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## Homework Statement

Let's say I'm given two vectors

[tex] v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \in \mathbb R^4[/tex].

Let W be subspace spanned by these vectors, and define [itex] G = v_1 v_1^T + v_2 v_2^T [/itex] a matrix mapping [itex]\mathbb R^4 \to \mathbb R^4 [/itex]. Find [itex] G' [/itex] such that [itex] \left. G' = G^{-1} \right|_{W} [/itex].

## Homework Equations

## The Attempt at a Solution

Since [itex] v_1, v_2 [/itex] are linearly independent, the dimension of W is 2. Furthermore, since G is composed of these vectors, we can be guaranteed that that an inverse exists on W. Let W' be the image of W under G. That is, since [itex] v_1, v_2 [/itex] generate W, then [itex] G v_1, Gv_2 [/itex] should generate W'.

I've computed that

[tex] G = \begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, Gv_1 = \begin{pmatrix} 2\\ 3\\ 1 \\0 \end{pmatrix}, Gv_2 = \begin{pmatrix} 1 \\ 3 \\ 2 \\ 0 \end{pmatrix} [/tex].

Now maybe it's because it's been so long since I did any linear algebra, but I can't for the life of me figure out how to "extract" the restriction of G to W. Given this information, it should then be simple to construct the inverse and hence make G'.