Finding the Inverse of a Matrix Mapping on a Linear Subspace

[Kreizhn]

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}.

Homework Equations

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'.

 

[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.

 

[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?

 

[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?

 

[Kreizhn]

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

 

[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?

 

[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.

 

[Kreizhn]

Sorry, I wrote that backwards. It should be

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

 

[Dick]

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.

 

[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.

 

[Dick]

Sorry, I really don't know that formalism.