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Let $A$ be a positive definite symmetric matrix. Show that there exists an invertible matrix $B$ such that $A=B^TB$. [Hint: Use the Spectral Theorem to write $A = QDQ^T$. Then show that D can be factored as $C^TC$ for some invertible matrix $C$.]

I can't seem to get to the correct answer. I'm not entirely sure what is meant by the last line of the hint too. Could anyone please help me out?

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    $\begingroup$ Consider the diagonal matrix whose diagonal elements are the square roots of the diagonal elements of $\mathbf D$. $\endgroup$ Commented Apr 9, 2013 at 16:16
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    $\begingroup$ $B=Q^T D^{\frac{1}{2}}$ $\endgroup$
    – Learner
    Commented Apr 9, 2013 at 16:17
  • $\begingroup$ @Learner, how did you get to this answer? $\endgroup$
    – dreamer
    Commented Apr 9, 2013 at 16:18
  • $\begingroup$ cruise: start from my hint. $\endgroup$ Commented Apr 9, 2013 at 16:19
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    $\begingroup$ That isn't it. You have the decomposition from the "spectral theorem"; replace $\mathbf D$ with $\sqrt{\mathbf D}\sqrt{\mathbf D}$. Remember that $(\mathbf A\mathbf B)^\top=\mathbf B^\top\mathbf A^\top$... $\endgroup$ Commented Apr 9, 2013 at 16:23

3 Answers 3

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$A$ is symmetric, so it can be written as $A=QDQ^T$ (Eigenvalue decomposition of a symmetric matrix), where $D$ is a diagonal matrix with real elements on main diagonal.

$A$ is PSD, so elements on main diagonal of $D$ are nonnegative, so they have square root. Let $D= D_1D_1^T$, where $D_1$ is also diagonal.

You can write $C^T= QD_1$. You will have $A=C^TC$.

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  • $\begingroup$ Thanks a lot for your help, that's excactly what I needed :). $\endgroup$
    – dreamer
    Commented Apr 9, 2013 at 16:26
  • $\begingroup$ You are welcome. :) $\endgroup$
    – user25004
    Commented Apr 9, 2013 at 17:49
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Do you know Cholesky Decomposition? You can even chose $B$ as a triangular matrix.

If you don't like an abstract proof you just can compute it, for example $$\begin{pmatrix} l_{11} & 0 \\ l_{21} & l_{22} \\ \end{pmatrix}\cdot \begin{pmatrix} l_{11} & l_{21}\\ 0 & l_{22} \\ \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12}\\ a_{12} & a_{22} \\ \end{pmatrix}$$ when you chose $l_{ii}>0$ you will always have $\frac{n(n+1)}{2}$ equations and unknowns, hence a unique solution exists.

To get the Cholesky decomposition from the eigenvalue decomposition we use an lu dcomposition, as $Q$ is invertible we know that $Q=L\cdot U$ for some normed lower triangular matrix $L$, and some upper triangular matrix $U$. So $$A= Q^T D Q = U^T L^T D L U $$ As $U^T$ and $U$ are invertible we have $$U^{T^{-1}} A U^{-1}=L^T D L $$ We see that $$U^{T^{-1}} A U^{-1}$$ is symmetric and positiv definit iff $A$ is symmetric and positiv definit, and we can get any positive definite matrix, hence we get the cholesky decomposition.

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  • $\begingroup$ No, I don't. Could you show me how that would be done? $\endgroup$
    – dreamer
    Commented Apr 9, 2013 at 16:18
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    $\begingroup$ I'd sure like to see somebody derive the Cholesky triangle from an eigendecomposition... $\endgroup$ Commented Apr 9, 2013 at 16:20
  • $\begingroup$ @J.M. oh from an eigendecomposition. that is not so obvious, but let me try it $\endgroup$ Commented Apr 9, 2013 at 16:34
  • $\begingroup$ Thats a whole other way of thinking. Thanks for the insight :). $\endgroup$
    – dreamer
    Commented Apr 9, 2013 at 16:35
  • $\begingroup$ @J.M. added it, hope everything works $\endgroup$ Commented Apr 9, 2013 at 17:01
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Let $$ D=\mbox{diag}(\lambda_1, \ldots,\lambda_n) \quad\mbox{and}\quad U=\mbox{diag}(\sqrt{\lambda_1}, \ldots,\sqrt{\lambda_n}). $$ So check it yourself $B=QU$.

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