Two linear algebra questions?
1. if you have two vectors, how would you theoretically find the equation of the plane they span (you're not given a point in the plane) 2. if you have two planes that intersect at a line, how do you find the equation of the orthogonal vector from that line to a point? thanks!!
Math, Linear Algebra question:?
T is a rotation by +90 degrees. Sorry you will just have to guess what I was trying to write below since the miserable editor available rejects all the blank spaces I introduced to try to write matrices and entries all run together. I'll put in periods that you should think of as just invisible so the matrices look square. The usual rules of matrix multiplication (row of first matrix times column of second) apply. Let F be the 2 x 2 matrix for the reflection in the y axis f1 f2 = -1 0 f3 f4 . . 0 1 which gives -x = -1 0 * x +y . . 0 1 . y as required. Similarly, let G be the 2 x 2 matrix for the reflection in the x axis g1 g2 = 1 .0 g3 g4 . . 0 -1 which gives +x = 1 .0 * x -y . . 0 -1 . y as required. Let H be the matrix for the -90 degree rotation. Recall the angle addition formulas sin(A+B) = sinAcosB + cosAsinB cos(A+B) = cosAcosB - sinAsinB We need for any position vector r with angle A from the origin r cos(A-90degrees) = h1 h2 * r cosA r sin(A-90degrees) .. h3 h4 .. r sinA By angle addition this is r [cosAcos(-90degrees) - sinAsin(-90degrees)] r [sinAcos(-90degrees) + cosAsin(-90degrees)] = r [h1cosA + h2sinA] r [h3cosA + h4sinA] which we can see is satisfied by h1 h2 = cos(-90degrees) -sin(-90degrees) = 0 1 h3 h4 . . sin(-90degrees) cos(-90degrees) .. -1 0 We can find T by applying the transformation matrices successively, i.e., by multiplying their matrices: T = F * G * H or t1 t2 = -1 0 * 1 0 * 0 1 t3 t4 . .0 1 .. 0 -1 . -1 0 which multiplies out to t1 t2 = -1 0 * 0 1 = -1 0 * h1 h2 = 0 -1 t3 t4 . .0 -1 . -1 0 . . 0 -1 . h3 h4 . 1 0 From the identities above we see that T boils down to a negative identity matrix applied to H or, in other words, T is a rotation matrix for the same amount as H but in the opposite direction. Since H is a -90 degree rotation, T is a +90 degree rotation.
3 Linear Algebra Questions?
1. suppose that v1, ... vk are linearly dependent, so a1v1+....+akvk=0, for some scalars a1,...,ak not al equal to cero. then A(a1v1+....+akvk)=0 but A(a1v1+....+akvk) = a1A(v1) +.... + akA(vk) =a1b1+...+akbk so we found a linear combination of b1, ... , bk that equals cero.... this contradicts the fact that b1, ... , bk are linearly independent, which means that v1,... vk must be linearly independent also... 2. (x_1,x_1,x_3,x_3,x_5) =x_1(1,1,0,0,0)+x_3(0,0,1,1,0)+x_5(0,0... so the basis is: (1,1,0,0,0), (0,0,1,1,0), (0,0,0,0,1) this set is clearly linearly independent and also it generates the subspace.
Linear Algebra question?
Hi, yes, V is a vector space over F, since F is a sub-field of |R. In order to prove this, consider the following statement: "If V is a vector space over a field K and F is a sub-field of K then V is a vector space over F, with addition the same as before and scalar multiplication given by the restriction to F of thee scalar multiplication defined over K". If we want to prove this, first of all note that if (V, +) is a commutative group if we view V as a K-vector space, it will also be so if we view it as an F-vector space, since this is an operation defined inside V, regardless the base field. All we need to prove then is that given a and b in F and u and v in V, we get: a*(b*v) = (a*b)*v; 1*v = v; a*(u + v) = a*u + a*v; (a + b)*v = a*v + b*v. But of course this is true, since a, b can be seen as elements in K, where these equalities hold by hypothesis (in fact F was suposed to be a sub-field of K). I hope I helped you. Bye...
