WHAT IS MEANT BY ROW OR COLUMN VECTOR(in linear algebra)? I understand that if a row of ( or column) of a matrix represent a point in a plane or in space (in 2D or in 3D) then the row or column is said to be row or column vector.

For two equation having two unknown, \begin{equation} AX=B \hspace{2 in} (1) \end{equation} is the matrix form of the system of equation.Here X denotes the solution(x and y) which represents a point in a plane .hence it is a column vector.

But my QUESTION is that how the right hand side of (1),B, which represent the constant term in of the system of equations,is said to be a column vector. The B (RHS of(1)) which doesn't represent any point on a plane or on a space is said to be column vector. But HOW IS THIS POSSIBLE? Is there any ambiguity in my understanding? Please help me. This doubt arise when I saw the lecture of professor gilbert strang.

  • $\begingroup$ I'm not sure what you mean by "C" -- in Equation (1), you simply have AX=B. The right hand side of (1) would be B? $\endgroup$ – nukeguy Aug 31 '16 at 19:14
  • $\begingroup$ @nukeguy ooh sorry i corrected $\endgroup$ – Sathasivam K Aug 31 '16 at 19:31

Assuming I understand your question correctly, here goes.

First, I'd like to point out that a huge part of mathematics is drawing parallels and representing one thing as something else. So, even if one draws up a system of equations without thinking about what it means in terms of column vectors/points in space, it doesn't mean it cannot be represented as such. Likewise, just because something can be represented in a certain way, it doesn't mean that it has to be represented that way. So to answer your question "HOW IS THIS POSSIBLE?" in a very general and philosophical sense: It all depends on the context, and how you choose to view the problem. Given any object, one can choose to view it however he/she pleases.

Having said that, there is a very natural connection between all of what you've mentioned and, in my opinion (as well as the opinion of many others), linear algebra/matrix-vector notation is very useful for studying systems of equations. First, a matrix $M$ is, first and foremost, a linear transformation from one vector space to another vector space. In most examples, a matrix $M$ with $m$ columns and $n$ rows will map from $\mathbb{R}^m$ to $\mathbb{R}^n$ where $m$ is the length of $X$ (i.e., the number of "variables" in your linear system) and $n$ is the length of $B$ (i.e., the number of equations you have). One way to think about it is that the coefficients of your linear system (i.e., the entries of $M$) are taking linear combinations of your variables $x_i$ (i.e., the entries of $X$) and producing a new set of numbers $b_i$ (i.e., the entries of $B$). When you seek a solution to $MX = B$, you are essentially given (1) the transformation, $M$, and (2) the image of the transformation, $B$, and trying to reverse the transformation to figure out the original vector $X$.

The benefit of viewing linear systems in such a manner is that it becomes very easy to figure out certain properties about your linear system. For example, one might ask whether the system has a unique solution, or any solutions at all. And, to answer that question, one can use all the tools present in linear algebra to analyze the matrix/transformation $M$.

If the matrix $M$ is not square, then either $m > n$ or $n < m$. Recall that $m$ is the number of unknowns and $n$ is the number of equations. If $m > n$, then we do not have enough "information" to solve the problem uniquely (i.e., if there is a solution, there's more than one solution). But there is an elegant parallel when we view $M$ as a transformation; when $m > n$, $M$ maps from a larger space to a smaller space, and so there cannot be a 1-to-1 transformation; $B$ cannot be uniquely mapped back to $X$.

If $n > m$, then we have too much information and there may be no solution. It depends on one's choice of $B$. In the context of transformations, one could ask "Is $B$ in the image of $M$?" If the answer is yes, then there is a solution. If not, then there is no solution.

In your case, $m = n = 2$. However, it is still possible for $MX=B$ to yield no solutions or many solutions. For example, if your system is

$$ x_1 + x_2 = b_1 \\ 3x_1 + 3x_2 = 3b_1 $$

then you have a redundant equation and you don't have enough information to determine a unique solution. When viewed as a linear system $MX=B$, one could say instead that the columns of $M$ are not independent. Notice that, if I change the right hand side, it is possible for no solution to exist. That is, I can pick a $B$ that is not in the image of $M$.

To summarize my response:

(1) Mathematics is all about representation. Your professor is simply trying to show you how a system of equations can be represented using linear algebra/transformations/matrices and vectors. You, of course, are free to view the system of equations however you like. (But it would behoove you to see it from his perspective since you are taking his class.)

(2) By using matrix-notation and the coefficients on the left hand side of the system of equations as a transformation, we can make use of many well-studied well-documented linear algebra properties/theorems/features to obtain information, such as whether or not there exists a unique solution, about the linear system. It facilitates discussion and understanding of the system of equations, especially when they become complex (i.e., much larger than 2 variables/2 equations).

  • $\begingroup$ I think you are saying that we are assuming the RHS B as a vector because A IS ASSUMED AS TRANSFORMATION .Am I right sir? $\endgroup$ – Sathasivam K Sep 1 '16 at 17:33
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    $\begingroup$ "assume" is a strong word. I like the word "represent" better; your math professor is simply choosing to describe the coefficients of your system of equations as a transformation. What they actually "represent" is ambiguous/irrelevant; each person can represent/view something however they like. (See point (1) above.) But yes, I think we're on the same page now. And sorry, I just realized I used $M$ instead of $A$. My $M$ is your $A$. $\endgroup$ – nukeguy Sep 1 '16 at 18:15

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