linear algebra and real analysis 1

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span ( ) ={0 } S 1 ={( 1,0) , ( 0,1) } →span ( S 1 ) =R 2 i.exyplane S 2 ={( 1,1) } →span ( S 2 ) ={( x,y)∨x=y } . …………………………………………………………………………. if V={0 } ,thendimV =0 …………………………………………………………… S 1 ⊆S 2 →span ( S 1 ) ⊆span( S 2 )

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Linear Algebra and Real Analysis 1

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Similar matrices have same eigenvalues but different eigenvectors

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.Darboux theorem

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Krishna series page 20 are basic feasible solutions cannot be basic feasible solution as it fails to be non-negative

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same eigenvalues (with same algebraic multiplicity) A and B need not be similar.

x2=2, x3=0, x4=0 ; x1=1 u1 =(1, 2, 0, 0)x2=0, x3=2, x4=0 ; x1= -3 u2 =(-3, 0, 2, 0)x2=0, x3=0, x4=1 ; x1=0 u3 =(0, 0, 0, 1)they are linearly independent

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Thomas 855..

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Usually transportation problem is a minimization problem, where the objective is to minimize the transportation cost.When the objective is to maximize then subtract each element of the transportation matrix by the largest element.

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4 x 1

4 2 3

x 3 x

x : epsilon cannot be assigned as they form a closed loop with other assigned cells..

is uniformly continuous

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..or

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.uniform continuity need not imply differentiability

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Second method:

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is discontinuous everywhere

..Example of a function which is derivable but its derivative is not derivable

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use Dinis Theorem fn f pointwise each fn is continuous and f is continuous fn is monotonic fn is monotonically increasing

to s.t fn is bounded above squaring both sides

fn is monotonically increasing and bounded above , hence convergent (pointwise)to find the limit

and f(x) is continuous.by Dinis theorem : fn converges to f uniformly..

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both f and g are integrable, but which is Dirichlets function, is not integrable

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.Rank of a Skew-Symmetric Matrix is always even.Rank of a skew-symmetric matrix is atleast 2.

the smallest minor with non-zero determinant

hence rank is atleast 2... i.e

composition of two linear transformations is again linear.

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(try for 2 x 2 case)

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Diagonalise RHS matrix

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apply log test

..second log test

replace a by x

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by cauchys nth root test

this is equivalent to

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solution 1:

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non-homogenous equations

find the condition for this to be consistent.

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.f is R integrable f is boundedso, if f is not bounded f cannot be R integrable.

However, bounded functions need not be R integrable. (ex. Dirichlets function).