square origin of symmetrical matrix and also transposition

I have a symmetrical matrix A. How do I calculate a matrix B such that $B^tB=A$ where $B^t$ is the transpose of $B$. I can not identify if this goes to all pertaining to the square origin of $A$.

I've experienced wikimedia web links of square origin of a matrix.

2019-05-09 11:36:20
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Answers: 2

What you evidently desire below is the Cholesky disintegration, which variables a matrix A right into $BB^T$ where $B$ is a triangular matrix. Nonetheless, this just functions if your matrix declares precise.

2019-05-10 09:49:16

As J. M. claims, you require your matrix $A$ to be favorable precise. Given that $A$, being symmetrical, is constantly diagonalizable, this coincides as claiming that it has non - adverse eigenvalues. If this holds true, you can adjust alex's comment virtually essentially for the actual instance : as we've claimed, $A$ is diagonalizable, yet, additionally, there exists an orthonormal base of eigenvectors of $A$. That is, there is an invertible matrix $S$ and also an angled matrix $D$ such that

$$ D = SAS^t , \quad \text{with} \quad SS^t = I \ . $$


$$ D = \mathrm{diag} (\lambda_1, \lambda_2, \dots , \lambda_n) \ , $$

is an angled matrix and also has just non - adverse eigenvalues $\lambda_i$, you can take its square origin

$$ \sqrt{D} = \mathrm{diag} (\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots , \sqrt{\lambda_n} ) \ , $$

and afterwards, on one hand, you have :

$$ \left( S^t \sqrt{D} S \right)^2 = \left( S^t \sqrt{D} S\right) \left(S^t \sqrt{D} S \right) = S^t \left( \sqrt{D}\right)^2 S = S^t D S = A \ . $$

On the various other hand, $S^t \sqrt{D} S$ is a symmetrical matrix also :

$$ \left( S^t \sqrt{D} S \right)^t = S^t (\sqrt{D})^t S^{tt} = S^t \sqrt{D^t} S = S^t \sqrt{D} S \ , $$

so you have your $B = S^t \sqrt{D} S$ such that $B^t B = A$.

2019-05-10 07:57:55