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The Line messaging app that spent last year pushing beyond its home market of Japan has inked a deal aiming to bolster its presence in markets in Latin America. Today it’s announcing a partnership with telecoms giant Telefónica focused on building traction via exclusivity on the lower cost Firefox OS (FFOS) platform. The partnership gives Telefónica subscribers exclusive access to the Line app across what is described as “key Telefónica Firefox OS markets”, including Venezuela, Peru, Colombia, Uruguay, Brasil and Mexico. The deal also includes Spain. The pair said additional markets will be announced “shortly”. This could be a big win for Line — if FFOS takes off. Telefónica has 320 million global subscribers using its network infrastructure, including a significant presence in the LatAm region where Line believes its cuddly cartoon messaging style, which features a range of cute sticker characters and games, can find a Hispanic home from home. An attempt by the company to crack the U.S. market, this time last year, has evidently proven harder, as Line appears to be reconfiguring its strategy around other regions where growth has been easier to come by, such as LatAm and Spain. Turning to the Mozilla flavoured portion of this partnership, the platform launched with much fanfare and a raft of operator supporters at last year’s Mobile World Congress. It’s since been filtering into various markets via carrier launches, with a focus on developing regions where the network operators see a gap to compete with Android at the lower end of smartphones. Since June, Telefónica has launched FFOS in seven countries, while other carriers such as Telenor, Deutsche Telekom and TIM have also started ranging phones running the HTML5-based platform. However it’s unclear exactly how many users there are of FFOS at this point as its carrier backers have not disclosed sales figures. It seems unlikely growth has been rampant or the carriers would have been shouting about it. Still, FFOS’s backers are evidently continuing to push the platform — and to seek ways to differentiate on it via the likes of today’s messaging partnership between Line and Telefónica. Telefónica is presumably hoping to drive interest in its FFOS devices by leveraging Line’s regional popularity, and the popularity of over-the-top mobile messaging in general. So, while the Line application is now available globally via the Firefox OS Marketplace its FFOS app will be exclusive to Telefónica subscribers in the aforementioned (mostly LatAm) markets. That may seem

Messaging app Line began the limited rollout of its new C2C e-commerce platform Line Mall today (link to TechCrunch Japan via Google Translate). The app, which is now available for download in Japan's Google Play store, is a sign that the service is doubling down on its efforts to increase engagement among users as it competes WeChat, WhatsApp and other popular messaging apps. Line Mall will officially launch next spring with an iPhone app and more features.

Line has announced it's added another 100 million people to its registered user count in just four months, pushing it past the 300 million user mark globally. It revealed the new numbers at a press conference today in Tokyo, attended by TechCrunch Japan. CEO Akira Morikawa said it's targeting a 500 million registered user milestone next year.

According to Nikkei, Line, a popular messaging and call service, intends to go public in 2014 on the Tokyo stock exchange. The company is expected to go public at a valuation between $800 million USD and $1 billion. According to Serkan Toto, Line has 270 million registered users. It isn’t clear what percentage of that user base is active on a monthly basis. Twitter, a company that is also set to go public shortly, has around 230 million monthly active users. The comparison matters are both mobile-focused communications applications. And Twitter could go public at an IPO valuation of as high as $20 billion, if scuttlebutt is to be believed. We can’t be too perfectly sure, but it would seem that Twitter is being valued much higher on a per-user basis. Line as a company has been at the forefront of the rising profile of so-called ‘over-the-top’ (OTT) messaging services that have quickly grown in recent years. Others include Viber and KakaoTalk. Essentially, they replace the native SMS and call capabilities that come in smartphones. It has become a massive market. Line’s 270 million users are only a fraction of the larger OTT communications market. When Line goes public, it could clear the way for other OTT applications to follow suit. Top Image Credit: Joi Ito

Mobile messaging app Line, which has been on a marketing push in India, has added 10 million Indian users to its registered users count in around three months, as it prepares to kick off a Bollywood actress-backed TV ad campaign in the country. Since officially launching in India, Line said voice calls on its service have grown 3,200 %, use of stickers is up 1,800%, and messages have grown by 700%.

