Questions and Answers #2 Forms of Linear Equations

Questions and Answers Sheet 2 Forms of Linear Equations Question #1 Consider the equation 𝑥 + 3 = 3(𝑥 + 1) − 2𝑥. Is this equation a linear equation? I would say to my students that an equation in 1 variable is linear when it can be simplified to the form 𝑎𝑥 = 𝑏 where 𝑎 ≠ 0 . Also, I would mention that a linear equation in 1 variable has only 1 solution. Therefore, I would say that the equation 𝑥 + 3 = 3(𝑥 + 1) − 2𝑥 is not linear because it can be reduced to 0𝑥 = 0 which has infinite solutions. However, a colleague whose research area is algebra says that a linear equation is one that only involves polynomials of degree 1. Therefore, the equation 𝑥 + 3 = 3(𝑥 + 1) − 2𝑥 is linear. Then, according to the definition of my colleague, the equation 𝑥 − 𝑥 = 0 is linear as well. I was thinking that maybe a third possibility is that according to the definition of "linear equation" only equations of the form "expression=0" can be classified as linear or non-linear. In this case, the equation 𝑥 + 3 = 3(𝑥 + 1) − 2𝑥 is not linear nor non-linear, but the equation 𝑥 − 𝑥 = 0 would be linear. What should be the definition for "linear equation" in 1 variable suitable for a math course? Answer: However, a colleague whose research area is algebra says that a linear equation is one that only involves polynomials of degree 1. Therefore, the equation 0 = 0(on simplification) is linear. Then, according to the definition of my colleague, the equation is linear as well. The problem is "The degree of a zero polynomial is not defined or −∞ ". So saying that zero polynomial has degree 1 is little vague as 0 ≡ 0𝑥 ≡ 0𝑥 2 ≡ 0𝑥 3 ≡ ⋯.. What should be the definition for "linear equation" in 1 variable suitable for a math course? A linear polynomial 𝑝(𝑥) = 𝑎𝑥 + 𝑏, 𝑎 ≠ 0 when equated to a real number 𝛼("{𝑠𝑎𝑦} i.e. 𝑝(𝑥) = 𝛼 is called a linear equation. Question #2 Consider the following system of linear equations 𝑥+𝑦+𝑧+𝑤 = 2 𝑥 − 𝑦 + 2𝑧 + 3𝑤 = −10 𝑥 − 3𝑦 + 3𝑧 + 5𝑤 = −7 𝑥 + 3𝑦 − 𝑤 = −1 If we add the first two (solve for y) or the last two (solve for y), then both are same: 2𝑥 + 3𝑧 + 4𝑤 = −8 Does it mean that all the points (𝑎, 𝑦, 𝑏, 𝑐) are the solutions for the system of equations? Here (𝑎, 𝑏, 𝑐) is the point lying on the plane 2𝑥 + 3𝑧 + 4𝑤 = −8 and y is arbitrary. Whereas, if we compute the row echelon form it says the rank of the coefficients matrix is 2 and the augmented matrix is 3. This implies the solution does not exist. Any help will be good! Answer: Here is a smaller toy example which, I believe, captures the essence of your problem (it is a system of four equations in two unknowns which is clearly inconsistent): 𝑥+𝑦 =1 𝑥+𝑦=2 𝑥+𝑦=1 𝑥+𝑦=2 Adding either the first or the last pair of equations together yields the line 2𝑥 + 2𝑦 = 3 . But you can't conclude anything about the solution space from this single equation. Really all you are saying is that if (x,y) solves a pair of original equations, then it solves the new equation, but