document.write( "Question 1183382: Let n ∈ N and B be a n × n matrix that has determinant 1. Show that there exist n × n matrices
\n" ); document.write( "K, A and N with the following four properties:
\n" ); document.write( "(a) B = KAN,
\n" ); document.write( "(b) K is an orthogonal matrix with det(K) = 1,
\n" ); document.write( "(c) N is an upper triangular matrix with all diagonal entries 1; and
\n" ); document.write( "(d) A is a diagonal matrix whose all diagonal entries are positive and det(A) = 1 \n" ); document.write( "

Algebra.Com's Answer #813695 by robertb(5830) $\"\"$ $\"About$

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The LU decomposition of B states that B = LU where L is a lower triangular matrix and U is an upper triangular matrix.
\n" ); document.write( "An even better decomposition is the LU decomposition with partial pivoting, which says that \r
\n" ); document.write( "\n" ); document.write( " $\"PB+=+LU\"$ \r
\n" ); document.write( "\n" ); document.write( "for some permutation matrix P. One property of permutation matrix is that it is always orthogonal, i.e., $\"P%5ET%2AP+=+P%2AP%5ET++=+I\"$ .
\n" ); document.write( "In other words, \r
\n" ); document.write( "\n" ); document.write( " $\"P%5E%28-1%29+=+P%5ET\"$ , or its transpose is its own inverse.\r
\n" ); document.write( "\n" ); document.write( "Hence, $\"B+=+P%5E%28-1%29%2ALU+=+P%5ET%2ALU\"$ \r
\n" ); document.write( "\n" ); document.write( "Note that by symmetry, $\"P%5ET\"$ is also orthogonal.\r
\n" ); document.write( "\n" ); document.write( "Now none of the diagonal elements of U is equal to 0, since the determinant of B is 1. This implies that U can be decomposed into\r
\n" ); document.write( "\n" ); document.write( " $\"U+=+%28diag%28U%29%29%2AU%5B1%5D\"$ , \r
\n" ); document.write( "\n" ); document.write( "where diag(U) is a diagonal matrix whose diagonal elements and their locations are the same as those of U,
\n" ); document.write( "and $\"U%5B1%5D\"$ is an upper triangular matrix where all diagonal elements are 1.\r
\n" ); document.write( "\n" ); document.write( "Now $\"QL+=+U%5B2%5D\"$ , or $\"L+=+Q%5ET%2AU%5B2%5D\"$ where Q is a permutation matrix that turns L into an upper triangular matrix $\"U%5B2%5D\"$ .\r
\n" ); document.write( "\n" ); document.write( "==>

.\r
\n" ); document.write( "\n" ); document.write( "Similarly, $\"U%5B2%5D+=+diag%28U%5B2%5D%29%2AU%5B3%5D\"$ , where $\"diag%28U%5B2%5D%29\"$ is a diagonal matrix whose diagonal elements and their locations are the same as those of $\"U%5B2%5D\"$ ,
\n" ); document.write( "and $\"U%5B3%5D\"$ is an upper triangular matrix where all diagonal elements are 1.\r
\n" ); document.write( "\n" ); document.write( "==>

.\r
\n" ); document.write( "\n" ); document.write( "where $\"U%5B4%5D\"$ is still an upper triangular matrix with diagonal elements equal to 1.\r
\n" ); document.write( "\n" ); document.write( "==> $\"B+=+%28QP%29%5ET%2A%28diag%28U%5B2%5D%29%2Adiag%28U%29%29%2A%28U%5B4%5D%2AU%5B1%5D%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Now $\"det%28QP%29%5ET=1\"$ , because its rows are just rearrangements of the rows of I;
\n" ); document.write( " $\"U%5B4%5D%2AU%5B1%5D\"$ is an upper triangular matrix whose diagonal elements are all 1, hence its determinant is equal to 1;
\n" ); document.write( "and finally $\"diag%28U%5B2%5D%29%2Adiag%28U%29\"$ has the same determinant as LU, and therefore equal to 1.\r
\n" ); document.write( "\n" ); document.write( "Hence, let $\"K+=+%28QP%29%5ET\"$ , $\"A+=+diag%28U%5B2%5D%29%2Adiag%28U%29\"$ , and $\"N+=+U%5B4%5D%2AU%5B1%5D\"$ , and the statement is proved. \n" ); document.write( "

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