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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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2 Answers 2

As the map given by $x\mapsto Tx$ is linear, it suffices to prove that the image of the spanning vectors is again in the span of the two vectors. This is done by the following computation:

\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\0\\1\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\3\cdot 1+(-1)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\(-2)\cdot 1+2\cdot 1+2\cdot 0+0\cdot 1\\1\cdot 1+(-1)\cdot 1+0\cdot 0+1\cdot 1\end{bmatrix}=\begin{bmatrix}1\\1\\0\\1\end{bmatrix}\\ &=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+0\cdot\begin{bmatrix}1\\0\\2\\0\end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\0\\2\\0\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\3\cdot 1+(-1)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\(-2)\cdot 1+2\cdot 0+2\cdot 2+0\cdot 0\\1\cdot 1+(-1)\cdot 0+0\cdot 2+1\cdot 0\end{bmatrix}=\begin{bmatrix}2\\1\\2\\1\end{bmatrix}\\ &=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+1\cdot \begin{bmatrix}1\\0\\2\\0\end{bmatrix} \end{align*}

Another way to see this (but unnecessary complicated is to see that the spanned set is the kernel of $(T-I)^2$ and as such is $T$-invariant. (This is suggested by the theory of Jordan normal form.

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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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