the converse is not true at all. What you can do with such a combination of two previous equations is substitute it in the original system (this is one of the elementary row operations), but then you still have to solve the system. Question #3 During some computations I came up with the following system of linear recurrences: 𝐵𝑛+2 = 3𝐵𝑛 + 𝐴𝑛 𝐴𝑛 = 𝐴𝑛−1 + 𝐵𝑛−1 Here I am trying to find the solution for B (hoping to get some sort of homogeneous equation, find it roots and get the closed form formula). But I can't solve it. The only thing I can do is 𝐵𝑛+2 = 3𝐵𝑛 + ∑𝑛−1 𝑖=0 𝐵𝑖 , which will not help me to solve recurrence. So is there a way I can find B without the summation through n? Answer: You can transform the above equations into equations only in A and only in B: The second equation gives 𝐵𝑛 = 𝐴𝑛+1 − 𝐴𝑛 applying this to the first equation gives 𝐴𝑛+3 − 𝐴𝑛+2 = 3(𝐴𝑛+1 − 𝐴𝑛 ) + 𝐴𝑛 = 3𝐴𝑛+1 − 2𝐴𝑛 ⇒ 𝐴𝑛+3 = 𝐴𝑛+2 + 3𝐴𝑛+1 − 2𝐴𝑛 Similar 𝐴𝑛 = 𝐵𝑛+2 − 3𝐵𝑛 𝐵𝑛+2 − 3𝐵𝑛 = 𝐵𝑛+1 − 3𝐵𝑛−1 + 𝐵𝑛−1 ⇒ 𝐵𝑛+2 = 𝐵𝑛+1 + 3𝐵𝑛 − 2𝐵𝑛−1 Question #4 Establish the conditions under which the equations 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑞 − 𝑟; 𝑏𝑥 + 𝑐𝑦 + 𝑎𝑧 = 𝑟 − 𝑝; 𝑐𝑥 + 𝑎𝑦 + 𝑏𝑧 = 𝑝 − 𝑞 , are consistent. I am aware that by taking the system to echelon can get me the rank of the coefficient matrix and the augmented matrix. But that turns out to be a very cumbersome process and the terms in the echelon form are very big. Is there other ways to establish the conditions for consistency which are less cumbersome and time consuming? Answer: The right-hand sides of the three equations suggest that adding them all together might be fruitful. When you do this, you get, after a little bit of fiddling, (𝑎 + 𝑏 + 𝑐)(𝑥 + 𝑦 + 𝑧) = 0 . This equation can be satisfied for any values of a, b and c, so the original system is also consistent for all values of these parameters, as well as all values of p and q, since the latter don’t appear in this equation at all. Another approach is to use Cramer’s rule, just as in an answer to your previous question. det((a b c), (b c a), (c a b)) = 3abc − (a 3 + b3 + c 3 ) This is equal to zero when 𝑎 + 𝑏 + 𝑐 = 0 , so there are two cases to consider. If 𝑎 + 𝑏 + 𝑐/𝑛𝑒0 , there is a unique solution to this system of equations for any values of p and q. If it’s zero, you’ll need to look a little more closely to determine consistency. One way to proceed in that case is to substitute 𝑐 = −𝑎 − 𝑏 in the augmented matrix of the system and row-reduce. The computation is a bit messy, but not as bad a row-reducing the original augmented matrix. The system is consistent (thought indeterminate) iff all of the zero rows on the left (unaugmented) side of the rref have zeros in the right-hand column. (You’ll end up with denominators of 𝑎2 + 𝑎𝑏 + 𝑏 2 , but that can’t be zero.) Another way to go is to perform this substitution in the three equations, and then add them together as in the first part of this answer. You end up with 0=0, which is obviously