Math, asked by Cosmicexplorer4515, 1 year ago

Prove that the rank of transpose of a matrix is same as that of the original matrix

Answers

Answered by pulakmath007
9

SOLUTION

TO PROVE

The rank of transpose of a matrix is same as that of the original matrix

PROOF

Let A be a non zero matrix of order m × n. The Rank of A is defined to be the greatest positive integer r such that A has at least one non-zero minor of order r

For a non-zero m × n matrix A

0 < rank of A ≤ min {m, n}

For a non-zero matrix A of order n,

rank of A < , or = n according as A is singular or non-singular

 \sf{Now \:  transpose \:  of  \: A = { A }^{t} }

 \sf{Since \:  \:  A \:  \: and \:  {A}^{t}   \:  \: have \:  identical  \: minors}

 \sf{ \therefore \:  \: Rank \:  of \:  A  = Rank \: of \:  {A }^{t} }

Hence the rank of transpose of a matrix is same as that of the original matrix

Hence the proof follows

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Answered by dilkhushbhai9958
0

Answer:

show that the rank of the transpose of a matrix is the same as that if the original matrix

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