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I read here that if a matrix is positive definite then it's invertible.

If I take some vector nx1 and multiply it by another one that's 1xn then I get an nxn matrix. But It doesn't seem to be invertible according to a short sympy script I wrote.

a1, a2, a3, a4, a5, a6 = sp.symbols('a1 a2 a3 a4 a5 a6')
va = sp.Matrix([a1, a2, a3, a4, a5, a6])
pos_def_matrix = va*va.T
pos_def_matrix.inv()

which gives me the error.

Matrix det == 0; not invertible.

So am I misunderstanding something?

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The operation performed in a paper I’m trying to understand is this

$w_t = E_t[R^e_{t+1} {R^e_{t+1}}^T]^{-1} E_t[R^e_{t+1}]$

But it seems we showed $R^e_{t+1} {R^e_{t+1}}^T$ is not possible.

financial_physician
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    The matrix you get with your operation is only positive semidefinite. – Michal Adamaszek Jun 11 '20 at 05:43
  • Also note that an outer product of 2 vectors has rank at most 2, so your matrix is rank deficient and obviously not invertible – Ben10 Jun 11 '20 at 05:53
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    if you do SVD on positive definite matrix, you will get all singular values positive. If you do SVD on positive semi-definite, you will possible get some singular values zero. if there any zero singular value, it means matrix loose rank, thus the matrix is not invertible. You have rank 1 matrix, you will get only 1 nonzero singular value, others will be zero. – Lee Jun 11 '20 at 05:56
  • Thanks for the answer! I ask because I’m confused about part of a paper I’m reading because it doesn’t seem like the operation performed is possible. I’ve appended it onto the question but I’m happy to make a new post if that’d be more appropriate. – financial_physician Jun 12 '20 at 05:42

1 Answers1

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$uu^*$, where $u\in\mathbb{C}^n$ is not positive definite. Although $x^* uu^* x\geq0$ for all $x\in \mathbb{C}^n$, there are vectors $y\neq0$ for which $y^*uu^*y=0$.

Take for instance any $y\in\{x\in \mathbb{C}^n: xu^*=0\}\setminus\{0\}$.

For example, for $u=[1\,-1\,1\, -1]^\top$ and $y=[1\,1\,1\, 1]^\top$ you have $y^*uu^*y=0$.

Mittens
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