diagnol matricies

calchere said:
Give 2 different diagonal matricies that are similar to the following matrix.
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What is "the following matrix"?

calchere said:
(i couldn't get texaide to post correctly, so i just posted a picture of it.)
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You could also try formatting LaTeX directly:

. . . . .\(\displaystyle \left[ \begin{array}{cc} 1&4\\0&-3 \end{array} \right]\)

. . . . .\(\displaystyle =\, \left[ \begin{array}{cc} {1\,-\, \lambda}&4\\0&{-3\, -\, \lambda} \end{array} \right]\)

. . . . .\(\displaystyle =\, (1\, -\, \lambda)(-3\, -\, \lambda)\)

. . . . .\(\displaystyle \lambda_1\, =\, -3\, \mbox{ and } \, \lambda_2\, =\, 1\)

. . . . .\(\displaystyle =\, \left[ \begin{array}{cc}-3&0\\0&1 \end{array} \right]\)

calchere said:
That is what I got for the first diagonal matrix, but I don't know how to get a second one.
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It appears that you are switching between matrices, related determinants, quadratic equations, and variable values...? Have you computed eigenvectors for whatever matrix...?

Eliz.
 
The matrix i am trying to get the diagonal matricies from is:

\(\displaystyle \left[ \begin{array}{cc} 1&4\\0&-3 \end{array} \right]\)

I got:
\(\displaystyle \[\lambda _1 = \left[ {\begin{array}{*{20}c} 4 & 4 \\ 0 & 0 \\\end{array}} \right] = \left[ {\begin{array}{*{20}c} { - 1r} \\ r \\\end{array}} \right] = \left[ {\begin{array}{*{20}c} { - 1} \\ 1 \\\end{array}} \right]\]\)

\(\displaystyle \[\lambda _2 = \left[ {\begin{array}{*{20}c} 0 & 4 \\ 0 & { - 4} \\\end{array}} \right] = \left[ {\begin{array}{*{20}c} 0 \\ s \\\end{array}} \right] = \left[ {\begin{array}{*{20}c} 0 \\ 1 \\\end{array}} \right]\]\)

\(\displaystyle \left[ {\begin{array}{*{20}c} 1 & { - 1} \\ 0 & 1 \\\end{array}} \right]\left[ {\begin{array}{*{20}c} { - 3} & 0 \\ 0 & 1 \\\end{array}} \right]\left[ {\begin{array}{*{20}c} 1 & 1 \\ 0 & 1 \\\end{array}} \right]\)

\(\displaystyle \left[ {\begin{array}{*{20}c}{ - 3}&0\\0 & 1\\\end{array}} \right]\)

which is circled correct on my homework, but i couldn't figure out how to get another diagnoal matrix different from the one i got.
 
Can you use scalars to create another matrix, similar to the one you've found...?

Eliz.
 
I suppose i could. The question doesn't specify how to create the second one, so i wasn't sure. I just now noticed the following question asks:

Prove that, if a matrix is diagonalizable, so is its transpose.

So, maybe the second one could come from the transpose of the original matrix.

Thanks.
 
calchere said:
which is circled correct on my homework, but i couldn't figure out how to get another diagnoal matrix different from the one i got.
Click to expand...
The choice of which eigenvector was in which column was arbitrary; the eigenvalues can occur in the diagonal matrix in any order.
 
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