not all simultaneously zero such that:$ \\ \sum_{i=1}^n \lambda_i T(u_i) = 0 \iff T(\sum_{i=1}^n \lambda_i u_i) = 0 \iff \sum_{i=1}^n \lambda_i u_i = 0$ as $T(0)= 0$ for linear transformation$.$

Thus, {u_i} are linearly independent. Converse uses the same ideas.

He-Mann · Oct 30, 2016

2) Hope this makes sense, I can't Engrish!

$\begin{align*}Av = \lambda v & \iff (A - I \lambda)v = 0 \\&\iff \mathrm{det}(A - I \lambda) = 0\\ & \iff \mathrm{det}[(A - I \lambda)^T] = 0 \\ &\iff \mathrm{det}(A^T - I \lambda) = 0 \end{align*} \\\\ $This means the the polynomials $\mathrm{det}(A - I \lambda)$ and $ \mathrm{det}(A^T - I \lambda) $ are the same and when set to zero (to solve for eigenvalues) will give same solutions. Hence, eigenvalues of $A$ is equal to eigenvalues of $A^T$.$

You could also start with A^T v = lambda v and find the characteristic equation (ah, this is the term I was looking for) of A^T and realise it's the same as the characteristic equation as A.

InteGrand · Oct 30, 2016

3)

$\noindent An easy induction! If $k=1$, it's true by definition. Then assuming it's true for some particular $k\in \mathbb{Z}^{+}$, i.e. that $A^k \mathbf{v} = \lambda^{k}\mathbf{v}$ ($\mathbf{v}$ a non-zero vector), we have$

$\begin{align*}A^{k+1}\mathbf{v} &= A\left(A^k \mathbf{v}\right)\\ &= A\left(\lambda^k \mathbf{v}\right) \quad (\text{inductive hypothesis}) \\ &= \lambda^k A\mathbf{v} \\ &= \lambda^k \lambda \mathbf{v} \\ &= \lambda^{k+1}\mathbf{v},\end{align*}$

$\noindent and so the result is true by induction.$

InteGrand · Oct 30, 2016

He-Mann · Oct 30, 2016

Thanks for the solution, InteGrand! Do you know if there is another way to do it? I have a solution but it requires A to be diagonalizable. I had another idea that does this:

$WRONG$ A^kv = \lambda^k v, $ then try to prove: $ \mathrm{det}(A^k - I \lambda^k) = 0 $WRONG$

Ignore the above TeX, the following is what I meant:

$Aw = \lambda w, $ then try to prove: $ \mathrm{det}(A^k - I \lambda^k) = 0$
Using factorization:

$\begin{align*} \mathrm{det}(A^k - I \lambda^k) &\iff \mathrm{det}[(A - I \lambda)(\text{poly degree k - 1})] \\ &\iff \mathrm{det}(A - I \lambda)\mathrm{det}(\text{poly degree k- 1}) \\ &\iff 0 \times \mathrm{det}(\text{poly degree k - 1}) = 0\end{align*} \\\\ $This means $\lambda^k$ is an eigenvalue of $A^k$.$

Not sure about it.

InteGrand · Oct 30, 2016

$\noindent To prove 1) by contrapositive, assume the $\mathbf{u}_{k}$ are linearly dependent and show the $T\left(\mathbf{u}_{k}\right)$ are too. Assuming that the $\mathbf{u}_{k}$ are dependent, we have a non-trivial solution $\left(\alpha_1,\ldots, \alpha_{n}\right)$ to the equation $\alpha_{1}\mathbf{u}_1 + \cdots + \alpha_n \mathbf{u}_n = \mathbf{0}_{U}$. By relabelling we can assume $\alpha_{1}\ne 0$. Then applying $T$ to both sides, using linearity, and using $T\left(\mathbf{0}_U\right) = \mathbf{0}_V$, we have $\alpha_1 T\left(\mathbf{u}_1\right) + \cdots + \alpha_n T\left(\mathbf{u}_n\right) = \mathbf{0}_V$, so the $T\left(\mathbf{u}_{k}\right)$ are dependent, as we have expressed the zero vector as a non-trivial linear combination of them.$

Reading He-Mann's solution more carefully for this now, I think this is what he was doing (wasn't sure on first reading).

He-Mann · Oct 30, 2016

My mistake on the not being precise with where each zero-vector belonged to which vector space.