Messaging app Line, which recently announced it has passed 150 million users, has withdrawn a function from its iOS app that allowed users to gift paid-for stickers to each other by paying for them with a virtual currency. The change, spotted earlier by The Next Web, was made at Apple's request, said the company in a blog post today.

Line’s animated characters, which helped it become one of Asia’s top messaging apps with more than 120 million users, now have their own cartoon series. The TV show, called “LINE TOWN,” premiered on Tokyo TV Channel 6 last week (h/t Asiajin). “Line Town” features half-hour-long episodes and a prime 6:30PM to 7:30PM slot, and has already been picked up for syndication on BS Japan. According to Asiajin, the series also has the distinction of having a theme composed by Japanese pop idol and actress Shoko Nakagawa, the former presenter of Pokemon Sunday. Tech In Asia found a “Line Town” clip on YouTube: This is not the first time the characters from the NHN Japan messaging app have popped up on television. In January, a 5-minute long animated short featuring the characters called “Line Offline: Salaryman” was broadcast on Tokyo TV. Line’s characters, which users can add in the form of animated “stickers” to their messages, are among a host of innovative features, including integrated video, images, and doodles, meant to increase user engagement. Found in top Asian messaging apps including Korea’s KakaoTalk, China’s WeChat, and Taiwan’s Cubie, these features are now making their way to American apps like MessageMe. Line is keen, however, to be known as more than just a messaging and free calls app. Last month, Line’s U.S. CEO Jeanie Han told TechCrunch’s Natasha Lomas that the app wants to position itself as a social media alternative, and sees itself as an entertainment–not technology–company. Line has its own main characters (including Brown the bear, Cony the bunny and highly emotive Moon), who appear in the new cartoon series, but it also relies on cultural research to create new stickers for different markets like Spain and the U.S. Of course, Line is not the first app that has spun-off its characters into an animated series. Most notably, Rovio Entertainment announced last month that it is launching a cartoon based on its megahit “Angry Birds” characters. The Finnish company said last week that it doubled its revenue to $197.8 million in 2012 thanks in large part to Angry Birds titles like “Angry Birds Star Wars” and “Bad Piggies.” It will be interesting to see if Line’s characters enjoy the same success and eventually join Pokemon, Totoro, and Astro Boy on the roster of Japanese hit animations that have achieved international fame.

This is the sixth post in an article series about MIT’s course “Linear Algebra”. In this post I will review lecture six on **column spaces** and **null spaces** of matrices. The lecture first reviews vector spaces and subspaces and then looks at the result of intersect and union of vector subspaces, and finds out when Ax=b and Ax=0 can be solved.

Here is a list of the previous posts in this article series:

- Lecture 1: Geometry of Linear Equations
- Lecture 2: Elimination with Matrices
- Lecture 3: Matrix Multiplication and Inverse Matrices
- Lecture 4: A=LU Factorization
- Lecture 5: Vector Spaces and Subspaces

Lecture six starts with a reminder of what the vector space requirements are. If vectors **v** and **w** are in the space, then the result of adding them and multiplying them by a number stays in the space. In other words, all linear combinations of **v** and **w** stay in the space.

For example, the 3-dimensional space **R**^{3} is a vector space. You can take any two vectors in it, add them, and multiply them by a number and they will still be in the same **R**^{3} space.

Next, the lecture reminds subspaces. A subspace of some space is a set of vectors (including the **0** vector) that satisfies the same two requirements. If vectors **v** and **w** are in the subspace, then all linear combinations of **v** and **w** are in the subspace.

For example, some subspaces of **R**^{3} are:

- Any plane P through the origin (0, 0, 0).
- Any line L through the origin (0, 0, 0).

See the previous, lecture 5, on more examples of spaces on subspaces.

Now suppose we have two subspaces of **R**^{3} - plane P and line L. Is the union P∪L a subspace? **No.** Because if we take some vector in P and some vector in L and add them together, we go outside of P and L and that does not satisfy the requirements of a subspace.

What about intersection P∩L? Is P∩L a subspace? **Yes.** Because it’s either the zero vector or L.

In general, given two subspaces S and T, union S∪T is a not a subspace and intersection S∩T is a subspace.