LINEAR ALGEBRA QUESTION, PLEASE HELP!?
The first four Laguerre polynomials are 1, 1 − ?, 2 − 4? + ?^2, 6 − 18? + 9^2 − ?^3 . a) Show that they for a basis for ℙ3? b) Find the change of coordinate matrix from ? to the standard basis ? = {1, ?, ?^2, ?^3} : ?B→S c) Find ?S→B by using ?B→S d) Find the polynomial ?(?) such that [?]B = [1/2/3/4] e) Using ?S→B and/or ?B→S find [?]S
Markov Chain (Linear Algebra) Question?
The annual transition probability matrix, P, is [0.1 0.9] [0.25 0.75] where the first row represents having the flu in a given year, the second row represents not having the flu in a given year, the first column represents having the flu in the following year, and the second column represents having the flu in the following year. a) The 2-year probability transition matrix is P^2, which (from matrix-multiplying P by P) equals [0.235 0.765] [0.2125 0.7875] Therefore, if a person does not have the flu one year, then the probability that he/she will also not have the flu 2 years later is 0.7875. b) Let v represent the two-component row vector [x y], where x is the long-run proportion of the years of getting the flu and y is the long-run proportion of the years of not getting the flu. Then we have x+y = 1 vP = v. The equation vP = v becomes [0.1x+0.25y 0.9x+0.75y] = [x y] and therefore becomes the two equations 0.1x+0.25y = x 0.9x+0.75y = y. Therefore, we have the three equations 1) x+y = 1 2) -0.9x+0.25y = 0 3) 0.9x-0.25y = 0. Equations 2) and 3) are algebraically equivalent (redundant) so equation 3) provides no new information. So we solve equations 1) and 2) to get x = 0.25/(0.25+0.9) = 5/23 and y = 1-(5/23) = 18/23. Therefore, in the long run, a person does not have the flu 18/23 of the years. Have a blessed, wonderful day!
How do solve this linear algebra question?
You have more indeterminates than equations, so you have to say that one of the indeterminates is in fact a parameter, and let's go for [math]t[/math] (with [math]X = (x, y, z, t)[/math])[math]AX=B \iff \begin{pmatrix}1&1&2\\2&1&1\\1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1-t\\1-t\\1-t\end{pmatrix}[/math]You see that this always has solutions, the determinant of our new square matrix is [math]1[/math]. In fact what we've done here is saying we force [math]t[/math] to be some parameter, we do it by adding a new line to the matrix [math]A[/math], thus becoming a square matrix:[math]\begin{pmatrix} 1 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\z\\t\end{pmatrix} = \begin{pmatrix}1\\1\\1\\\lambda\end{pmatrix}[/math].Then substracting the last line from the first, the second and the third yields:[math]\begin{pmatrix} 1 & 1 & 2 & 0 \\ 2 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\z\\t\end{pmatrix} = \begin{pmatrix}1-\lambda\\1-\lambda\\1-\lambda\\\lambda\end{pmatrix}[/math]Then you should remember properties of diagonal-by-blocks matrices and you end up with what I wrote at the very beggining of this answer.
Which is the best website to ask linear algebra questions?
If you are looking for QUESTIONS on linear algebra-This is surely best website for that:Course Overview (in Hindi) | (Hindi) Linear Algebra (All Branch GATE Questions) - Unacademy
From what topics in linear algebra are questions asked in the GATE?
Topics which cover, properties of the roots are most likely to be asked in GATE exam.For example - They will provide the algaebric equation and they can ask the following things -1 . Roots2. Product of the roots taken ‘n’ at a time3. Sum of the roots taken ’n’ at a time4. Eigen values 5. Properties of Eigen values like Multiplication, addition,inverse of the eigen values.6. Eigen Vectors and their properties.Most probable questions include direct implementation of the properties so that they can check your application ability. These questions will mostly be of 1 marks which can be solved within 2–3 minutes of time.If u dont remember these properties, the question can take u 5–7 minutes of time. Make sure u remember most of these properties to save your time in GATE exam.
Where can I practice linear algebra questions online?
MyOpenMath. You can register for a free account, then look for a linear algebra course.