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For a positive definite matrix $n\times n$ prove that for its any $(n-1)\times (n-1)$ submatrix all eigen-values lie within the range of the eigenvalues of the original matrix.

Please provide some intuition as of why this is the case.

Juniven Acapulco
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Dnaiel
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2 Answers2

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It seems to me that this is not true. The identity matrix

$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

has eigenvalue $1$. However the submatrix

$$\left( \begin{array}{cc} 0 & 1 \\ 0 & 0\end{array} \right)$$

is nilpotent and so only has eigenvalue $0$.

Matt
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As the other answer shows, this is not true (and the eigenvalues of an arbitrary submatrix of a positive definite matrix are not necessarily real in the first place). The correct statement should be that if $A$ is a positive definite matrix of size $n$ and $B$ is a principal submatrix of $A$ of size $n-1$, then $\lambda_\min(A)\le\lambda_\min(B)\le\lambda_\max(B)\le\lambda_\max(A)$. This is simply because the maximum eigenvalue of a Hermitian matrix $H$ is given by $\max_{\|x\|=1}x^\ast Hx$, and $\{y^\ast Ky: y\in\mathbb C^{n-1},\,\|y\|=1\}$ is a subset of $\{x^\ast Hx: x\in\mathbb C^n,\,\|x\|=1\}$ whenever $K$ is a principal submatrix of $H$.

We actually have a stronger result, known as Cauchy's interlacing inequality, which says that if $A$ is a Hermitian matrix of size $n$ and $B$ is a principal submatrix of $A$ of size $n-1$, then $\lambda_j(A)\le\lambda_j(B)\le\lambda_{j+1}(A)$ for each $j$, when the eigenvalues of each matrix are arranged in ascending order. This is a consequence of Courant-Fischer minimax principle.

user1551
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  • Is this answer directed at Hermitian matricies or merely positive definite but not necessarily symmetric (or Hermitian) ones? – BadAtLaTeX May 19 '21 at 08:39
  • @gr4nt3d The answer assumes that a real positive definite matrix is symmetric. (No analogous assumption is needed for complex matrices, because complex positive definite matrices are necessarily Hermitian.) – user1551 May 19 '21 at 08:48
  • Thank you, this led me to believe it could be more general. Can we estimate the eigenvalues of leading principal submatrices also for matrices that only have $\Re (x^T A x) > 0$ (ie. as positive definiteness if defined there)? – BadAtLaTeX May 19 '21 at 09:37
  • @gr4nt3d I don't think there are similar inequalities for non-Hermitian matrices. E.g. the three eigenvalues of $A=\pmatrix{3&5&0\ -3&4&-2\ -2&3&1}$ are $2.4667$ and $2.7667\pm4.3240i$ and the three eigenvalues of its symmetric part are $0.3787,,3.0000$ and $4.6213$. All these six eigenvalues have magnitudes $\le5.1333$. However, the eigenvalues of the leading principal $2\times2$ submatrix of $A$ are $3.5000\pm3.8406i$, whose magnitudes (namely, $5.1962$) are strictly greater than $5.1333$. – user1551 May 19 '21 at 11:06