The lecture now turns to **column spaces of matrices**. The notation for a column space of a matrix A is C(A).

For example, given this matrix,

The column space C(A) is all the linear combination of the first (1, 2, 3, 4), the second (1, 1, 1, 1) and the third column (2, 3, 4, 5). That is, C(A) = { a·(1, 2, 3, 4) + b·(1, 1, 1, 1) + c·(2, 3, 4, 5) }. In general, **the column space C(A) contains all the linear combinations of columns of A**.

A thing to note here is that C(A) is a subspace of **R**^{4} because the vectors contain 4 components.

Now the key question, does C(A) fill the whole **R**^{4}? **No.** Because the third column (2, 3, 4, 5) is actually twice the first column (1, 2, 3, 4)!

From this question follows another question, the most important question in the lecture - Does Ax=b have a solution for every right-hand side vector b? **No.** Because the columns are not linearly independent (the third can be expressed as a multiple of the first)! Therefore the column space C(A) is actually a two-dimensional subspace of **R**^{4}.

Another important question arises - For which right-hand sides b can this system be solved? The answer is: **Ax=b can be solved if and only if b is in the column space C(A)**! It’s because Ax is a combination of columns of A. If b is not in this combination, then there is simply no way we can express it as a combination.

That’s why we are interested in column spaces of matrices. They show when can systems of equation Ax=b be solved.

Now the lecture turns to **null spaces of matrices**. The notation for a null space of a matrix A is N(A).

Let’s keep the same matrix A:

The null space N(A) contains something completely different than C(N). **N(A) contains all solutions x’s that solve Ax=0**. In this example, N(A) is a subspace of **R**^{3}.

Let’s find the null space of A. We need to find all x’s that solve Ax=0. The first one, obviously, is x = (0, 0, 0). Another one is x = (1, 1, -1). In general all x’s (c, c, -c) solve Ax=0. The vector (c, c, -c) can be rewritten as c·(1, 1, -1).

Note that the null space c·(1, 1, -1) is a line in **R**^{3}.

The lecture ends with a proof that solutions x to Ax=0 always give a subspace. The first thing to show in the proof is that if x is a solution and x’ is a solution, then their sum x + x’ is a solution:

We need to show that if Ax=0 and Ax’ = 0 then A(x + x’) = 0. This is very simple. Matrix multiplication allows to separate A(x + x’) into Ax + Ax’ = 0. But Ax=0 and Ax’=0. Therefore 0 + 0 = 0.

The second thing to show is that if x is a solution, then c·x is a solution:

We need to show that if Ax=0 then A(c·x)=0. This is again very simple. Matrix multiplication allows to bring c from A(c·x) outside c·A(x) = c·0 = 0.

That’s it. We have proven that solutions x to Ax=0 always form a subspace.

Here is the video of the sixth lecture:

Direct link: http://www.youtube.com/watch?v=8o5Cmfpeo6g

Topics covered in lecture six:

- [01:00] Vector space requirements.
- [02:10] Example of spaces R
^{3}. - [02:40] Subspaces of spaces.
- [03:00] A plane P is a subspace of R
^{3}. - [03:50] A line L is a subspace of R
^{3}. - [04:40] Union of P and L.
- [07:30] Intersection of P and L.
- [09:00] Intersection of two subspaces S and T.
- [11:50] The column space C(A) of a matrix A.
- [16:20] Does Ax=b have a solution for every b?
- [19:45] Which b’s allow Ax=b to be solved?
- [23:50] Can solve Ax=b exactly when b is in C(A).
- [28:50] The null space N(A) of a matrix A.
- [37:00] Why is the null space a vector space?
- [37:30] A proof that the null space is always a vector space.
- [41:50] Do the solutions to Ax=b form a subspace?

Here are my notes of lecture six:

Have fun with this lecture! The next post is going to be about general theory of solving equations Ax=0, pivot variables and special solutions.

PS. This course is taught from Introduction to Linear Algebra textbook. Get it here:

This is the fifth post in an article series about MIT’s course “Linear Algebra”. In this post I will review lecture five that finally introduces real linear algebra topics such as **vector spaces** their **subspaces** and **spaces from matrices**. But before it does that it closes the topics that were started in the previous lecture on **permutations**, **transposes** and **symmetric matrices**.