consistent. Question #5 From a text book: "The general form of a linear equation in two variables is 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 or, 𝑎𝑥 + 𝑏𝑦 = 𝑐 where a,b,c are real numbers such that 𝑎 ≠ 0, 𝑏 ≠ 0 and x,y are variables. (we often denote the condition a and b are not both zero by 𝑎2 + 𝑏 2 ≠ 0 .)" I don’t understand this last condition. How can we say that 𝑎2 + 𝑏 2 ≠ 0 represents the condition that a and b are not both zero. Let 𝑎 = 0, 𝑏 = 1, then also this condition fulfills. Answer: a and b are both zero ⟺ 𝑎 = 𝑏 = 0 , so a and b are NOT both zero ⟺ at least one of 𝑎, 𝑏 is not 0 which is equivalent to 𝑎2 + 𝑏 2 = 0 in the case that a,b are both real numbers. Answering to your comment, yes, you are right. Maybe a better way to write in an inequality is (𝑎, 𝑏) ≠ (0,0) instead of " 𝑎 ≠ 0, 𝑏 ≠ 0 " Question #6 General form of linear equation using two variables I always read can be written: 𝐴𝑥 + 𝐵𝑦 + 𝑐 = 0 where A and B cannot both be 0. So does it means like 1𝑥 + 0𝑦 + 𝑐 = 0 is a linear equation of two variables i.e basically what I am asking is that both A and B cannot be zero but can either A or B be 0 like: 0𝑥 + 4𝑦 − 5 = 0 or 3𝑥 + 0𝑦 − 15 = 0 Are the above equation a Linear equation of 2 variables? Need your help! Answer: Yep, that's still valid. One way to see this is to check that the equation e.g. 3𝑥 − 15 = 0 specifies a line in the xy -plane -- a vertical one. (What line do we specify if 𝐵 = 0 and 𝐶 = 0?) x=o is the equation of the y -axis; and every pt. on it is (0,t), where t is a real number... This is indeed a linear equation in two variables. Question #7 Given a linear equation pass through the point(6,4) and two axes, and formed a triangle with area 6. I want to find the equation of that line. My attempt is letting the equation be the equation 1 2 𝑦−4 𝑥−6 = 𝑚 . Then using the fact that the area of triangle is 6, forming base ⋆ height = 6 . I got the base and height of the triangle from the axes, base = 6 − 4 𝑚 and height = 4 − 6𝑚 Then substitute the equation \frac{1}{2}\text{base}\star\text{height}=6 and get 36m^2-36m+16=0 which leads to no solution. Can anyone let me know what's wrong? Answer: You get if calculated correctly 36𝑚2 − 60𝑚 + 16 = 0 Question #8 Given Simultaneous linear equations of the form, 𝑎11 𝑥1 + 𝑎12 𝑥2 + 𝑎13 𝑥3 + ⋯ 𝑎1𝑛 𝑥𝑛 = 𝑏1 𝑎21 𝑥1 + 𝑎22 𝑥2 + 𝑎23 𝑥3 + ⋯ 𝑎2𝑛 𝑥𝑛 = 𝑏2 𝑎31 𝑥1 + 𝑎32 𝑥2 + 𝑎33 𝑥3 + ⋯ 𝑎3𝑛 𝑥𝑛 = 𝑏3 ⋮ 𝑎𝑛1 𝑥1 + 𝑎𝑛2 𝑥2 + 𝑎𝑛3 𝑥3 + ⋯ 𝑎𝑛𝑛 𝑥𝑛 = 𝑏𝑛 Can the Newton Raphson's Method use to solve this system of linear equations? Please mention the reasons and possible arguments. Given Simultaneous linear equations of the form, 𝑎11 𝑥1 + 𝑎12 𝑥2 + 𝑎13 𝑥3 + ⋯ 𝑎1𝑛 𝑥𝑛 = 𝑏1 𝑎21 𝑥1 + 𝑎22 𝑥2 + 𝑎23 𝑥3 + ⋯ 𝑎2𝑛 𝑥𝑛 = 𝑏2 𝑎31 𝑥1 + 𝑎32 𝑥2 + 𝑎33 𝑥3 + ⋯ 𝑎3𝑛 𝑥𝑛 = 𝑏3 ⋮ 𝑎𝑛1 𝑥1 + 𝑎𝑛2 𝑥2 + 𝑎𝑛3 𝑥3 + ⋯ 𝑎𝑛𝑛 𝑥𝑛 = 𝑏𝑛 Can the Newton Raphson's Method use to solve this system of linear equations? Please mention the reasons and possible arguments. Answer: That just leads you back to solving the same system in the (only) iteration. No