I have a new one, and I am not sure about a few things. I hoped you might help me in understanding them.
For matrix $$ A=\left[ \begin{matrix} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{matrix} \right] $$ Using Sylvester's law of inertia and $LDL^T$ decomposition find number of eigenvalues of matrix $A$ in the interval $[-2, 3>$. Determine and sketch the location of the eigenvalues using Gerschgorin theorem. Are you able to refine the results of the Gerschgorin theorem considering properties of the matrix $A$. Explain.
Sylvesters law of inertia (from what I understand of it, I am by no means close to an expert), claims that signs of the diagonal elements of the $D$ matrix will be identical to the signs of the eigenvalues of the matrix $A$.
First, I introduced $\lambda$ as eigenvalue.
$$ A=A-\lambda I=\left[\begin{matrix} 4-\lambda & 4 & 0 \\ 4 & 6-\lambda & 2 \\ 0 & 2 & 1-\lambda \\ \end{matrix} \right] $$
Second, I performed partial pivoting, in order to perform $LU$ decomposition of the matrix $A$.
$$ PA=LU $$
$$ \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{matrix} \right] \left[ \begin{matrix} 4-\lambda & 4 & 0 \\ 4 & 6-\lambda & 2 \\ 0 & 2 & 1-\lambda \\ \end{matrix} \right] = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1-\frac14 \lambda & -1+\frac54 \lambda-\frac18 \lambda^2 & 1 \end{matrix} \right] \left[ \begin{matrix} 4 & 6-\lambda & 2 \\ 0 & 2 & 1-\lambda \\ 0 & 0 & -\frac18 \lambda^3+{11\over8}\lambda^2-{14\over8}\lambda-1 \end{matrix} \right] $$ From here, I used diagonal elements of the matrix $U$ to create matrix $D$, and matrix $L$ stays the same. $$\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -\frac18 \lambda^3+{11\over8}\lambda^2-{14\over8}\lambda-1 \end{matrix} \right] $$ Of course, I immediately checked if $A=LDL^T$ is true, and it held true.
From here, I recognized one third degree polynomial, and found null points of that polynomial: $$-\frac18 \lambda^3+{11\over8}\lambda^2-{14\over8}\lambda-1=0 $$ Solving this equation yields the eigenvalues of $A$ $$\lambda_1 = -0.424429$$ $$\lambda_2 = 2$$ $$\lambda_3 = 9.42443$$ Finally, I can say that two of the eigenvalues lie in the defined interval$[-2,3>$, but I am not certain if I was supposed to actually calculate the eigenvalues and compare the to the interval, or there exists a check that I can perform using Sylvester's law of inertia and $LDL^T$ decomposition.
Second part of the question is to use Gerschgorin theorem to determine and sketch the location of the eigenvalues. I created three disks, with centers on real axis, coordinates 4, 6, and 1, and assigned them radii 4, 6 and 2, , as can be seen on the picture. Red circles represent Gerschgorin disks, and black dots represent calculated eigenvalues.
I do not know how to refine the results. The matrix $A$ is hermitian, symmetric and indefinite, from what I can see, but I do not know how can this help me.
Thank you in advance.
Here is the outcome I commented about. You should repeat this task for $H+2I$ and $H-3I.$ I think starting from there is necessary as $P^TP$ is likely not diagonal.
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 1 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
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Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 1 & 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 1 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 4 & 0 \\ 4 & 6 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
Out comes for $H+2I, H-2I, H-3I.$ The middle one confirms that $2$ is an exact eigenvalue of $h$ itself
H+2I
$$ P^T (H+2I) P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 2 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 3 }{ 8 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 6 & 4 & 0 \\ 4 & 8 & 2 \\ 0 & 2 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 2 }{ 3 } & \frac{ 1 }{ 4 } \\ 0 & 1 & - \frac{ 3 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & \frac{ 16 }{ 3 } & 0 \\ 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) $$
H-2I
$$ P^T (H-2I) P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 1 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 4 & 0 \\ 4 & 4 & 2 \\ 0 & 2 & - 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 1 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
H-3I
$$ P^T (H-3I) P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 1 & 0 \\ - \frac{ 8 }{ 13 } & \frac{ 2 }{ 13 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & 0 \\ 4 & 3 & 2 \\ 0 & 2 & - 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & - \frac{ 8 }{ 13 } \\ 0 & 1 & \frac{ 2 }{ 13 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 13 & 0 \\ 0 & 0 & - \frac{ 22 }{ 13 } \\ \end{array} \right) $$