Math 2270 Summary, Fall 2014
September 27, 2014
Test 1
Introduction to vectors
• Vectors, vector addition, scalar multiplication
• Algebraic properties (zero vector, commutativity, distributivity, etc.)
• Geometric interpretation (mainly as points, but also as arrows based at the origin)
• Linear combinations and spans of vectors
• Length of vectors, unit vectors, perpendicular vectors, cosine formula
Introduction to matrices and A~x = ~b
• A an n ⇥ n square matrix
• Row picture: system of linear equations (intersections of hyperplanes)
• Column picture: linear combination of the columns of A
• Elimination for square matrices: find an upper triangular U with pivots on the diagonal,
then use back substitution to find the solution
• Elimination succeeds when there are exactly n pivots, possibly after row exchanges
(then A is invertible!)
Test 2
Key notation
• For A an m ⇥ n matrix: m rows, n columns, r pivots.
• Multipliers `ij ; elimination matrix Eij ; lower-triangular square matrix L; upper-triangular
square matrix or non-square echelon matrix U ; row-reduced echelon
h
imatrix R; diagonal
square matrix D; permutation matrix P ; augmented matrix A ~b ; special solutions
~si ; particular solution ~xp .
Key definitions to look up
• Matrix multiplication AB; inverse matrix A 1 ; transpose AT ; symmetric matrix.
• Pivots; rank r; pivot columns; free columns; pivot variables; free variables.
• Multipliers `ij ; elimination matrix Eij ; permutation matrix P ; echelon matrix U ; rowreduced echelon matrix R.
• Vector space (usually Rn ); subspace of Rn ; span of a collection of vectors.
A
/ Rm by the rule ~
• An m⇥n matrix A yields a function Rn
x 7! A~x; nullspace N (A)
n
is a subspace of the domain R ; column space C(A) is a subspace of the codomain Rm .
Algebraic properties
• Properties of matrix multiplication: A(BC) = (AB)C, AB 6= BA in general, etc.
• Properties of inverse matrices: (AB)
1
= B 1A
1
when A and B are invertible, etc.
• Properties of transpose: (A + B)T = AT + B T , (AB)T = B T AT , etc.
Uses of elimination for A an m ⇥ n matrix
• Compute the nullspace N (A) of A (all solutions to A~x = ~0): elimination on A with row
swaps yields the echelon matrix U or (also using row-scaling and upward elimination)
the row-reduced echelon matrix R. Then find the special solutions by repeatedly setting
one free variable to 1 and all other free variables to 0 and solving for the pivot variables
by back substitution. Each free column yields one special solution. N (A) is the span
of the special solutions.
h
i
⇥
⇤
• Solve A~x = ~b: elimination on A ~b with row swaps yields U ~c or (also using rowh
i
~
scaling and upward elimination) R d . Find one particular solution ~xp by setting all
free variables to 0 and solving for the pivot variables by back substitution. The set of
all solutions is ~x = ~xp + ~xn , where ~xn ranges over all vectors in N (A).
2
Uses of elimination for A an n ⇥ n matrix
• Both uses from the previous section (m = n).
1
• Compute
⇥
⇤ the inverse matrix A (only possible if A has n pivots!): elimination
⇥
⇤ on
1
A I (with row swaps, row scaling, and upward elimination) produces I A .
• Compute the factorization A = LU (only works if rows of A are in the right order and
A is invertible): elimination on A finds the upper triangular U . The lower triangular
L contains all the multipliers `ij .
• Compute the factorization A = LDU (only works if rows of A are in the right order
and A is invertible): same as above, but factor the diagonal pivot matrix out of U .
• Compute the factorization P A = LU or P A = LDU (works as long as A is invertible):
same as above, but begin with row swaps P .
Some connections
• The only subspaces of R3 are the origin {~0} (dimension 0), the whole vector space R3
(dimension 3), any line through the origin (dimension 1), and any plane through the
origin (dimension 2). Subspaces of Rn can be classified in a similar manner.
• N (A) and C(A) are subspaces. The set of solutions to A~x = ~b for ~b 6= ~0 is not a
subspace: it is obtained by shifting the subspace N (A) away from the origin using the
particular solution ~xp .
• dim N (A) = n
r and dim C(A) = r, so dim N (A) + dim C(A) = n.
• Every free column of A is a linear combination of the earlier pivot columns. The special
solution for which you set that free variable to equal 1 gives this linear combination.
• No pivot column of A is a linear combination of earlier columns.
• When A is invertible, A~x = ~b has a unique solution for every ~b.
• When r = m (full row rank), A~x = ~b has at least one solution for every ~b, which means
that C(A) = Rm (the function Rn
A
/
Rm is onto).
• When r = n (full column rank), A~x = ~b has at most one solution for every ~b, which
means that N (A) = {~0} (the function Rn A / Rm is one-to-one).
3