Here is a list of the previous posts in this article series:

- Lecture 1: Geometry of Linear Equations
- Lecture 2: Elimination with Matrices
- Lecture 3: Matrix Multiplication and Inverse Matrices
- Lecture 4: A=LU Factorization

Lecture starts with reminding some facts about permutation matrices. Remember from the previous lecture that permutation matrices P execute row exchanges and they are identity matrices with reordered rows.

Let’s count how many permutation matrices are there for an nxn matrix.

For a matrix of size 1x1, there is just one permutation matrix - the identity matrix.

For a matrix of size 2x2 there are two permutation matrices - the identity matrix and the identity matrix with rows exchanged.

For a matrix of size 3x3 we may have the rows of the identity matrix rearranged in 6 ways - {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2}, {3,2,1}.

For a matrix of size 4x4 the number of ways to reorder the rows is the same as the number of ways to rearrange numbers {1,2,3,4}. This is the simplest possible combinatorics problem. The answer is 4! = 24 ways.

In general, for an nxn matrix, there are n! permutation matrices.

Another key fact to remember about permutation matrices is that their inverse P^{-1} is their transpose P^{T}. Or algebraically P^{T}·P = I.

The lecture proceeds to **transpose matrices**. The transpose of a matrix exchanges its columns with rows. Another way to think about it that it flips the matrix over its main diagonal. Transpose of matrix A is denoted by A^{T}.

Here is an example of transpose of a 3-by-3 matrix. I color coded the columns to better see how they get exchanged:

A matrix does not have to be square for its transpose to exist. Here is another example of transpose of a 3-by-2 matrix:

In algebraic notation transpose is expressed as (A^{T})_{ij} = A_{ji}, which says that an element a_{ij} at position ij get transposed into the position ji.

Here are the rules for matrix transposition:

- The transpose of A + B is (A + B)
^{T}= A^{T}+ B^{T}. - The transpose of A·B is (A·B)
^{T}= B^{T}·A^{T}. - The transpose of A·B·C is (A·B·C)
^{T}= C^{T}·B^{T}·A^{T}. - The transpose of A
^{-1}is (A^{-1})^{T}= (A^{T})^{-1}.

Next the lecture continues with **symmetric matrices**. A symmetric matrix has its transpose equal to itself, i.e., A^{T} = A. It means that we can flip the matrix along the diagonal (transpose it) but it won’t change.

Here is an example of a symmetric matrix. Notice that the elements on opposite sides of the diagonal are equal:

Now check this out. If you have a matrix R that is not symmetric and you multiply it with its transpose R^{T} as R·R^{T}, you get a symmetric matrix! Here is an example:

Are you wondering why it’s true? The proof is really simple. Remember that matrix is symmetric if its transpose is equal to itself. Now what’s the transpose of the product R·R^{T}? It’s (R·R^{T})^{T} = (R^{T})^{T}·R^{T} = R·R^{T} - it’s the same product, which means that R·R^{T} is always symmetric.

Here is another cool fact - the inverse of a symmetric matrix (if it exists) is also symmetric. Here is the proof. Suppose A is symmetric, then the transpose of A^{-1} is (A^{-1})^{T} = (A^{T})^{-1}. But A^{T} = A, therefore (A^{T})^{-1} = A^{-1}.

At this point lecture finally reaches the fundamental topic of linear algebra - **vector spaces**. As usual, it introduces the topic by examples.

Example 1: Vector space **R**^{2} - all 2-dimensional vectors. Some of the vectors in this space are (3, 2), (0, 0), (π, e) and infinitely many others. These are all the vectors with two components and they represent the xy plane.

Example 2: Vector space **R**^{3} - all vectors with 3 components (all 3-dimensional vectors).

Example 3: Vector space **R**^{n} - all vectors with n components (all n-dimensional vectors).

What makes these vectors vector spaces is that they are closed under multiplication by a scalar and addition, i.e., vector space must be closed under linear combination of vectors. What I mean by that is if you take two vectors and add them together or multiply them by a scalar they are still in the same space.

