There is raw echelon form and reduced raw echelon form:    The row echelon form is used to solved the system of linear equations. What is the practical use of reduced raw echelon form?  Previous question in the forum was related to row echelon form (and not to reduced row echelon form):   Thanks
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I have just started auditing Linear Algebra, and the first thing we learned was how to solve a system of linear equations by reducing it to row echelon form (or reduced row echelon form). I know how to solve a system of linear equations using determinants, and so the using row echelon form seems to be very inefficient and an easy way to make mistakes.  The lecturer seemed to indicate that this will be useful in the future for something else, but a friend who has a PhD in mathematics said it is totally useless.  Is this true? Is this technique totally useless or is there some "higher mathematics" field/ technique where row echelon form will be useful?
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Consider A $\Rightarrow$ B, A $\models$ B, and A $\vdash$ B.  What are some examples contrasting their proper use? For example, give A and B such that A $\models$ B is true but A $\Rightarrow$ B is false. I'd appreciate pointers to any tutorial-level discussion that contrasts these operators.  Edit: What I took away from this discussion and the others linked is that logicians make a distinction between $\vdash$ and $\models$, but non-logicians tend to use $\Rightarrow$ for both relations plus a few others. Points go to Trevor for being the first to explain the relevance of completeness and soundness.
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(a) Show that any two-dimensional vector can be expressed in the form $s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},$ where $s$ and $t$ are real numbers.  (b) Let $\vec{u}$ and $\vec{v}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form $s \vec{u} + t \vec{v},$ where $s$ and $t$ are real numbers, if and only if of the vectors $\vec{u}$ and $\vec{v}$, one vector is not a scalar multiple of the other vector.  I've already gotten part (a) but (b) is a challenge. I have solved part a hough:  Let's say that $ \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}$ is equal to $\binom{a}{b}$. \begin{align*}  \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix} &amp;= \binom{a}{b}\\ \binom{3s}{-s}+\binom{2t}{7t}&amp;=\binom{a}{b}\\ \binom{3s+2t}{-s+7t}&amp;=\binom{a}{b}\\ \Rightarrow 3s+2t&amp;=a\\ -s+7t&amp;=b \end{align*}
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I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.  In the proof of the theorem, they wrote:     $$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$  ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$  [...]   I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?  Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so?
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If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
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I have come across the term "closed-form solution" quite often. What does a closed-form solution mean? How does one determine if a close-form solution exists for a given problem? Searching online, I found some information, but nothing in the context of developing a statistical or probabilistic model / solution.   I understand regression very well, so if any one can explain the concept with reference to regression or model-fitting, it will be easy to consume. :)
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I am trying to understand reduction of an unsymmetric real square matrix to Hessenberg form from Numerical Recipes Vol. 3.  In it, the author states that one does not use Gaussian elimination for reducing to Hessenberg form because the Gaussian elimination does not preserve similarity and hence ends up changing the eigenvalues of the matrix, which is undesirable.  Why does Gaussian elimination not preserve similarity?   Also, what conditions decide whether a particular matrix transformation preserves similarity?  Heres a  to where I found it. In case you cant view that, it's on page 594 of Numerical Recipes in C++ (Volume 3).
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Given $m$ rows with $n$ variables where $m &lt; n$, and given that there exists a combination of some of the rows that gives a zero vector. Is it promised when computing the reduced-row echelon form, with Gaussian elimination to find a zero vector?  Update: I clearly don't know, how to pick up the right words to ask my question. So let me give you the full context and please help me say it right, so the question will make sense. I am solving step 3 in the quadratic sieve algorithm, here is a  description of it.  Suppose the composite number N is being factored. Bound $B$ is chosen, and the factor base is identified (which is called $P$), the set of all primes less than or equal to $B$. Next, positive integers $z$ are sought such that $z_2$ mod $N$ is B-smooth. Therefore it can be written, for suitable exponents $a_i$, $${\displaystyle z^{2}{\text{ mod }}N=\prod _{p_{i}\in &gt; P}p_{i}^{a_{i}}}$$ When enough of these relations have been generated (it is generally sufficient that the number of relations be a few more than the size of P), the methods of linear algebra can be used (for example, Gaussian elimination) to multiply together these various relations in such a way that the exponents of the primes on the right-hand side are all even: $${\displaystyle {z_{1}^{2}z_{2}^{2}\cdots z_{k}^{2}\equiv \prod &gt; _{p_{i}\in P}p_{i}^{a_{i,1}+a_{i,2}+\cdots +a_{i,k}}\ {\pmod {N}}\quad ({\text{where }}a_{i,1}+a_{i,2}+\cdots +a_{i,k}\equiv 0{\pmod &gt; {2}})}}$$  Now, let's say I found $m$ relations and the number of primes in $P$ is $n$, such that $m&lt;n$. It is given that there exists a combination of some of the $m$ relations that their product is a square number. My question is: When I perform the Gaussian elimination over F2 to find a null space combination, is it promised that I will find one? I successfully implemented the algorithm and the matrix part of my implementation works. But what I see is that if I found a null space(a combination of making a null space) and latter I add more relations(still $m &lt; n$), I lose the null space solution and not able to solve the matrix again. Each time I add a new relation I start solving it from the beginning, I don't remember any previous state. So what I want to know, could it be that there is a limitation with the Gaussian elimination algorithm, that it does not promise to find the null space if $m &lt; n$?
