Question

Suppose A and B are matrices in lower triangular form, show that A ˙ ×B [ Kronecker product of A and B] is also in lower triangular form. Furthermore,

every eigenvalue of A ˙ ×B [Kronecker product of A and B] has the form of αβ, where α is an eigenvalue of A and β is an eigenvalue of B.

Answer 1

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AS FOR GIVEN DATA..

Let A and B be the two input upper triangular matrices and C be the product matrix. We have to prove that C is an upper triangular matrix.

Cij = Aik*Bjk

ij element of C is calculated by inner product of i th row of A and jth coulmn of B.

For given i,ji,j this sum will only be nonzero if there are k with i ≤ k ≤j (or at least one of Aik,Bkj will vanish) which requires i ≤ j Therefore ,C is upper triangular.

Now to prove,  every eigenvalue of A×B has the form of αβ, where α is an eigenvalue of A and β is an eigenvalue of B

Let Bx = βx ;

multiply by matrix A on both sides : ABx = Aβx

= βAx

we know that  Ax = αx

So, ABx = β(αx)

ABx = (βα)x

which means that eigen value of product of AxB is αβ

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