Line 22: Line 22:
 
##<math>\text{ker}(A)=\{\vec 0\}</math>
 
##<math>\text{ker}(A)=\{\vec 0\}</math>
  
==Section 3.2: Subspaces of <math>\mathbb R^n</math>.==
+
 
*
+
  
 
----
 
----
 
[[2010_Spring_MA_35100_Kummini|Back to Course Page]]
 
[[2010_Spring_MA_35100_Kummini|Back to Course Page]]

Latest revision as of 06:14, 29 March 2010

Exam 2 Review

Exam 2 covers Sections 3.1, 3.2 3.3, 3.4, and 4.1. The following is a list of the sections, the topics covered in them, and some basic review

Section 3.1

  • Image
    • Definition: All the vectors you can get by multiplying the matrix of a linear transformation by an arbitrary vector. See also: Image
    • It is a subspace
    • If the matrix is to be invertible, then the image must be the codomain
    • The span of the image consists of the columns of the matrix. This makes sense since the image is really just the linear combination of the columns of A, the matrix for the transformation.
  • Kernel
    • Definition: All the vectors $ \vec v $ in which $ A\vec v =\vec 0 $. See also: Kernel
    • It is a subspace
    • If the matrix is to be invertible, then the kernel will consist only of $ \vec 0 $
    • Find the kernel by finding rref of augmented matrix for $ A\vec v =\vec 0 $. Then substitute one for a free variable and zeros for all the other. Repeat for each other free variable. You now have the vectors that span the kernel.
  • Invertible Matrices list:
    • The following are equivalent and each proves the other for the case where A is an nxn matrix
    1. A is invertible
    2. The linear system $ A\vec x=\vec b $ has a unique solution for all $ \vec b $ in $ \mathbb R^n $
    3. $ \text{rref}(A)=I_n $
    4. $ \text{rank}(A)=n $
    5. $ \text{im}(A)=\mathbb R^n $
    6. $ \text{ker}(A)=\{\vec 0\} $



Back to Course Page

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood