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I am modeling the effect of neural activity on synaptic strength. My question, though, is mathematical.

I have the following differential equation:

$ \tau_{W} \frac{d\mathbf{W}}{dt}=\mathbf{KWQ}-\alpha\mathbf{R_{\infty}}\mathbf{W} \quad\mathbf{Q}=\left\langle \vec{u}\vec{u}\right\rangle ,\,\mathbf{R_{\infty}}=\left\langle \vec{v_{\infty}}\vec{v_{\infty}}\right\rangle $

$\tau_{W}$ and $\alpha$ are constants. $\mathbf{W}$ is $n \times m$. $\mathbf{R_{\infty}} $ is $n \times n$. $\mathbf{Q}$ is $m \times m$. $\mathbf{K}$ is $ n \times n$.

  1. If $\mathbf{Q}$ and $\mathbf{R_{\infty}}$ come from the outer products of vectors, how do I know that they are invertible?

  2. How would I find the value for $\mathbf{W}$ for which the derivative vanishes? Especially if neither $\mathbf{Q}$ nor $\mathbf{R_{\infty}}$ are invertible.

  3. How would I find $\mathbf{X}$ such that $ \mathbf{X}\mathbf{W} = \mathbf{Q}$ if neither $\mathbf{Q}$ nor $\mathbf{W}$ are invertible?
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1 Answer 1

up vote 4 down vote accepted
  1. The outer product of two vectors is a rank 1 matrix, and so it's never invertible (as long as the matrices are more than 1 dimensional.)

  2. Your differential equation matrix isn't square, so it doesn't have a determinant. The matrices formed with outer products have determinant $0$.

  3. Your third question needs some clarification. You mean you need a matrix $X$ such that $XW=Q$? This is possible but it really depends on a lot more information than you have given.

"Neither is antisymmetric so I know that they may be invertible."

Not sure what this means either, because invertibility does not bear much relationship to antisymmetry (for example $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ has inverse $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, and of course there are noninvertible antisymmetric matrices).

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I thought the outer product of a vector isn't invertible- it wouldn't make sense. I wanted to be sure, though. –  mac389 Sep 1 '12 at 20:53
    
What more information is needed for the third point? (Neither matrix is invertible. But, I could consider the case where all matrices are square.) Thanks for your help. –  mac389 Sep 1 '12 at 20:55
1  
@mac389 You can find an $X$ iff the rows of $Q$ are in the rowspace of $W$. So, that requires some knowledge of the matrices. –  rschwieb Sep 1 '12 at 22:06
    
Let's assume that all matrices have real entries, that $n = m$ and that the rank of $\mathbf{W}$ is not less than the rank of $\mathbf{Q}$. Is that enough to place $\mathbf{Q}$ in the rowspace of $\mathbf{W}$? –  mac389 Sep 2 '12 at 15:30
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@mac389 You will not be able to decide anything with just a rank argument. Unless $W$ has rank 0, $W$ is always going to have higher rank than (the rank 1 matrix) $Q$. Since the rowspace of $Q$ is one dimensional, things are easier, because any nonzero vector in it generates it. So if one of those vectors is in the rowspace of $W$, then you can find $X$. –  rschwieb Sep 2 '12 at 17:28

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