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I'm a student in an elementary linear algebra course. Without bashing on my professor, I must say that s/he is very poor at answering questions, often not addressing the question itself. Throughout the course, there have been multiple questions that have gone unanswered, but I must have the answer to this one question.  "Why are some vectors written as row vectors and others as column vectors?"  I understand that if I transpose one, it becomes the other. However, I'd like to understand the purpose behind writing a vector in a certain format.  Taking examples from my lectures, I see that when I'm trying to prove linear independence of a group of vectors, the vectors are written as column vectors in a matrix, and the row reduced form is found.  Other times, like trying to find the cross product or just solving a matrix, I see the vectors written as row vectors.  My professor is very vague on the notations and explanations, and it bugs me as a person who needs to know the reason behind every small thing, why this variation occurs in the format. Any input is greatly appreciated.
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I have matrices for my syllabus but I don't know where they find their use. I even asked my teacher but she also has no answer. Can anyone please tell me where they are used? And please also give me an example of how they are used?
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Is there any general and systematic way of reducing the higher order linear ODE to a system of first order ODE?  For example, assume we have $a_3x^{(3)}+a_2x^{(2)}+a_1x^{(1)}+a_0x=0$, then how do we convert this into matrix form(a system of first order ODE). And after we solve the system of equation, how to combine them into our final solution $x(t)$?  Thanks for helping me out!
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I'm having troubling finding an example/proof that the following statement is true.     Row operations on a matrix A can change the linear   dependence relations among the rows of A.   Any tips on understanding this would be greatly appreciated.
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Given is a linear regression problem, where we have one training point, which is 1-dimensional: $x \in R_{&gt;0}$ and the corresponding output, $y \in R$. We duplicate the feature, such that we have one training point with two (identical) features.   For this, we have to determine if we can apply the closed form solution $\beta = (X^TX)^-1*X^T*y$. Then we have to solve the linear regression problem by taking into account that $f(X) = ||y-X*\beta||^2$ is convex.   Application of the closed form solution: For this I want to determine if $X^TX$ has full rank. Given is $X = (1,x_{11},x_{12})$. Hence $X^T*X$ results in:  \begin{bmatrix}     1 &amp; x_{11} &amp; x_{12}  \\     x_{11} &amp; x_{11}^2 &amp; x_{11}x_{12}  \\     x_{12}  &amp; x_{11}x_{12} &amp; x_{12}^2   \end{bmatrix}   This matrix seems to have full rank (independent column vectors), so the closed form solution is applicable. Correct? Is there another way to determine the rank without explicitly calculating $X*X^T$?   For minimizing $f(\beta)$ I would derive $||y-X*\beta||^2$ and set it equal to zero.
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I have seen a definition of orthogonality between two vectors something like this:  $$&lt;\vec{u}, \vec{v}&gt; = u^T \cdot v = 0$$  I am wondering what is the purpose of using a transpose of a vector (in this case and in general). I have also seen this in the formula to find the projection of a vector over another, but I have used just the normal vector instead of its transpose and everything seems to work.
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states:  "In practice, explicit right outer joins are rarely used, since they can always be replaced with left outer joins and provide no additional functionality."  Can anyone provide a situation where they have preferred to use the RIGHT notation, and why? I can't think of a reason to ever use it. To me, it wouldn't ever make things more clear.  Edit: I'm an Oracle veteran making the New Year's Resolution to wean myself from the (+) syntax. I want to do it right
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Show that $n$ lines separate the plane into  $\frac{n^2+n+2}{2}$ regions if no two of these lines are parallel and no three pass through a common point.   I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0.  But I really have no idea how to begin the inductive step.  How do we know what k+1 we're supposed to arrive at?  Thanks!
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In general inference, why orthogonal parameters are useful, and why is it worth trying to find a new parametrization that makes the parameters orthogonal ?   I have seen some textbook examples, not so many, and would be interested in more concrete examples and/or motivation.