Recent content by blackbear

is the L(x) operator be as follows:L(x)= \begin{pmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} Please confirm...thanks

Yes...

Got it...thank you;

so L(x)= aL(X1) + bL(X2) + cL(X3) + dL(X4) + eL(X5) + fL(X6) but the first 3 terms will always be zero...if that's true I just tried to multiply with a Y matrix posted below and did not get the zero diagonal matrix. Y= \begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 5 \\ 0 & 5 & 3\end{pmatrix}...

In that case L(X4) = X4; L(X5)= X5; L(X6)=X6 since all the elements have zero diagonals.

L multiplies with a matrix (X) and get a matrix (Y) which has diagonal elements zero. L*X=Y so L(X4)*X4 = X4 so I had to compute a matrix L(X4) which multiplies with X4 matrix so the result becomes same matrix X4. Guess this is not the way to go then...

L(X2), L(X3) \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} L(X4) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix} Having difficulty to compute L(X5) & L(X6)...

LX1, LX2, LX3= \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} LX4, LX4, LX6= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} Is this correct?

I am still not sure how to get L(X1) but logically speaking is L(X1)= X1 the 1st matrix from the basis set?

Question: I know the 6 basis set so x in the equation is given. But I am not sure how get L for the basis set since "Y" is not given. I only found LX1 using the Y matrix but that is not a arbitrary matrix. L = X Y-1 The basis matrix is: B= \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1...

let Y = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 1 & 5 \\ 0 & 5 & 3\end{pmatrix} then LX1= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} So, the transformation to brings back X1, X2 and so on...but...

Homework Statement Let L(x) a linear operator defined by setting the diagonal elements of x to zero. What will be the representation of this operator to the following basis set? x E X. X denote the set of all real symmetric 3x3 matrices. Homework Equations L*y=x L=x*inv(y)...

How can I represent L(x) to the 6 symmetric basis set? I am to use the trivial basis set {1} for the range-space (R,R) to solve the problem. The symmetric matrix is: \begin {pmatrix} a & e & d \\ e & b & f \\d & f & c \end{pmatrix} Based o...

a=b=c=d=e=0 f=2

There is none since elements a23 & a32 is not related to any values of a,b,c,d,e & f. So, following is not the basis matrix: \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix} The following set of basis are: \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0...