# Homework Help: Matrices on Linear Subspaces

1. Sep 29, 2009

### Kreizhn

1. The problem statement, all variables and given/known data
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 $\left. G' = G^{-1} \right|_{W}$.

2. Relevant equations

3. 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'.

2. Sep 29, 2009

### Dick

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.

3. Sep 30, 2009

### Kreizhn

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?

4. Sep 30, 2009

### Dick

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?

5. Sep 30, 2009

### Kreizhn

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

6. Sep 30, 2009

### Kreizhn

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?

7. Sep 30, 2009

### Dick

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.

8. Sep 30, 2009

### Kreizhn

Sorry, I wrote that backwards. It should be

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

9. Sep 30, 2009

### Dick

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

10. Sep 30, 2009

### Kreizhn

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.

11. Sep 30, 2009

### Dick

Sorry, I really don't know that formalism.