Kernel & Image of Linear Transformation Homework

  • Thread starter Thread starter jeffreylze
  • Start date Start date
  • Tags Tags
    Image Kernel
jeffreylze
Messages
43
Reaction score
0

Homework Statement



38) Determine whether or not v1 = (-2,0,0,2) and v2 = (-2,2,2,0) are in the kernel of the linear transformation T:R^4 > R^3 given by T(x) = Ax where

A = [1 2 -1 1;
1 0 1 1;
2 -4 6 2]

39) Determine whether or not w1 = (1,3,1) or w2 = (-1,-1,-2) is in the image of the linear transformation given in question 38?

Homework Equations


The Attempt at a Solution



I row-reduced it to rref then i let matrix = 0 and then solve for x1, x2, x3 and x4 . Which gave me x3 (-1,1,1,0) + x4 (-1,0,0,1) . v1 and v2 are not in that kernel i found but the answer states otherwise. is it because v1 and v2 are just scalar multiples of x3 and x4 ?For question 39 , i am stuck too. Please help.
 
Last edited:
Physics news on Phys.org
To determine whether or not given vectors are in the kernel of a linear transformation, simply take the product of the matrix which represents the linear transformation and the vector in question.

What should this product be if the vector is in the kernel?
 
The product should give me a zero vector. What about for the image, question 39 ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top