point in that. There are iterative methods to approximate the solution to (large, sparse) linear systems, like relaxation. Question #9 Help me with solving the following system of linear equations by using Gauss-Seidel method: 𝑥1 − 8𝑥2 + 3𝑥3 = −9 𝑥1 − 5𝑥2 − 6𝑥3 = 2 𝑥1 − 3𝑥2 + 7𝑥3 = −5 (0) Where 𝑥1 = 0.2, 𝑥(0)2 = 0.6, 𝑥(0)3 = 0.4 , use one loop? Answer: The gauss-seidel method is used for finding the iteration of 𝑥1 , 𝑥2 , 𝑥3 in the first equation isolate the variable 𝑥1 and in the second equation isolate the variable 𝑥2 and in the third equation isolate the variable 𝑥3 and in the first iteration substitute 𝑥1 = 0.2, 𝑥2 = 0.6, 𝑥3 = 0.4 and in the second iteration substitute the first iteration values of 𝑥1 , 𝑥2 , 𝑥3 .similarly proceeding the same way to find the values of 𝑥1 , 𝑥2 , 𝑥3 To find 𝑥1 in the first equation write the variables 𝑥2 and 𝑥3 on the right-hand side and also change the signs and divide it by the coefficients of 𝑥1 . and for 𝑥2 in the second equation write the variables 𝑥1 and 𝑥3 on the right-hand side and also change the signs and divide it by the coefficients of 𝑥2 . and for 𝑥3 in the third equation write the variables 𝑥1 and 𝑥2 on the right-hand side and also change the signs and divide it by the coefficients of 𝑥3 𝑥1 − 8𝑥2 + 3𝑥3 = −9 𝑥1 − 5𝑥2 − 6𝑥3 = 2 𝑥1 − 3𝑥2 + 7𝑥3 = −5 𝑥1 = −9 + 8𝑥2 − 3𝑥3 −5𝑥2 = 2 + 6𝑥3 − 𝑥1 𝑥2 = 𝑥1 − 6𝑥3 − 2 5 7𝑥3 = −5 + 3𝑥2 − 𝑥1 𝑥3 = −𝑥1 +3𝑥2 −5 7 To find the first iteration substitute 𝑥1 = 0.2, 𝑥2 = 0.6, 𝑥3 = 0.4 in the 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠𝑥1 = −9 + 8𝑥2 − 3𝑥3 , 𝑥2 = 𝑥1 −6𝑥3 −2 5 and 𝑥3 = −𝑥1 +3𝑥2 −5 7 𝑥1 = −9 + 8(0.6) − 3(0.4) 𝑥1 = −5.4 𝑥2 = 0.2 − 6(0.4) − 2 5 𝑥2 = −0.84 𝑥3 = −0.2 + 3(0.6) − 5 7 𝑥3 = −0.485714285 Therefore 𝑥1 = −5.4, 𝑥2 = −0.84, 𝑥3 = −0.485714285 Question #10 Help with solving the system of linear equations for all variables^ 2𝑥 + 6𝑦 − 𝑧 = 4 4𝑥 + 5𝑧 = 7 3𝑦 − 2𝑧 = 0 Answer: Given system of equation are 2𝑥 + 6𝑦 − 𝑧 = 4 → 1 4𝑥 + 5𝑧 = 7 → 2 3𝑦 − 2𝑧 = 0 → 3 from 2 4𝑥 + 5𝑧 = 7 4𝑥 = 7 − 5𝑧 𝑥= 7 − 5𝑧 →4 4 7 − 5𝑧 2( ) + 6𝑦 − 𝑧 = 4 4 14 − 10𝑧 + 24𝑦 − 4𝑧 =4 4 24𝑦 − 14𝑧 + 14 = 16 24𝑦 − 14𝑧 = 2 12𝑦 − 7𝑧 = 1 → 5 from 3 3𝑦 − 2𝑧 = 0 𝑦= 2𝑧 3 2𝑧 12 ( ) − 7𝑧 = 1 3 8𝑧 − 7𝑧 = 1 ⇒ 𝑧 = 1 𝑦= 2(1) 2 2 = ⇒ �= 3 3 3 𝑥= 7−5 2 1 = = 4 4 2 𝑥= 1 2 1 2 (𝑥, 𝑦, 𝑧) = ( , , 1) 2 3 Question #11 Here is the example I encountered : A matrix 𝑀(5 × 5) is given and its minimal polynomial is determined to be (𝑥 − 2)3 . So considering the two possible sets of elementary divisors {(𝑥 − 2)3 , (𝑥 − 2)2 } and {(𝑥 − 2)3 , (𝑥 − 2), (𝑥 − 2)} we get two possible Jordan Canonical forms of the matrix , namely 𝐽1 and 𝐽2 respectively. So 𝐽1 has 2 and 𝐽2 has 3 Jordan Blocks. Now we are to determine the exact one from these two. From the original matrix M, we determined the Eigen vectors and 2 eigen vectors were linearly independent. So the result is that 𝐽1 is the one .So, to determine the exact one out of all possibilities , we needed two information 1) the minimal polynomial ,together with 2) the