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 May 18th, 2017, 06:39 AM #1 Newbie   Joined: May 2017 From: Essex Posts: 1 Thanks: 0 Computing a matrix according to inner product Hi I've got an answer to this question, but I'm not sure if it's actually right. If someone could read through and see what they get, that would be a massive help. Thanks. Question: consider the matrix K = (2 0 1; 5 7 0; 3 1 1). Recall the definition of projector operator: P(A)B = A /||A||^2 on a matrix space with inner product <.,.> Compute the matrix: M = (1-P(1))K, according to the inner product = Tr(A'S'SB) Here 1 is the identity matrix, S is the diagonal matrix with diagonal entries (1,2,1). Also A' means A transpose, I wasn't sure how to write it. And P(A)B for the projector operator is P subscript A, B. So I expanded the brackets and used the projector operator to come to K - 1(Tr(1'S'SK))||1||^2 From there I worked the trace part to be 31, and then norm of the identity to be 6. This meant the matrix M came out to: M = (-184 0 1; 5 -179 0; 3 1 -185) Please help and let me know if I've gone wrong. Last edited by skipjack; May 18th, 2017 at 06:23 PM.
May 22nd, 2017, 02:56 AM   #2
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Quote:
 Originally Posted by ptda Hi I've got an answer to this question, but I'm not sure if it's actually right. If someone could read through and see what they get, that would be a massive help. Thanks. Question: consider the matrix K = (2 0 1; 5 7 0; 3 1 1).
So K is $\begin{bmatrix} 2 & 0 & 1 \\ 5 & 7 & 0 \\ 3 & 1 & 1\end{bmatrix}$
and its transpose is $\begin{bmatrix}2 & 5 & 3 \\ 0 & 7 & 1 \\ 1 & 0 & 1 \end{bmatrix}$

Quote:
 Recall the definition of projector operator: P(A)B = A /||A||^2 on a matrix space with inner product <.,.> Compute the matrix: M = (1-P(1))K, according to the inner product = Tr(A'S'SB) Here 1 is the identity matrix, S is the diagonal matrix with diagonal entries (1,2,1).
S is $\begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and, since S is a diagonal matrix, S' is the same.

Quote:
 Also A' means A transpose, I wasn't sure how to write it. And P(A)B for the projector operator is P subscript A, B. So I expanded the brackets and used the projector operator to come to K - 1(Tr(1'S'SK))||1||^2
1'S'SK= S^2K which is $\begin{bmatrix}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 5 & 7 & 0 \\ 3 & 1 & 1\end{bmatrix}$= $\begin{bmatrix}2 & 0 & 1 \\ 20 & 28 & 0 \\ 3 & 1 & 1 \end{bmatrix}$

Yes, that has trace 21.

Quote:
 From there I worked the trace part to be 31, and then norm of the identity to be 6.
The "norm" is the square root of the inner product of a matrix with itself so, using this inner product, The inner product of the identity with itself is trace of IS'SI= S^2, 1+ 4+ 1= 6.

Quote:
 This meant the matrix M came out to: M = (-184 0 1; 5 -179 0; 3 1 -185) Please help and let me know if I've gone wrong.
Looks to me like you are correct.

Last edited by Country Boy; May 22nd, 2017 at 02:58 AM.

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