I am struggling with proving the theorem that if $A$ and $B$ are $n\times n$ matrices, then:     $$\operatorname{rank}(AB)\leq \operatorname{rank}(B)$$   Could anyone suggest me a hint? Any help is greatly appreciated, thank you very much.
Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$   I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.  How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?
About the continuity of the function $f(x) = \sum\limits_k2^{-k}\mathbf 1_{q_k \leq x}$ Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)? mysql_insert_id alternative for postgresql is there an alternative for mysql_insert_id() php function for PostgreSQL? Most of the frameworks are solving the problem partially by finding the current value of the sequence used in the ID. However, there are times that the primary key is not a serial column.... I am currently making a map in the Minecraft 1.13 snapshots and have encountered a problem. I want to use the title command to give the player the number of resources they have for a certain resource. For example Wood: 150. I looked this up and found you could do this using the title command. Example:   title @a[r=1] actionbar [{"text":"Wood: "},{"score":{"name":"@p","objective":"Wood"},"bold":true}]}}   This works in 1.12 but not in 1.13. It seems that the JSON format has changed in 1.13 but I can't find anything on the wiki or the changelog.  Could someone please help me to work this out? What was L's real name? I watched through Death Note once, and I never remember hearing L's name, even after he died.  What was L's real name, or did he never have one (it was never said)? 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Replace Builtin commands with Custom Commands for CMD.exe Why doesn't backpropagation work when you initialize the weights the same value? Why doesn't backpropagation work when you initialize all the weight the same value (say 0.5), but works fine when given random numbers?   Shouldn't the algorithm calculate the error and work from there, despite the fact that the weights are initially the same?