number of linearly independent eigen vectors. Now this was a question-answer book so not much theoretical explanations are given . From the given result , I assume the number of linearly independent eigen vectors -which is 2 in this case - decided 𝐽1 to be the exact one because it has 2 Jordan Blocks. So the equation "Number of linearly independent eigen vectors=Number of Jordan Blocks" must be true for this selection to be correct . Now this equation is not proved in this book or the text book I have read says nothing of this sort So, that is my question here : How to prove the equation "Number of linearly independent eigen vectors=Number of Jordan Blocks"? Answer: Let A be a matrix reduced in Jordan form. Notice that for every Jordan block relative to the eigenvalue 𝛼 there is an eigenvector with respect to 𝛼 and these vectors ar independent. To see this just look at the first column of a generic block. So Jordan blocks independent eigenvectors. To prove the other inequality, consider the matrix 𝐴 − 𝛼𝐼 for every eigenvalue 𝛼 and calculate its rank. Question #12 How can we derive the standard form of the linear equation: 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 ? What do "A", "B" and "C" in the standard form of the linear equation mean? As in the point-slope form of the linear equation: 𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 ) , m is the slope, and x, y are the coordinates of any point. Likewise, 𝑥0 , 𝑦0 , are the coordinates of the given point. Answer: Let's say we start with the point-slope form 𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 ), where (𝑥, 𝑦) represents any point on the line, and (𝑥0 , 𝑦0 ) represents a given point on the line, and m is the given slope of the line. Now, 𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 ) ⟺ 𝑦 − 𝑦0 = 𝑚𝑥 − 𝑚𝑥0 ⟺ 𝑦 − 𝑦0 − 𝑚𝑥 − 𝑚𝑥0 = 0 ⟺ (−𝑚)𝑥 + 𝑦 + (−𝑦0 − 𝑚𝑥0 ) = 0 Now, we can simply let 𝐴 = −𝑚, 𝐵 = 1, 𝐶 = (−𝑚𝑥0 − 𝑦0 ) , which means that 𝑦 + (−𝑚)𝑥 + (−𝑦0 − 𝑚𝑥0 ) = 0 ⟺ 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 Question #13 How do I craft a linear equation so that it is in the form of 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 where 𝑎2 + 𝑏 2 = 1 if I have two points? I know how to get it into the form 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 but I can't figure out the algorithm for satisfying the second condition. Answer: I'm assuming you meant 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 so that this equation is that of a line. If you have 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 , then calculate 𝑚 = √𝑎2 + 𝑏 2 . Then dividing by 𝑚, 𝑎 𝑚 𝑥+ 𝑏 𝑚 𝑦+ 𝑐 𝑚 =0 satisfies the condition. Question #14 How to solve the system of linear differential equation of the form 𝑥 ′ = 𝐴𝑥 + 𝑏 I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Answer: Being b a fixed constant (suppose det 𝐴 ≠ 0 ), we can choose a new variable 𝑥∗ that solves 𝐴𝑥∗ + 𝑏 = 0 ⇒ 𝑥∗ = −𝐴−1 𝑏 The now we can solve the equation in the fixed