InteGrand · Oct 30, 2016

Drsoccerball said:
$1) T is a linear transformation in $ L(U,V) $ and $ u_1, u_2,..u_n \in U. $ Given that $ T(u_1), T(u_2)...T(u_n) $ are linear independent elements of V, prove that $u_1,u_2...u_n $ are also linear independent. Also prove the converse is false.$

$2) Prove that the Eigenvalues of an n x n matrix is equal to the Eigenvalues of its transpose.$

$3) A is an n x n matrix with real elements. Prove that if $ \lambda $ is an Eigenvalue of A, then $ \lambda^k $ is an Eigenvalue of $ A^k $ where k is a positive integer.$

$$ 4) $ V_1 , V_2, V_3 $ are vector spaces over the field F. $ T_1 \in L(V_1,V_2) $ and $ T_2 \in L(v_2,v_3), $ where $ T_2(V_2) = V_3 $ and $ Ker(T_2) = T_1(V_1) $ Prove that: $ \\ Dim(V_1) - Dim(V_2) + Dim(V_3) = Dim(Ker(T_1))$

4) let r_k be the rank of T_k and n_k the nullity of T_k (k = 1,2). Let d_j be the dimension of V_j (j = 1,2,3).

We have im(T_2) = T_2 (V_2) = V_3, so r_2 = d_3 (taking dimensions). By Rank-Nullity, we thus have n_2 = d_2 - d_3.

We also have im(T_1) = T_1 (V_1) = ker(T_2). Taking dimensions, we have r_1 = n_2. Hence by Rank-Nullity Theorem, d_1 = n_1 + r_1 = n_1 + n_2.

Now, d_1 - d_2 + d_3 = (n_1 + n_2) - (n_2) (using our established relationships above)

= n_1,

As required.

Drsoccerball · Oct 30, 2016

Thanks guys

! Will be back for more soon, I pretty much skipped 70-80 % of my lectures so I need your help !

InteGrand · Oct 30, 2016

He-Mann said:
Thanks for the solution, InteGrand! Do you know if there is another way to do it? I have a solution but it requires A to be diagonalizable. I had another idea that does this:

$WRONG$ A^kv = \lambda^k v, $ then try to prove: $ \mathrm{det}(A^k - I \lambda^k) = 0 $WRONG$

Ignore the above TeX, the following is what I meant:

$Aw = \lambda w, $ then try to prove: $ \mathrm{det}(A^k - I \lambda^k) = 0$
Using factorization:

$\begin{align*} \mathrm{det}(A^k - I \lambda^k) &\iff \mathrm{det}[(A - I \lambda)(\text{poly degree k - 1})] \\ &\iff \mathrm{det}(A - I \lambda)\mathrm{det}(\text{poly degree k- 1}) \\ &\iff 0 \times \mathrm{det}(\text{poly degree k - 1}) = 0\end{align*} \\\\ $This means $\lambda^k$ is an eigenvalue of $A^k$.$

Not sure about it.

$\noindent This is a valid method too, e.g. because the factorisation $A^{k} - B^{k} = \left(A -B\right) \left(A^{k-1} + A^{k-2} B + A^{k-3}B^2 + \cdots + AB^{k-2} + B^{k-1}\right)$ for same-sized square matrices holds \emph{if $A$ and $B$ commute ($AB = BA$)}. (If they don't commute, this won't generally hold.) So using $B = \lambda I$, since $A$ and $\lambda I$ commute, the factorisation is valid.$

He-Mann · Oct 30, 2016

InteGrand said:
$\noindent This is a valid method too, e.g. because the factorisation $A^{k} - B^{k} = \left(A -B\right) \left(A^{k-1} + A^{k-2} B + A^{k-3}B^2 + \cdots + AB^{k-2} + B^{k-1}\right)$ for same-sized square matrices holds \emph{if $A$ and $B$ commute ($AB = BA$)}. (If they don't commute, this won't generally hold.) So using $B = \lambda I$, since $A$ and $\lambda I$ commute, the factorisation is valid.$

Thanks for the verification. I didn't think of commutativity of A and B, thanks for that aswell.

Drsoccerball · Oct 30, 2016

$$ If $ u(x,t) = f(x+tu(x,t)) $ prove that $ u(x,t) $ is a solution to $ \frac{du}{dt} = u \frac{du}{dx}$

EDIT: nevermind got it

Drsoccerball · Oct 30, 2016

If we are given the eigenvalues of the matrix with corresponding eigenvectors what is the simplest way to find the original matrix without doing A = MDM^(-1)?

