Chapter 4, Problem 6
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Prove Sylvester's theorem: A Hermitian matrix $\mathbf{Q}$ is positive definite if and only if all principal minora det $\mathbf{Q}_{\mathbf{n}}>0$.
If $Q$ is positive deflnite, $Q=(x, Q x)=0$ for any $x$. Let $x_m$ be the vector of the flrat $m$ elements of $x$. For those $x^0$ whoge last $m-n$ elements are zero, $\left(x_w, Q_m x_m\right)=\left(x^0, Q x^0\right)=0$. Therefore $Q_{: n}$ is positive definite, and all its eigenvalues are positive. From Problem 4.1, the determinant of any matrix equals the product of its eigenvalues, 80 det $\mathbf{Q}_{\mathrm{x}}>0$.
If det $Q_m>0$ for $m=1,2, \ldots, n$, we proceed by induction. For $n=1$, $\operatorname{det} Q=\lambda_1>0$.
Arzume now that if $\operatorname{det} \mathbf{Q}_1>0, \ldots$, det $\mathbf{Q}_{n-1}>0$, then $\boldsymbol{Q}_{n-1}$ is positive deflnite and must possess an inverse. Partition $\mathbf{Q}_n$ ns
$$
Q_a=\left(\begin{array}{c|c}
Q_{n-1} & \mathbf{q} \\
\hline \mathbf{q}^{+} & q_{n n}
\end{array}\right)=\left(\begin{array}{c|c}
\mathbf{I} & \mathbf{0} \\
\hline q+Q_{n-1}^{-1} & 1
\end{array}\right)\left(\begin{array}{c|c}
Q_{n-1} & 0 \\
\hline 0 & q_{s=}-q^{+} Q_{n-1}^{-1} \mathbf{q}
\end{array}\right)\left(\begin{array}{c|c}
\mathbf{I} & \mathbf{Q}_{n-1}^{-1} \mathbf{q} \\
\hline 0 & 1
\end{array}\right)
$$
We are also given det $\mathbf{Q}_x>0$. Then use of Problem 3.5 gives $\operatorname{det} \mathbf{Q}_0=\left(q_{\mathbf{k e}}-\mathbf{q}+\mathbf{Q}_{n-1}^{-1} \mathbf{q}\right) \operatorname{det} \mathbf{Q}_{n-1}$, so that $q_{n n}-q^{+} \mathbf{Q}_{n-1}^{-1} Q>0$. Hence
for any vector $\left(x_{n-1}^{+} \mid x_n^*\right)$. Then for any vectors $y$ defined by
$$
y=\left(\begin{array}{c|c}
\mathbf{I} & \mathbf{Q}_{\mathrm{n}-1}^{-1} \mathrm{q} \\
\hline 0 & 1
\end{array}\right)^{-1}\binom{\mathbf{x}_{\mathrm{s}-1}}{\hline x_{\mathrm{n}}}
$$
substitution into $(x, Q \mathbf{Q})$ and use of (4,28) will give $\left(y, Q_e y\right)>0$.
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Prove Sylvester's theorem: A Hermitian matrix $\mathbf{Q}$ is positive definite if and only if all principal minora det $\mathbf{Q}_{\mathbf{n}}>0$.If $Q$ is positive deflnite, $Q=(x, Q x)=0$ for any $x$. Let $x_m$ be the vector of the flrat $m$ elements of $x$. For those $x^0$ whoge last $m-n$ elements are zero, $\left(x_w, Q_m x_m\right)=\left(x^0, Q x^0\right)=0$. Therefore $Q_{: n}$ is positive definite, and all its eigenvalues are positive. From Problem 4.1, the determinant of any matrix equals the product of its eigenvalues, 80 det $\mathbf{Q}_{\mathrm{x}}>0$.If det $Q_m>0$ for $m=1,2, \ldots, n$, we proceed by induction. For $n=1$, $\operatorname{det} Q=\lambda_1>0$.Arzume now that if $\operatorname{det} \mathbf{Q}_1>0, \ldots$, det $\mathbf{Q}_{n-1}>0$, then $\boldsymbol{Q}_{n-1}$ is positive deflnite and must possess an inverse. Partition $\mathbf{Q}_n$ ns$$Q_a=\left(\begin{array}{c|c}Q_{n-1} & \mathbf{q} \\\hline \mathbf{q}^{+} & q_{n n}\end{array}\right)=\left(\begin{array}{c|c}\mathbf{I} & \mathbf{0} \\\hline q+Q_{n-1}^{-1} & 1\end{array}\right)\left(\begin{array}{c|c}Q_{n-1} & 0 \\\hline 0 & q_{s=}-q^{+} Q_{n-1}^{-1} \mathbf{q}\end{array}\right)\left(\begin{array}{c|c}\mathbf{I} & \mathbf{Q}_{n-1}^{-1} \mathbf{q} \\\hline 0 & 1\end{array}\right)$$We are also given det $\mathbf{Q}_x>0$. Then use of Problem 3.5 gives $\operatorname{det} \mathbf{Q}_0=\left(q_{\mathbf{k e}}-\mathbf{q}+\mathbf{Q}_{n-1}^{-1} \mathbf{q}\right) \operatorname{det} \mathbf{Q}_{n-1}$, so that $q_{n n}-q^{+} \mathbf{Q}_{n-1}^{-1} Q>0$. Hencefor any vector $\left(x_{n-1}^{+} \mid x_n^*\right)$. Then for any vectors $y$ defined by$$y=\left(\begin{array}{c|c}\mathbf{I} & \mathbf{Q}_{\mathrm{n}-1}^{-1} \mathrm{q} \\\hline 0 & 1\end{array}\right)^{-1}\binom{\mathbf{x}_{\mathrm{s}-1}}{\hline x_{\mathrm{n}}}$$substitution into $(x, Q \mathbf{Q})$ and use of (4,28) will give $\left(y, Q_e y\right)>0$.
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Schaum's Outline of Theory and Problems of State Space and Linear Systems
Chapter 4
โ Chapter 1
โ Chapter 2
โ Chapter 3
โ Chapter 4
โ Chapter 5
โ Chapter 6
โ Chapter 7
โ Chapter 8
โ Chapter 9
โ Chapter 10
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