point 𝑧 = 𝑥 − 𝑥∗ = 𝑥 + 𝐴−1 𝑏 . This variable solves the following equation 𝑧̇ = 𝐴𝑧 In fact, form the definition of z 𝑧̇ = 𝑥̇ 𝐴𝑧 = 𝐴(𝑥 + 𝐴−1 𝑏) = 𝐴𝑥 + 𝑏 So we're saying that, if x is a solution to 𝑥̇ = 𝐴𝑥 + 𝑏 then the variable 𝑧 = 𝑥 + 𝐴−1 𝑏 is a solution to the simpler equation 𝑧̇ = 𝐴𝑧 The solution to this simpler equation is just 𝑧(𝑡) = [𝑒 𝐴𝑡 ]𝑧(0) where [e^{At}] is the exponential of the matrix A. At this point you can go back to find x(t) simply by using the definition of z: 𝑥(𝑡) = 𝑧(𝑡) − 𝐴−1 𝑏 = [𝑒 𝐴𝑡 ](𝑥(0) + 𝐴−1 𝑏) − 𝐴−1 𝑏 = [𝑒 𝐴𝑡 ]𝑥(0) + 𝐴−1 𝑏([𝑒 𝐴𝑡 ] − 1) where we have used the fact that 𝑧(0) = 𝑥(0) + 𝐴−1 𝑏 Easer way You can even use this simpler method: given the system of equations 𝑥̇ = 𝐴𝑥 + 𝑓(𝑡) the solution to this is as follows 𝑡 ′ 𝑥(𝑡) = [𝑒 𝐴𝑡 ]𝑥(0) + ∫0 𝑒 𝐴(𝑡−𝑡 ) 𝑓(𝑡 ′ )𝑑𝑡 ′ in your case the function is just a constant function f(t)=b for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix Question #15 How would you solve the following system of linear equations: 2𝑥1 + (3 + 𝑎)𝑥2 + 2𝑥3 = 2 + 𝑎 𝑥1 + 𝑎𝑥2 + 2𝑥3 = 𝑎 𝑎𝑥1 + 2𝑥2 + 2𝑎𝑥3 = 0 assuming that 𝑎 ≠ ±√2 ? I feel confident in solving linear equation systems with just constants as the coefficients but the variable coefficient a is what gives me problems. Here is the solution I find: solution as augmented matrix but how would the assumption about a make the reduced echelon form any different compared to an assumption about e.g. 𝑎 ≠ ±√2? Answer: With 𝑥1 = 𝑎 − 𝑎𝑥2 − 2𝑥3 can we eliminate 𝑥1 : 𝑥_2(3 − 𝑎) − 2𝑥_3 = 2 − 𝑎 𝑥2 (2 − 𝑎2 ) + 4𝑎𝑥3 = −𝑎2 And then using 𝑥3 = 𝑥2 (3−𝑎)−(2−𝑎) 2 to eliminate 𝑥3 . Then you will get 𝑥2 (2 − 𝑎2 ) + 2𝑎(3 − 𝑎)𝑥2 − 2𝑎(2 − 𝑎) = −𝑎2 Question #16 I already know how the set of solutions of system of linear equations over real numbers infinite field PP is expressed. When there is only single solution then it is just a vector of scalars, where each scalar is a real number. When there are more than 1 solution, and actually infinite solutions, then it is just a parametric linear vector space in the form: ∀𝑡 ∈ 𝑅: 𝑝⃗ + 𝑣⃗ ⋅ 𝑡 where 𝑝⃗ ∈ 𝑅𝑛 ∧ 𝑣⃗ ∈ 𝑅𝑛 where n denotes the number of real variables in each linear equation thus 𝑛 ∈ 𝑁 But my question is how the set of solutions of system of linear equations over the finite field 𝑍𝟚 or galois field GF(2) is expressed? I already know that when there is only single solution then it is also just a vector of scalars, but where each scalar is a binary number either zero or one in 𝑍𝟚 where 𝑍𝟚 = {0,1} , but when there are more than 1 solution, but always finite number of solutions, then how are they expressed? Is this similar to how real solutions are expressed by parametric linear vector space by modulo 2? Or something else? I don't know. I am trying to google the answer for this question for days but I don't find the answer anywhere. It seems like nobody talks about this topic. How? Answer: Theory of linear equations works exactly the same way no matter the field. Let k a field and 𝐴 ∈ 𝑀𝑛 (𝑘) The most relevant result is Kronecker-Capelli theorem: Linear system 𝐴𝑥 = 𝑏 has solution if and only if rank𝐴 = rank[𝐴|𝑏] where [𝐴|𝑏]denotes augmented system matrix. In that case, let 𝑥0 be a particular solution of the system, i.e. 