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show that the span of \begin{bmatrix}1\\1\\0\\1\end{bmatrix} and \begin{bmatrix}1\\0\\2\\0\end{bmatrix} is a T invariant subspace of the linear map given by

\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}


i tried to take some general vector in the span and multiply it by the matrix in the hope of getting something that was clearly a linear combination of my two original spanning vectors, but this did not work, that is, the vector was clearly not in the span.

so how am i meant to show T-invariance?

note: apologies for the formatting, my first time

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1 Answer

Apply the matrix (this is what you call $T$, no?) to both given vectors, call them $a$ and $b$, say (applying means matrix multiplication), and try to find $\lambda,\kappa$ numbers such that the result is $\lambda a+\kappa b$, in both cases.

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yes i did something similar. I took a vector v in the span, and applied T to it. since T is linear, and the vector v can be represented as v' + v'' (where v' is an element of the span of <1, 1, 0, 1>, and v'' is an element of the span of <1, 0, 2, >) i can break it up. so Tv = T(v' + v'') = Tv' + Tv''. Then I can apply the matrix representing T to each which will return something in the span, so the span is T invariant. I think this is sound – Michael Davis Oct 1 '12 at 13:01

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