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Let $\mathbf{A}\in\mathbb{R}^{N \times N}$ be symmetric positive definite. For some $1\leq k<N$, partition $$\mathbf{A}=\begin{pmatrix}\mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{12}^T & \mathbf{A}_{22}\end{pmatrix},$$ where $\mathbf{A}_{11}$ is $k\times k$ and $\mathbf{A}_{22}$ is $(N-k)\times (N-k)$.

I'm trying to show that the principal submatrices $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are also symmetric positive definite.

I've been able to show that

$$\left [ \begin{array}{cc} \mathbf{A}_{11} & \mathbf{A}_{12}\\ \mathbf{A}_{12}^T & \mathbf{A}_{22} \end{array} \right ] = \mathbf{A} = \mathbf{A}^T = \left [ \begin{array}{cc} \mathbf{A}_{11}^T & \mathbf{A}_{12}\\ \mathbf{A}_{12}^T & \mathbf{A}_{22}^T \end{array} \right ]$$

This implies that $\mathbf{A}_{11} = \mathbf{A}_{11}^T$, and $\mathbf{A}_{22} = \mathbf{A}_{22}^T$. Thus, $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are symmetric.

I'm struggling to show that $\mathbf{A}$ is positive definite. Does anyone have any ideas?

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  • $\begingroup$ You can self-study the term of leading principal submatrices, really useful because it points out A11 and A22 are symmetric positive definite by definition. Good luck on you! $\endgroup$
    – Zixiao Xu
    Commented Jul 17, 2018 at 21:06

1 Answer 1

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You know that positive definiteness means $v^TAv>0$ for all nonzero vectors $v$. Choose $v$ to be vectors with non-zero entries only at the first $k$ positions. (And then do the opposite).

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    $\begingroup$ well that is delightfully simple. Thank you. $\endgroup$
    – Drew
    Commented Apr 6, 2015 at 0:58

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