InteGrand · Oct 30, 2016

Drsoccerball said:
If we are given the eigenvalues of the matrix with corresponding eigenvectors what is the simplest way to find the original matrix without doing A = MDM^(-1)?

In some special cases there may be some shortcuts, but in general I don't think there's much we can do that's simpler than just doing A = MDM^-1.

InteGrand · Oct 30, 2016

By the way, a nice and easy check you should do after doing such a question (finding the matrix A when given its eigenvalues and eigenvectors): check that the sum of diagonal elements of the matrix A you found equals the sum of the given eigenvalues. (It should be the case, because for any square matrix, the sum of diagonal elements (called the trace of the matrix) is equal to the sum of its eigenvalues.)

Another test you could do (which is more tedious but would give further confidence in your answer) would be to multiply your matrix A with a given eigenvector v and check that it satisfies the relation Av = λv (where λ is the given eigenvalue associated with that eigenvector v).

Drsoccerball · Oct 30, 2016

Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:

$\sum_{k=0}^{\infty}a_k < \sum_{k=0}^{\infty}b_k $ If we know $ \sum_{k=0}^{\infty}a_k $ converges $$

2) To prove that $\ \sum_{k=0}^{\infty}a_k$ converges we can prove that $\lim_{k \rightarrow \infty}a_k = 0 $ and that $ \lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k} = 1 $ If this is true then how come following the same logic, $ \sum_{k=1}^{\infty}\frac{1}{k} $ diverges? $$

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?

Drsoccerball · Oct 30, 2016

Drsoccerball said:
Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:

$\sum_{k=0}^{\infty}a_k < \sum_{k=0}^{\infty}b_k $ If we know $ \sum_{k=0}^{\infty}a_k $ converges $$

2) To prove that $\ \sum_{k=0}^{\infty}a_k$ converges we can prove that $\lim_{k \rightarrow \infty}a_k = 0 $ and that $ \lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k} = 1 $ If this is true then how come following the same logic, $ \sum_{k=1}^{\infty}\frac{1}{k} $ diverges? $$

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?

From my knowledge 1 is completely wrong. If a_k diverges then that can prove that b_k diverges. Also from my knowledge of 2 if the ratio is 1 then we have to do integral test?

EDIT: and for 3 we just have to prove a_n > a_{n+1}, a_{n} -> 0 as n -> infty, a_n >= 0?

leehuan · Oct 30, 2016

InteGrand said:
By the way, a nice and easy check you should do after doing such a question (finding the matrix A when given its eigenvalues and eigenvectors): check that the sum of diagonal elements of the matrix A you found equals the sum of the given eigenvalues. (It should be the case, because for any square matrix, the sum of diagonal elements (called the trace of the matrix) is equal to the sum of its eigenvalues.)

Another test you could do (which is more tedious but would give further confidence in your answer) would be to multiply your matrix A with a given eigenvector v and check that it satisfies the relation Av = λv (where λ is the given eigenvalue associated with that eigenvector v).

Do you call it the trace or something? I don't quite remember.

Drsoccerball said:
Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:

$\sum_{k=0}^{\infty}a_k < \sum_{k=0}^{\infty}b_k $ If we know $ \sum_{k=0}^{\infty}a_k $ converges $$

2) To prove that $\ \sum_{k=0}^{\infty}a_k$ converges we can prove that $\lim_{k \rightarrow \infty}a_k = 0 $ and that $ \lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k} = 1 $ If this is true then how come following the same logic, $ \sum_{k=1}^{\infty}\frac{1}{k} $ diverges? $$

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?

$\text{For 1) look at your ordering. That's false.}$

$\text{With the ratio test, if }\lim_{k \to \infty}\frac{a_{k+1}}{a_k}=1\text{ then the test is inconclusive.}\\ \text{In some cases there's convergence, in others there's divergence.}$

$\text{Convergence is guaranteed iff }\lim_{k\to \infty}\frac{a_{k+1}}{a_k}<1$

For our purposes with 3) we must rely on the alternating series test.

InteGrand · Oct 31, 2016

leehuan said:
Do you call it the trace or something? I don't quite remember.

Yeah, I wrote that it was trace.

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