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Mathematics 28 Online
OpenStudy (anonymous):

For fun

OpenStudy (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 \[\chi \neq0\]

OpenStudy (anonymous):

\[\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\chi= \lambda \chi\]where \[\chi \neq0\]

OpenStudy (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\[(0,\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

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 \(\lambda = 0\)

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

i know the answers actually so just for fun :)

OpenStudy (anonymous):

pick the options which are eigenpairs

ganeshie8 (ganeshie8):

By definition, eigenvector cannot be zero, so the last option can be eliminated

OpenStudy (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!

OpenStudy (anonymous):

That's correct!

ganeshie8 (ganeshie8):

yaay!

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