---- Linear systems: 

-   write a system of linear equations in matrix form
-   decide if a given system has a solution;
-   solve a linear system by Gauss- Jordan elimination
       or by finding rref of the matrix
-   interpret the linear system geometrically in simple cases
    (2 or 3 variables only; equations give lines on the plane or
    planes in the 3-space); make conclusions on how many solutions 
    the system has

--- Matrices and vectors

  -- Compute  dot product of vectors (if defined); use dot product to 
     determine if two vectors are perpendicular; find orthogonal projection 
     of a given vector to a given line.
  -- Compute product of a matrix and a vector (if defined)
     express this product as a linear combination of the columns of the matrix 
  -- compute product of two matrices (if defined); interpret matrix 
     multiplication as composition of corresponding linear transformations  
  -- Determine if a given n x n matrix is invertible; find inverse matrix

--- Subspaces, linear independence, basis, dimension

 -- decide if given vectors are linearly independent   
    (SUBSPACES WILL TYPICALLY BE GIVEN AS A SPAN OF VECTORS or AS KER or 
    IM of a linear transformation.)
 -- Decide if a given vector belongs to a given subspace; 
 -- Decide if given collection of vectors forms a basis of a given subspace; 
 -- Find a basis of a given subspace   
 -- Find dimension of a given subspace
 -- Basis change: find coordinates of a given vector with respect to a (new) basis 
    of a (sub)space. 
 
---- Linear transformations

--- Write a matrix of a linear trasformation given geometrically (projections, 
    reflections, reflections and scaling in R^2 and R^3; simple cases ONLY)    
--- Interpret geometrically a linear transformation given by a matrix (simple cases 
    in R^2 and R^3 ONLY)
--- Find Kernel and Image for a given linear transformation 
--- Use relation between dimensions of kernel, image, and the domain space 
   (RANK-NULLITY THEOREM) to find dim Ker from Dim Im abd vice versa  
--- Find bases for Ker and Im
--- use information on kernel and image to make conclusions about the number of 
    solutions for corresponding linear systems
--- basis change: write a matrix of a given transformation with respect to a new basis;
   use a convenient basis & a basis change to write a matrix of a transformation 
   given geometrically (projections and reflections onto arbitrary line on the plane)


--- Rank of a matrix/linear trasnformation

-- interpret and compute rank from rref of a matrix
-- interpret and compute rank as dimension of image of linear transformation
-- understand possible values of rank for square and rectangular matrices
-- using rank to make conclusions about invertibility of a square matrix, 
   and about the number of solutions of a linear system (with a square or non-square
   coefficient matrix)   

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                        Some types of questions: 

(1) Write a given linear system in matrix form Ax=b. Solve the system. 
    Find rref of the coefficient matrix A (or of augmented matrix). 
    Find the rank of A. Determine if A is invertible (for n x n matrices). 
     Find its inverse (if exists). What can you say about the number of solutions for 
    the equation Ax=c? (Here c is a given vector different from b.)
    Interpret the system geometrically. What does your geometric interpretation 
    say about the number of solutions?  

(2) Find the product of two given matrices A and B (if defined). Interpret 
    this product geometrically (in simple cases only: describe the transformations 
    given by A and B and their composition). Determine whether AB=BA. 

(3) Find a projection of a given vector onto a given line in R^2 or R^3.
    (Or find a reflection of the vector about a line in R^2). Write down 
    the matrix of this projection (or given reflection, or rotation). 
    Using the geometry of the transformation, compute powers such as A^5 
    or the inverse of the matrix A. 
    Find the matrix of this transformation in a different basis. 

(4) Determine if given vectors are linearly independent. Find a basis 
    of the subspace spanned by given vectors; find dimension of the subspace.
    Determine if a given vector is in the span of the given vectors. 
    Find coordinates of this vector with 
    respect to the given basis of the subspace.

(5) Consider the linear transformation determined by a given matrix A. 
    If this is a transformation R^m --> R^n, find m and n.
    Determine if a given vector lies in the kernel of A. Find kernel of A.
    Find a basis for the kernel of A. Find the dimension of ker A.

(6) For a linear transformation determined by a given matrix A, 
    find the image of A. Find a basis for im A. Find dim Im A.  
    Use this to find dim Ker A. Find rk A. What can you say about 
    the number of solutions of Ax =0? Ax =b? (for a given b or in general?) 
==============================================================================

                        Comments: 

(1) for an arbitrary matrix A, rank can be found from rref A. 
An n x n  matrix  A is invertible if and only if rank A = n. 
The inverse matrix can be found via a process similar to finding rref A. 
For n x n matrix A, solution to any equation Ax =b is always unique if 
A is invertible; if A is not invertible, Ax=0 has infinitely many solutions, 
while Ax =b has infinitely many solutions or no solutions at all, depending on b.
(For an n x m matrix, number of solutions is also dictated by rank, see book).

*** NOTE THAT in some cases, questions about rank and invertibility 
can alternatively be answered geometrically, see (3) and (6). 
The number of solutions can be understood from the information 
about ker A and Im A, see (6).*** 

(2) Remember that when AB is interpreted as a composition of transformations
and applied to a vector, B is applied first, A second.

(3) The matrix of a linear transformation (given geometrically or otherwise) 
  consists of the column vectors that are the images of the standard basis vectors
   under the transformation. (The standard basis is  ([1,0], [0,1]) in dim 2; 
   [1,0,0], [0,1,0], [0,0,1] in dim 3, etc.) So if you understand where these vectors 
   go, the matrix is easy to write.
   The same applies to finding the matrix in a non-standard basis (v1, v2,..):
   the column vectors of the matrix are the images of the basis vectors v1, v2,..
   under the transformation, but you have to express these images in the coordinates 
   corresp. to basis (v1, v2,..)
   Invertibility and compositions can often be understood from geometry: for example, 
   projection is not invertible because many non-zero vectors project to 0. 
   If A is a rotation, A^5 is the rotation by 5 times the angle.

(4) Vectors v1, v2, ... v are linearly dependent if there is a relation 
   c1 v1+ c2 v2+ ..   =0, where not all of the coefficients are 0. 
   Such a relation can often be found by inspection. If inspection doesn't 
   help, try explaining why it can't exist (looking at zero entries of the vectors 
   often helps). If inspection still doesn't help, you have to solve the corresponding 
   linear system. Same applies for deciding whether a vector is in the given span
   and finding coordinates: most of the time you can find the required relation, 
   or a reason why it can't exist, by a simple inspection. Otherwise, solve the system.
  To find a basis, find the "most economical" way to make the same span, by removing 
  (one-by-one) the vectors that can be expressed via the remaining vectors.       


(5) To find ker A, you have to solve Ax =0. Then, proceed as in (4).  Note that 
   it is often easier to find ker A by using the information about Im A; see (6).
   To check if a given vector is in ker A, just plug in into Ax.

(6) Im A is spanned by the columns of A. Then, proceed as in (4). Rank A = dim Im A.
   The dimension of Ker A can be found from rank-nullity theorem. Each vector
   of the kernel gives a linear relation between the column vectors of A; 
   these relations can often be found by inspection, so you can find a few vectors 
   in ker  A. If these vectors are linearly independent and you got enough
   to make the correct  dimension, you've found a basis for ker A.  
   Number of solutions: if dim Ker A>0, it means that Ax=0 has infinitely many 
   solutions, and that any equation Ax=b has either infinitely many or no solutions
   at all.   The equation Ax =b has a solution if and only if the vector b is in Im A;
   if you understand the image, you can often figure it out by inspection.