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Consider a symmetric invertible nnmatrix Awhich admits an LDU-factorization A=LDU. See Exercises 90, 93, and 94 of Section 2.4. Recall that this factorization is unique. See Exercise 2.4.94. Show that

U=LT (This is sometimes called theLDLT - factorizationof a symmetric matrix A.)

Short Answer

Expert verified

Because of LDU factorizationU=LT andL=UT claimed.

Step by step solution

01

Properties of the transpose.

Consider the properties of the transpose below.

  1. (A+B)T=AT+BT For all mnmatrices Aand B.
  2. (kA)T=kAT For all mnmatrices Aand for all scalars k.
  3. (AB)T=BTAT For all mpmatrices Aand for all matrices B.
  4. rank(AT)=rank(A) For all matrices A.
  5. (AT)-1=(A-1)T For all invertible nnmatrices A.

By Exercise 2.4.94b, the given LDU factorization of A is unique.

And by the properties of transpose consider the term below.

A=AT=LDU=UTDTLT=UTDLT

This is another way to write the LDU factorization of A.

Since,UTis lower triangular and is an upper triangular.

Hence, by the uniqueness of the LDU factorizationU=LTandL=UTclaimed.

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