# Matrices on Linear Subspaces

## Homework Statement

Let's say I'm given two vectors
$$v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \in \mathbb R^4$$.
Let W be subspace spanned by these vectors, and define $G = v_1 v_1^T + v_2 v_2^T$ a matrix mapping $\mathbb R^4 \to \mathbb R^4$. Find $G'$ such that $\left. G' = G^{-1} \right|_{W}$.

## The Attempt at a Solution

Since $v_1, v_2$ 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 $v_1, v_2$ generate W, then $G v_1, Gv_2$ should generate W'.

I've computed that
$$G = \begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, Gv_1 = \begin{pmatrix} 2\\ 3\\ 1 \\0 \end{pmatrix}, Gv_2 = \begin{pmatrix} 1 \\ 3 \\ 2 \\ 0 \end{pmatrix}$$.

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'.

## Answers and Replies

Dick
Homework Helper
Just express Gv1 and Gv2 as linear combinations of v1 and v2. At some point you should figure out you really are doing this the long way around. Gv1=v1(v1)^Tv1+v2(v2)^Tv1. I see dot products in there.

So what I think you're saying is that I have

$$v_1' = G v_1 = 2 v_1 + v_2$$
$$v_2' = G v_2 = v_1 + 2v_2$$

So I should associate $v_1' = (2,1), v_2' = (1,2)$. In this case then, if G is determined by its action on the basis states $v_1, v_2$ then G acting on W can be written as

$$G_W = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$$

Which in turn has inverse

$$G_W^{-1} = \frac13 \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$$

But then how do I make this four-dimensional again?

Dick
Homework Helper
You can make it four dimensional any way you want. You know how (G_W)^(-1) acts on v1 and v2. You'll have to pick v3 and v4 so that {v1,v2,v3,v4} is a basis for R^4. Now you can define (G_W)^(-1)(v3) and (G_W)^(-1)(v4) to be anything you want. You know you can't extend it to be G^(-1) on R^4, since G isn't invertible, right?

Is there no loss of information or structure defined by the original G by choosing v3 and v4 as such?

Okay, let me see if I understand this then:

$$G_W^{-1} v_1 = \frac13 \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 2 v_1 -v_2$$
$$G_W^{-1} v_2 = \frac13 \begin{pmatrix} -1 \\ 2 \end{pmatrix} = -v_1 + 2 v_2$$

Thus we use our original definitions of
$$v_1 = \begin{pmatrix} 1 \\ 1\\ 0 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$
To find that the first two columns of G' should be

$$\frac13 \begin{pmatrix} 2 & - 1 \\ 1 & 1 \\ -1 & 2 \\ 0 & 0 \end{pmatrix}$$

But now you say that $\{ v_1, v_2, v_3, v_4 \}$ must form a basis for $\mathbb R^4$, but that I can choose $G_W^{-1} v_3 , G_W^{-1} v_4$ to be anything I want. This seems contradictory, since I could certainly choose these such that $v_3, v_4$ did not help form a basis. So should I choose them such that they form a basis? Or choose them arbitrarily?

Dick
Homework Helper
I'm not really sure what you are trying to do. $\left. G' = G^{-1} \right|_{W}$, doesn't make much sense because $G^{-1}$ doesn't exist. G maps R^4->W. ${G \right|_W}^{-1}$ exists, but it doesn't have any unique extension to R^4.

Sorry, I wrote that backwards. It should be

$$\left. G_W^{-1} = G' \right|_{W}$$

Dick
Homework Helper
Sorry, I wrote that backwards. It should be

$$\left. G_W^{-1} = G' \right|_{W}$$

I still don't see how that would be unique if you want G' to be 4x4.

Perhaps some perspective will help. I have a series of density matrices $\{ \rho_i \}$ and my goal is to construct a positive operator valued measure from these states to form a measure on $(\mathbb C^3)^{\otimes 3} = \mathbb C^9$. This is done by defining $M = \sum_i p_i \rho_i$ where $p_i$ are the associated density matrix probabilities. Then the positive operator valued measure is defined as

$$E_i = p_i M^{-\frac12} \rho_i M^{-\frac12}$$

where $\left( M^{-\frac12} \right)^2 M$ is the projection operator onto the image of M.

Dick