# Every positive definite matrix can be written as $B^TB$ for some invertible $B$

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?

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

$A$ 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 $A=C^TC$.

• Thanks a lot for your help, that's excactly what I needed :). Commented Apr 9, 2013 at 16:26
• You are welcome. :) Commented Apr 9, 2013 at 17:49

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.

• No, I don't. Could you show me how that would be done? Commented Apr 9, 2013 at 16:18
• I'd sure like to see somebody derive the Cholesky triangle from an eigendecomposition... Commented Apr 9, 2013 at 16:20
• @J.M. oh from an eigendecomposition. that is not so obvious, but let me try it Commented Apr 9, 2013 at 16:34
• Thats a whole other way of thinking. Thanks for the insight :). Commented Apr 9, 2013 at 16:35
• @J.M. added it, hope everything works Commented Apr 9, 2013 at 17:01

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 $B=QU$.