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**Lecture 10 Dimensions, Independence, Basis and Complete Solution of Linear Systems**

Shang-Hua Teng

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**Linear Independence Linear Combination Linear Independence**

is linearly independent if only if none of them can be expressed as a linear combination of the others

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Examples

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**Linear Independence and Null Space**

Theorem/Definition is linearly independent if and only a1v1+a2v2+…+anvn=0 only happens when all a ’s are zero The columns of a matrix A are linearly independent when only solution to Ax=0 is x = 0

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2D and 3D v w u v

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**How do we determine a set of vectors are independent?**

Make them the columns of a matrix Elimination Computing their null space

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**Permute Rows and Continuing Elimination (permute columns)**

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**There must be free variables.**

Theorem If Ax = 0 has more have more unknown than equations (m > n: more columns than rows), then it has nonzero solutions. There must be free variables.

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Echelon Matrices Free variables

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**Reduced Row Echelon Matrix R**

Free variables

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**Computing the Reduced Row Echelon Matrix**

Elimination to Echelon Matrix E1PA = U Divide the row of pivots by the pivots Upward Elimination E2E1PA = R

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**Example: Gauss-Jordan Method for Matrix Inverse**

[A I] E1[A I] = [U, I] In its reduced Echelon Matrix A-1 [A I] = [I A-1]

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**A Close Look at Reduced Echelon Matrix**

The last equation of R x = 0 is redundant 0 = 0 Rank of A is the number of pivots rank(A).

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**What is the Rank of Outer Product**

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**Rank and Reduced Row Echelon Matrix**

Free variables Theorem/Definition Rank(A) = number of independent rows Rank(A) = number of independent columns

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**Dimension of the Column Space and Null Space**

The dimensions of the column space of A is equal to Rank(A). The dimension of the null space of A is equal to the number of free variables which is n – Rank(A) A is an m by n matrix

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**Rank and Reduced Row Echelon Matrix**

The Pivot columns are not combinations of earlier columns Pivot columns Free variables Free Columns

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**Reduced Echelon and Null Space Matrix**

Special Solutions

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Null Space Matrix Ax=0 has n-Rank(A) free variables and special solutions The Nullspace matrix has n-Rank(A) columns The columns of the nullspace matrix are independent The dimension of the Null space is n – rank(A)

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**Complete Solution of Ax = 0**

After column permutation, we can write r pivot columns n-r free columns Nullspace matrix Pivot variables Free variables Moreover: RN = [0]

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**Complete Solution to Ax = b**

A is an m by n matrix, and b is an n-place vector Unique solution Infinitely many solution No solution Suppose Ax = b has more then one solution, say x1, x2 then A x1 = b A x2 = b So A (x1 - x2 ) = 0 (x1 - x2 ) is in nullspace(A)

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**Complete Solution to Ax = b**

Suppose we found a particular solution xp to Ax = b i.e, A xp = b Let F be the indexes of free variables of Ax = 0 Let xF be the column vector of free variables Let N be the nullspace matrix of A Then defines the complete set of solutions to Ax = b xp

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**Example: Complete Solution to Ax = b**

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Augmented matrix [A b] Elimination to obtain [R d]

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**Set free variables to 0 to find a particular solution**

Compute the nullspace matrix Complete solution is

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**Full Rank Matrix Suppose A is an m by n matrix. Then**

A is full column if rank(A) = n columns of A are independent A is full row rank if rank(A) = m Rows of A are independent

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**Full Column Rank Matrix**

Columns are independent All columns of A are pivot columns There are non free variables or special solutions The nullspace N(A) contains only the zero vector If Ax=b has a solution (it might not) then it has only one solution n by n m-n rows of zeros

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**Full Row Rank Matrix Rows are independent**

All rows of A have pivots, R has no zero rows Ax=b has a solution for every right hand side b The column space is the whole space Rm There are n-m special solutions in the null space of A

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**The Whole Picture Rank(A) = m = n Ax=b has unique solution**

Ax=b has n-m dimensional solution Rank(A) = n < m Ax=b has 0 or 1 solution Rank(A) < n, Rank(A) < m Ax=b has 0 or n-rank(A) dimensions

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**Basis and Dimension of a Vector Space**

A basis for a vector space is a sequence of vectors that The vectors are linearly independent The vectors span the space: every vector in the vector can be expressed as a linear combination of these vectors

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**Basis for 2D and n-D (1,0), (0,1) (1 1), (-1 –2)**

The vectors v1,v2,…vn are basis for Rn if and only if they are columns of an n by n invertible matrix

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**Column and Row Subspace**

C(A): the space spanned by columns of A Subspace in m dimensions The pivot columns of A are a basis for its column space Row space: the space spanned by rows of A Subspace in n dimensions The row space of A is the same as the column space of AT, C(AT) The pivot rows of A are a basis for its row space The pivot rows of its Echolon matrix R are a basis for its row space

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**Important Property I: Uniqueness of Combination**

The vectors v1,v2,…vn are basis for a vector space V, then for every vector v in V, there is a unique way to write v as a combination of v1,v2,…vn . v = a1 v1+ a2 v2+…+ an vn v = b1 v1+ b2 v2+…+ bn vn So: 0=(a1 - b1) v1 + (a2 -b2 )v2+…+ (an -bn )vn

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**Important Property II: Dimension and Size of Basis**

If a vector space V has two set of bases v1,v2,…vm . V = [v1,v2,…vm ] w1,w2,…wn . W= [w1,w2,…wn ]. then m = n Proof: assume n > m, write W = VA A is m by n, so Ax = 0 has a non-zero solution So VAx = 0 and Wx = 0 The dimension of a vector space is the number of vectors in every basis Dimension of a vector space is well defined

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**Dimensions of the Four Subspaces Fundamental Theorem of Linear Algebra, Part I**

Row space: C(AT) – dimension = rank(A) Column space: C(A)– dimension = rank(A) Nullspace: N(A) – dimension = n-rank(A) Left Nullspace: N(AT) – dimension = m –rank(A)

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