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Chapter 3: Q21P (page 142)

Show that the transpose of a sum of matrices is equal to the sum of the transposes. Also show that(Mn)T=(MT)n. Hint: Use (9.11)and (9.8).

Short Answer

Expert verified

The sum of the transposes is equal to the transpose of the sum of the matrices.

Step by step solution

01

Given information.

The equation, MnT=MTnneeds to be proven.

02

Step 2: Transpose of a matrix.

A matrix's transpose is determined by converting its rows into columns or columns into rows.

03

Show that the transpose of the sum of the matrix is equal to the sum of their transposes.

Assume A and B are two matrices. Now,

(A+B)ijT=(A+B)ij(A+B)ij=Aij+Bij(A+B)ijT=AijT+BijTAijT+BijT=AT+BTij

Also,

(A+B)ijT=AT+BTij(A+B)T=AT+BT

Hence, from the above equation, the sum of the transposes is equal to the transpose of the sum of the matrices. Now,

MnT=(M.M鈥�M)T(forntimes)=(M(M鈥�M))T

SinceA(BC)=(AB)C=ABC,

(M鈥�M)TMT鈥�鈥�鈥�鈥�鈭�(ABCD)T=DTCTBTAT=(M(M鈥�.M))TMT=(M鈥�M)TMTMT

Solve further.

=(M鈥�M)TMTMT=(MM)TMT鈥�MT=MTMTMT鈥�鈥�MT(n-times)=MTn

Hence,

MTn=MnT

Hence, proved.

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Most popular questions from this chapter

Let each of the following matrices M describe a deformation of the(x,y)plane for each given Mfind: the Eigen values and eigenvectors of the transformation, the matrix Cwhich Diagonalizesand specifies the rotation to new axesrole="math" localid="1658833126295" (x',y')along the eigenvectors, and the matrix D which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.

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