# anyone can help me for this prove matrix

• Aug 2nd 2008, 11:03 PM
happystudent
anyone can help me for this prove matrix
1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
b) using only the definition of matrix multiplication , prove that B is upper triangular.

Thanks.
• Aug 3rd 2008, 12:01 AM
CaptainBlack
Quote:

Originally Posted by happystudent
1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
b) using only the definition of matrix multiplication , prove that B is upper triangular.

Thanks.

Let the columns of $\displaystyle \bold{A}$ be $\displaystyle \bold{A}_1, .. \bold{A}_n$, then by definition $\displaystyle (\bold{A}_i)_i \ne 0$.

Let the $\displaystyle k$-th row of $\displaystyle \bold{B}$ be $\displaystyle \bold{B}_k$.

Then the $\displaystyle l$-th element in the $\displaystyle k$-th row of $\displaystyle \bold{I}=\bold{B}\bold{A}$ is:

$\displaystyle \bold{I}_{k,l}=\bold{B}_k\bold{A}_l$

Now, as $\displaystyle \bold{A}$ is upper trianglar:

$\displaystyle \bold{I}_{k,1}=\bold{B}_k\bold{A}_1=(\bold{B}_k)_1 (\bold{A}_1)_1$

and if $\displaystyle k\ne 1$ this is zero and as by definition $\displaystyle (\bold{A}_1)_1 \ne 0$ for $\displaystyle k \ne 1 (\bold{B}_k)_1=0$

Now suppose that for some $\displaystyle j<k\ (\bold{B}_k)_r=0$ for all $\displaystyle r \le j$

Then:

$\displaystyle \bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r$

because $\displaystyle \bold{A}$ is upper triangular, and:

$\displaystyle \bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}$

because by hypothesis $\displaystyle (\bold{B}_K)_r=0$, for all $\displaystyle r \le j$.

Hence:

$\displaystyle \bold{I}_{k,j+1}=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}=0$

So $\displaystyle (\bold{B}_k)_{j+1}=0$

Which completes a proof by induction that $\displaystyle (\bold{B}_k)_j=0$ for all $\displaystyle j<k$, which proves that $\displaystyle \bold{B}$ is upper triangular.

(it is easier to see what is going on doing this element by element along the a row of $\displaystyle \bold{B}$)

RonL