𝐴𝑥0 = 𝑏 . Then the solution set S of homogeneous linear system 𝐴𝑥 = 0 is vector space of dimension 𝑛 − rank𝐴 and all solutions of the system 𝐴𝑥 = 𝑏 are given by {𝑥_0 + 𝑥 | 𝐴𝑥 = 0} = 𝑥_0 + 𝑆 . These results are elementary and will be found in any Linear Algebra textbook. However, it is important to 𝑍 note that ground field k is completely irrelevant in this context, be it 𝑅, 𝐶, or whatever. It becomes 𝑝𝑍 relevant in spectral theory, though. Question #17 I am currently working through a chapter in my textbook, and it is on solving a form of linear differential equation. I don't understand how 𝑑 𝑑𝑡 𝑑 𝑑𝑡 𝑒 𝑘(𝑡) 𝑥 = 0 , since it says that 𝑑 𝑑𝑡 𝑒 𝑘(𝑡) = 𝑝(𝑡)𝑒 𝑘(𝑡) . Surely this would mean that 𝑒 𝑘(𝑡) 𝑥 = 𝑥 ∗ 𝑝(𝑡)𝑒 𝑘(𝑡) . Answer: Start from 𝑒 𝑘(𝑡) 𝑑𝑥 𝑑𝑡 + 𝑝(𝑡)𝑒 𝑘(𝑡) 𝑥 = 0 Note that 𝑝(𝑡)𝑒 𝑘(𝑡) 𝑥 = 𝑑 𝑑𝑡 𝑒 𝑘(𝑡) Therefore, by the product rule 𝑑 𝑑𝑡 (𝑒 𝑘(𝑡) 𝑥) = 𝑑 𝑑𝑡 (𝑒 𝑘(𝑡) ) ∗ 𝑥 + 𝑑𝑥 𝑑𝑡 𝑒 𝑘(𝑡) = 𝑝(𝑡)𝑒 𝑘(𝑡) 𝑥 + 𝑒 𝑘(𝑡) 𝑑𝑥 𝑑𝑡 =0 Let me know if this clears up the confusion Question #18 I am looking for a numerical method that can solve a system of non-linear equations. The non-linear equations are polynomials that are in one of the following forms: 𝑐1 𝑥1 + 𝑐2 𝑥2 + ⋯ + 𝑐𝑛 𝑥𝑛 = 𝐶 𝑑1 𝑥12 + 𝑑2 𝑥22 + ⋯ + 𝑑𝑛 𝑥𝑛2 = 𝐷 (The variables are 𝑥1 , 𝑥2 , … 𝑥𝑛 .) I can't get initial guesses that are close to the root. Is there a good numerical method to solve the system of equations? Answer: So if I understood your question correctly you are trying to solve a non-linear system of equations of the form 𝐴𝑥 + 𝑥 𝑇 𝐵𝑥 = 𝑏. Restating it as (𝐴 + 𝑥 𝑇 𝐵)𝑥 = 𝑏 ⇔ 𝑥 = (𝐴 + 𝑥 𝑇 𝐵)−1 𝑏 = 𝑓(𝑥) ⇔ 𝑥 − 𝑓(𝑥) = 0 a fixed point is found. Any root-finding method can now be used. For example, the simplest approach would be a fixed-point iteration of the form 𝑥𝑖+1 = (𝐴 + 𝑥𝑖𝑇 𝐵)−1 𝑏 but alternatives include Newton's or any other higher order method of the Housholder family. Note 1: \𝐼𝑛𝑥𝑖+1 = (𝐴 + 𝑥𝑖𝑇 𝐵)−1 𝑏 the matrix (𝐴 + 𝑥𝑖𝑇 𝐵)−1 should not be interpeted as computing the inverse but as solving the linear system of equations (𝐴 + 𝑥𝑖𝑇 𝐵)𝑥𝑖+1 = 𝑏 . This system of equations will be solved multiple times until some criterion is met (e.g. the residual 𝑟 = |(𝐴 + 𝑥 𝑇 𝐵)𝑥 − 𝑏| being smaller than some prescribed tolerance 𝜖𝑡𝑜𝑙 . Note 2: this works assuming an unique solution exists. The closer your initial guess 𝑥0 is to the "exact" solution, the less number of iterations you will need. As such a first step towards improving your method is to get a better initial guess. Another possibility is to use a method with a larger convergence radious first (e.g. a fix-point iteration), and then switch to a method with better convergence properties when some residual criterion is met (e.g. switch to Newton's method if 𝑟 < 10−2 ). Question #19 I am looking for general solution of the form 𝐵 = 𝑋𝐴𝑋 −1 with the following