For example, take a vector (1,2,3) in **R**^{3}. If we multiply it by any number α, it’s still in **R**^{3} because α·(1,2,3) = (α, 2α, 3α). Similarly, if we take any two vectors (a, b, c) and (d, e, f) and add them together, the result is (a+d, b+e, f+c) and it’s still in **R**^{3}.

There are actually 8 axioms that the vectors must satisfy for them to make a space, but they are not listed in this lecture.

Here is an example of not-a-vector-space. It’s 1/4 of **R**^{2} (the 1st quadrant). The green vectors are in the 1st quadrant but the red one is not:

An example of not-a-vector-space.

This is not a vector space because the green vectors in the space are not closed under multiplication by a scalar. If we take the vector (3,1) and multiply it by -1 we get the red vector (-3, -1) but it’s not in the 1st quadrant, therefore it’s not a vector space.

Next, Gilbert Strang introduces **subspaces** of vector spaces.

For example, any line in **R**^{2} that goes through the origin (0, 0) is a subspace of **R**^{2}. Why? Because if we take any vector on the line and multiply it by a scalar, it’s still on the line. And if we take any two vectors on the line and add them together, they are also still on the line. The requirement for a subspace is that the vectors in it do not go outside when added together or multiplied by a number.

Here is a visualization. The blue line is a subspace of **R**^{2} because the red vectors on it can’t go outside of line:

An example of subspace of

And example of not-a-subspace of **R**^{2} is any line that does not go through the origin. If we take any vector on the line and multiply it by 0, we get the zero vector, but it’s not on the line. Also if we take two vectors and add them together, they are not on the line. Here is a visualization:

An example of not-a-subspace of

Why not list all the subspaces of **R**^{2}. They are:

- the
**R**^{2}itself, - any line through the origin (0, 0),
- the zero vector (0, 0).

And all the subspaces of **R**^{3} are:

- the
**R**^{3}itself, - any line through the origin (0, 0, 0),
- any plane through the origin (0, 0, 0),
- the zero vector.

The last 10 minutes of the lecture are spent on **column spaces of matrices**.

The column space of a matrix is made out of all the linear combinations of its columns. For example, given this matrix:

The column space C(A) is the set of all vectors {α·(1,2,4) + β·(3,3,1)}. In fact, this column space is a subspace of **R**^{3} and it forms a plane through the origin.

More about column spaces in the next lecture.

You’re welcome to watch the video lecture five:

Direct link: http://www.youtube.com/watch?v=JibVXBElKL0

Topics covered in lecture five:

- [01:30] Permutations.
- [03:00] A=LU elimination without row exchanges.
- [03:50] How Matlab does A=LU elimination.
- [04:50] PA=LU elimination with row exchanges
- [06:40] Permutation matrices.
- [07:25] How many permutation matrices are there?
- [08:30] Permutation matrix properties.
- [10:30] Transpose matrices.
- [11:50] General formula for transposes: (A
^{T})_{ij}= A_{ji}. - [13:06] Symmetric matrices.
- [13:30] Example of a symmetric matrix.
- [15:15] R·R
^{T}is always symmetric. - [18:23] Why is R·R
^{T}symmetric? - [20:50] Vector spaces.
- [22:05] Examples of vector spaces.
- [22:55] Real vector space
**R**^{2}. - [23:20] Picture of
**R**^{2}- xy plane. - [26:50] Vector space
**R**^{3}. - [28:00] Vector space
**R**^{n}. - [30:00] Example of not a vector space.
- [32:00] Subspaces of vector spaces.
- [33:00] A vector space inside
**R**^{2}. - [34:35] A line in
**R**^{2}that is subspace. - [34:50] A line in
**R**^{2}that is not a subspace. - [36:30] All subspaces of
**R**^{2}. - [39:30] All subspaces of
**R**^{3}. - [40:20] Subspaces of matrices.
- [41:00] Column spaces of matrices C(A).
- [44:10] Example of column space of matrix with columns in
**R**^{3}.

Here are my notes of lecture five:

Have fun with this lecture! The next post is going to be more about column spaces and null spaces of matrices.

PS. This course is taught from Introduction to Linear Algebra textbook. Get it here:

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