$a_k = \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^k)$ is increasing.   This is equivalent to  $$\mathrm{rank}(A^{k+1}) + \mathrm{rank}(A^{k-1}) \geq 2 \mathrm{rank}(A^k) $$  which is a consequence of the Sylvester inequality: $$\mathrm{rank}(XZY) + \mathrm{rank}(Z) \geq \mathrm{rank}(XZ) + \mathrm{rank}(ZY)$$ for $ X = Y = A, Z = A^{k-1}$.  Is there a simpler proof of this fact, without using the full power of Sylvester?
120705
Consider a finite-dimensional vector space $V$ over the field $\mathbb{F}$ with a linear operator $T.$ Prove that we have $\operatorname{rank}(T^3) + \operatorname{rank}(T) \geq 2 \operatorname{rank}(T^2).$   First, let us assume that $\dim_{\mathbb{F}}(V) = n.$ Our immediate observations include that $T^3(V) \subseteq T^2(V) \subseteq T(V)$ so that $\operatorname{rank}(T^3) \leq \operatorname{rank}(T^2) \leq \operatorname{rank}(T)$ and $\operatorname{rank}(T^3) \geq \operatorname{rank}(T) + \operatorname{rank}(T^2) - n$ by Sylvester's Rank Inequality. Can we do some manipulations with what we have to get the desired inequality, or do we require something else?
3416
Can you help me to prove that $$(x+y)^n\leq 2^{n-1}(x^n+y^n)$$ for $n\ge1$ and $x,y\ge0$.  I tried by induction, but I didn't get a result.
47646
Is the inequality $\lvert z_1 + z_2 \rvert \ge \lvert z_1 \rvert - \lvert z_2 \rvert$ incorrect, where $z_1$ and $z_2$ are any two complex numbers? I need an example to prove that it is. And in case it is correct, can you please give the proof? Thanks for any help.
84843
Prove that if $A$ is a square matrix with integer entries and $\det(A) = \pm 1$, then $A^{-1}$ contains all integer entries.  I'm really thrown off by this one, its unlike all the examples I've seen.. I just dont know where to start... how can I begin to prove this?
1837
Consider following inequality:     $|z_1 + z_2|\ge|z_1| - |z_2|  \  \ \ \ \ \  z_1,z_2 \in \mathbb{C}$   Is that true always? I tried many examples and it was correct but I don't know how to prove that if it's right.
88903
Prove that for positive integers $a_1,\dots,a_k$ (where $k\geq 1)$  are such that $a_1+\cdots+a_k &lt; n$ , then $a_1!\cdots a_k! &lt; n!$.   So far, i've tried adding the condition that suppose they were also in increasing order. I've thought about using induction to prove it, but it seems like this is not the right approach to go about this. Without induction, I am struggling to somehow use the fact that:  $$a_1+\cdots+a_k &lt; n \implies (a_1+\cdots+a_k)! &lt; n! $$ and relate $\displaystyle \left(\sum a_i\right)!$ to $\displaystyle \prod a_i!$.  How does one proceed?
83978
If $A^2 = I$ where $A$ is an $n\times n$ matrix, then why is $\operatorname{rank} (A+I) + \operatorname{rank} (A-I) =n$.   I am not sure why this seems to be true.
37435
Call $P(k): k!&lt;k^k$, for $k\geq 2$ Test it out with 2, and it's true ($2&lt;4$).   Assume that $P(k)$ is true for some $k\geq2$.  Then show that $P(k+1)$ is true.   $P(k+1): (k+1)!&lt;(k+1)^{k+1}$  Ministep: $(k+1)!=k!(k+1)$ Ministep: $(k+1)^{k+1}=(k+1)^{k}(k+1)$    $k!(k+1)&lt;(k+1)^{k}(k+1)$  Can I just pull out that (k+1) and call it a day? That will make $k!&lt;(k+1)^{k}$.   We're already assuming that $k!&lt;k^k$, for some integer $k\geq 2$, and I don't think I need to prove that $k^{k}&lt;(k+1)^{k}$. I mean, I guess I could go prove that as well, but let's assume for the moment that I don't have to. Since we're assuming the original $P(k)$ is true, I should be able to write $k!&lt;k^{k}&lt;(k+1)^{k}$ right? That would prove the $k+1$ case. Is this the correct way to go about things? It seems too easy, but makes sense.  Thanks for confirmation and/or pointing out mistakes!
73031
Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.  I want to prove the following inequality using mathematical induction:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?
27107
Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$.   I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions.  This is what I have so far: Let $m = nq + r$, then $a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1$. Since $0\le r\lt n$, $r = 0$ then $a^m - 1$ = $a^{nq} - 1$ so $a^m$ = $a^{nq}$ . Thus $m = nq$ hence $n\mid m$.
41998
Problem:  Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow  rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS} $ is the Hilbert Schmidt Norm.  Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.  Proof of the implication $\Leftarrow $:   If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$ If $rank(A)=1$, than: $ A=\begin{pmatrix} A_{1}\\  \alpha _{2}A_{1}\\  ...\\  \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$   and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$  Proof of the implication $\Rightarrow $   We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$  On the other hand: $ A=\begin{pmatrix} A_{1}\\  A_{2}\\  ... \\ A_{n}  \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\  \left \langle A_{2},x \right \rangle\\  ...\\   \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} $$  Where $ \left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$  Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} $ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$  Please let me know if my solution makes sense.
55204
Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$.    I don't want to use the AM-GM inequality to prove this (from which this statement would follow).   I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by  $$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$  Now suppose the statement holds for $n$, then $$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$  So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then  $$ x_1 + \dots + x_n x_{n+1} \ge n. $$  Then we also have $$ x_1 + \dots + x_{n-1} + x_n + x_{n+1}  \ge n - x_{n}x_{n+1} + x_n + x_{n+1}. $$  Now I'm not sure where else to go from here.
52353
I don't quite understand the proof my notes gave me.  Dimension of $n$ by $n$ matrix is $n^2$. Hence if $k \geq n^2$ then $\mathbf{ \{ I, A, A^2, ..., A^k \} }$ is linearly  dependent. So, there exist scalars $a_i$ not all zero such that $$a_k A^k + \cdots + a_0 I = 0$$  $f(x) = a_k x^k + \cdots + a_0$ is an annihilating polynomial.     I don't understand the bold bit. How do we know that $A, A^2, ...$ are all different? Surely there are matrices which give identity when raised to power of 2,3, etc right?  Thanks.
42565
I've been dealing with this problem for almost 2 hours now, with hardly any progress. I'm to prove the following inequality using induction   $$\sqrt[n+1]{n+1}&lt;\sqrt[n]{n}$$ where $n≥3$, $n∈\mathbb{N}$. How can this be done?
19620
I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:  Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$
116157
If $n\geq m$ show that: $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$, all numbers being positive real.   Obviously since $n\geq m$, for every $a\in \Bbb Z^+$: $a^{1/m}\geq a^{1/n}$, but does it help?
98424
Let $A$ and $B$ be $n\times n$ real matrices such that $A^2=A$ and $B^2=B$. Suppose that $I-(A+B)$ is invertible .    Show that $\operatorname{Rank}(A)=\operatorname{Rank}(B)$.     I proceed in this way: Note that $A(I-(A+B))=A-A^2-AB=A-A-AB=-AB$ and similarly $B(I-(A+B))=-AB$. So $A(I-(A+B))=B(I-(A+B))$. Since $I-(A+B)$ is invertible we get $A=B \Rightarrow \operatorname{Rank}(A)=\operatorname{Rank}(B)$.