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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Which of the following are eigenpairs (λ,x) of the 2×2 zero matrix: \[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]=\]where \[\chi \neq0\]

  2. anonymous
    • one year ago
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    \[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\chi= \lambda \chi\]where \[\chi \neq0\]

  3. anonymous
    • one year ago
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    A. \[(1,\left(\begin{matrix}0 \\ 0\end{matrix}\right))\]B.\[(0,\left(\begin{matrix}1 \\ 0\end{matrix}\right))\]C\[(0,\left(\begin{matrix}0 \\ 1\end{matrix}\right))\]D\[(0,\left(\begin{matrix}-1 \\ 1\end{matrix}\right))\]E\[(0,\left(\begin{matrix}1 \\ 1\end{matrix}\right))\]F\[(0,\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

  4. ganeshie8
    • one year ago
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    \[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\chi= \lambda \chi\]where \[\chi \neq0\] Notice that the left hand side evaluates to zero vector no matter what the vector \(\chi \) is, so it follows that \(\lambda = 0\)

  5. anonymous
    • one year ago
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    yea

  6. anonymous
    • one year ago
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    i know the answers actually so just for fun :)

  7. anonymous
    • one year ago
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    pick the options which are eigenpairs

  8. ganeshie8
    • one year ago
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    By definition, eigenvector cannot be zero, so the last option can be eliminated

  9. anonymous
    • one year ago
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    you mean first and last, both are 0

  10. ganeshie8
    • one year ago
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    Ahh right, first and last options are eliminated

  11. ganeshie8
    • one year ago
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    remaining all options look good to me!

  12. anonymous
    • one year ago
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    That's correct!

  13. ganeshie8
    • one year ago
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    yaay!

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