anonymous
  • anonymous
For fun
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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 \[\chi \neq0\]
anonymous
  • anonymous
\[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\chi= \lambda \chi\]where \[\chi \neq0\]
anonymous
  • anonymous
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\[(0,\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

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ganeshie8
  • ganeshie8
\[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\chi= \lambda \chi\]where \[\chi \neq0\] Notice that the left hand side evaluates to zero vector no matter what the vector \(\chi \) is, so it follows that \(\lambda = 0\)
anonymous
  • anonymous
yea
anonymous
  • anonymous
i know the answers actually so just for fun :)
anonymous
  • anonymous
pick the options which are eigenpairs
ganeshie8
  • ganeshie8
By definition, eigenvector cannot be zero, so the last option can be eliminated
anonymous
  • anonymous
you mean first and last, both are 0
ganeshie8
  • ganeshie8
Ahh right, first and last options are eliminated
ganeshie8
  • ganeshie8
remaining all options look good to me!
anonymous
  • anonymous
That's correct!
ganeshie8
  • ganeshie8
yaay!

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