constraints 𝑋 −1 = 𝑋 ′ where X is unknown matrix and A,B,X, are 3 × 3 matrices. The problem appears peculiar and thus I wish to simplify it further to more practical problem X is rotation matrix (R), and A is plane equation matrix, B is resulted homography in the sense of 𝐵 = 𝑋𝐴𝑋 −1 = 𝑋𝐴𝑋 𝑇 = 𝑅 (𝐼 − 𝑡𝑛𝑇 𝑑 ) 𝑅𝑇 , where t is translation vector, n is plane normal vector, and d is signed distance, all three are known My dirty solution so far is as follows : Finding the Jordan Canonical form: 𝑀𝐵𝑀−1 = 𝐽 = 𝑁𝐴𝑁 −1 and then X is obviously 𝑋 = 𝑁 −1 𝑀 but this works only if the canonical forms are the same, otherwise cannot be utilized for solution. The second dirty trick is to solve a system of Linear equations in terms of 𝐴𝑋 = 𝐵𝑋 but i cannot figure out how to program this in the software (e.g. Matlab linsolve). Could you help me to clarify how to write system of linear equations in the shape of 9 rows? The trick is how to notate them on the "right side" to use in ready to use functions for solving SLE. Answer: Case 1. A,B are not orthogonally similar; no solutions. Case 2. A,B are orthogonally similar; 𝐴 = 𝑃𝐵𝑃𝑇 , where 𝑃 ∈ 𝑂(𝑛) . Let 𝐶(𝐴) = {𝑋 ∈ 𝑀𝑛 (𝑅); 𝐴𝑋 = 𝑋𝐴} . Then, it is not difficult to show Proposition {𝑋 ∈ 𝑂(𝑛); 𝐴 = 𝑋𝐵𝑋 𝑇 } = (𝐶(𝐴) ∩ 𝑂(𝑛))𝑃 . It remains to obtain 𝐶(𝐴) ∩ 𝑂(𝑛) . When 𝑛 = 3, C(A) is a vector space of dimension 3,5 or 9 (when A is a scalar matrix). Generically (randomly choose A), A is cyclic and 𝐶(𝐴) = {𝑎𝐼 + 𝑏𝐴 + 𝑐𝐴2 ; 𝑎, 𝑏, 𝑐 ∈ 𝑅} . Generically, the eigenspaces of A are not orthogonal and, consequently, 𝐶(𝐴) ∩ 𝑂(𝑛) = ±𝐼𝑛 and 𝑋 = ±𝑃 . EDIT. 𝐴, 𝐵 are orthogonally similar iff the couples (𝐴, 𝐴𝑇 ) and (𝐵, 𝐵𝑇 ) are similar. Then, practically, one studies the linear system (𝑆)𝐴𝑋 = 𝑋𝐵, 𝐴𝑇 𝑋 = 𝑋𝐵𝑇 . If 𝐴, 𝐵 are not orthog. similar, then the sole solution is 𝑋 = 0. If 𝐴, 𝐵 are orthog. similar, then, for a generic A, (S) admits a vector space of solutions of dimension 1: 𝑋 = 𝑢𝑋0 . It remains to calculate u so that ||𝑢𝑋0 ||2 = 1. Question #20 I am stucked on the following challenge: "If the line determined by two distinct points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is not vertical, and therefore has slope (𝑦2 −𝑦1 ) , show that the point-slope form of its equation is (𝑥2 −𝑥1 ) the same regardless of which point is used as the given point." Okay, we can separate (𝑥0 , 𝑦0 )from the form to get: 𝑦(𝑥2 − 𝑥1 ) − 𝑥(𝑦2 − 𝑦1 ) = 𝑦0 (𝑥2 − 𝑥1 ) − 𝑥0 (𝑦2 − 𝑦1 ) But how exclude this point (𝑥0 , 𝑦0 ) and leave only 𝑥, 𝑦, 𝑥1 , 𝑦1 , 𝑥2 , 𝑦2 in the equation? UPDATE: There is a solution for this challenge: (𝑦1 − 𝑦2 )𝑥 + (𝑥2 − 𝑥1 )𝑦 = 𝑥2 𝑦1 − 𝑥1 𝑦2 From the answer I found that 𝑦2 (𝑥 − 𝑥1 ) − 𝑦1 (𝑥 − 𝑥2 ) = 𝑦(𝑥2 − 𝑥1 ) but why this is true? Answer: Thanks to saulspatz, the solution is to simply show, that whether we are using (𝑥1 , 𝑦1 ) or (𝑥2 , 𝑦2 )as the given point, the equation does not change. So both equations: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦 − 𝑦2 = 𝑚(𝑥 − 𝑥2 ) reduce to the (𝑦1 − 𝑦2 )𝑥 + (𝑥2 − 𝑥1 )𝑦 = 𝑥2 𝑦1 − 𝑥1 𝑦2