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Re: Discrete Maths Sem 2 2016 The 'holes' would be the parity (odd or even) and the 'pigeons' the four numbers. There are two holes (odd or even). Since ceil(4/2) = 2, there are at least two numbers of the same parity.

Yeah, I wrote that it was trace.

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...

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.

$\noindent Using $\frac{1}{w} = \frac{\overline{w}}{|w|^{2}}$, we have $z = -2\times \frac{1 - i\sqrt{2}}{1+2} = \frac{2}{3}\left(-1 + i \sqrt{2}\right)$. Hence $\mathrm{Arg}\left(z\right) = \tan^{-1} \left(\frac{\sqrt{2}}{-1}\right) + \pi = \pi - \tan^{-1} \left(\sqrt{2}\right)$. We can type...

Re: MATH1231/1241/1251 SOS Thread $\noindent Note that $I_n$ was defined as $\int_{0}^{\frac{\pi}{4}} \tan ^{n}x \, \mathrm{d}x$ (for $n=0,1,2,\ldots$). So by definition (putting $n=0$), we have $I_{0} =\int_{0}^{\frac{\pi}{4}} \tan ^{0}x \, \mathrm{d}x = \int_{0}^{\frac{\pi}{4}} 1 \...

Re: MATH1231/1241/1251 SOS Thread $\noindent Now that we have this formula, just keep using it to find $I_7$ and $I_8$ (noting the initial values of $I_0 = \frac{\pi}{4}$ and $I_1 = \frac{1}{2}\ln 2 $, which we need to find by actually computing those integrals for $n=0,1$). So we have$...

Re: MATH1231/1241/1251 SOS Thread $\noindent For any $n\geq 2$, we have$ $$\begin{align*}I_{n} &= \int _{0}^{\frac{\pi}{4}} \tan^{n}x \, \mathrm{d}x \\ &=\int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \tan^2 x\, \mathrm{d}x \\ &= \int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \left(\sec^2x -1\right)\...

Re: Discrete Maths Sem 2 2016 $\noindent The `holes' are pairs of values (like (Ace, King), (2, Jack), etc.), where the order of pairs doesn't matter. There are $13$ values to choose from for each entry in the unordered pair. The number of such pairs will be the $13$-th \textsl{triangular...

Re: Discrete Maths Sem 2 2016 Yeah, self-inverse functions are called involutions. You can read up about them and find examples of them here: https://en.wikipedia.org/wiki/Involution_(mathematics).

$\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...

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...

$\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...

$\noindent To prove false the converse of 1), easy example, let $T$ be the zero map on $\mathbb{R}^{2}$, then $\mathbf{e}_{1}$ and $\mathbf{e}_2$ are independent (standard basis vectors), but their images under $T$ are both $\mathbf{0}$ and hence \emph{de}pendent.$

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...

Re: First Year Linear Algebra Marathon $\noindent Let $A$ be a square matrix over $\mathbb{C}$ and suppose its eigenvalues are precisely its diagonal elements. Must $A$ be triangular? (Justify your answer.)$

Re: MATH1231/1241/1251 SOS Thread $\noindent Note that $A$ has columns that sum to 1. You can show as an exercise that this implies that the sequence $\mathbf{x}(0), \mathbf{x}(1), \mathbf{x}(2),\ldots$ has \emph{constant entry sum}$.$ $\noindent So basically the key result is \emph{if a...

Re: MATH1231/1241/1251 SOS Thread $\noindent By Cauchy-Schwarz, we have $\frac{a_1}{1} + \frac{a_2}{2} + \cdots + \frac{a_n}{n} \leq \sqrt{a_1 ^2 + a_2 ^2 + \cdots + a_n ^2}\sqrt{\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}}$. By the Basel problem series and assumption of...

Re: MATH1231/1241/1251 SOS Thread $\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 $A^k \mathbf{v} = \lambda^{k}\mathbf{v}$, we have$ $$\begin{align*}A^{k+1}\mathbf{v} &= A\left(A^k...

Re: MATH1231/1241/1251 SOS Thread $\noindent If $a>1$, the series converges using the comparison test and `$p$-test', since $0\leq \frac{1}{k^a \ln k} \leq \frac{1}{k^a}$ for all $k$ sufficiently large (say $k\geq 3$) and $\sum_{k=2}^{\infty} \frac{1}{k^a}$ converges if $a>1$.$ $\noindent If...