\(\text{ Normālai matricai izpildās nosacījums: } A^* A = AA^* \text{, kur } A^* \text{ir konjugēti transponēta matrica.}\)
\[\left(\begin{matrix} 1 & 0 \\ 0 & 3 + 2i\end{matrix}\right) \cdot \left(\begin{matrix} 1 & 0 \\ 0 & 3 - 2i\end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 0 &13\end{matrix}\right)\]
\(\text{Tātad šīs matrica ir normāla, taču nav Ermita jo } a_{22} \neq \overline{a}_{22}.\)
\(z = \left[ \begin{matrix}a \\ b \\ c \end{matrix} \right].\)
\[z^T \cdot M = \left[ \begin{matrix} 2a + c & 3b & a +2c \end{matrix}\right] \\ z^T \cdot M \cdot z = \left[ \begin{matrix} 2a + c & 3b & a +3c \end{matrix}\right] \cdot \left[ \begin{matrix} a \\ b \\ c\end{matrix}\right] = \left[ \begin{matrix}2a^2 + ac + 3b^2 + ac + 2c^2 \end{matrix}\right] = [3b^2 + 2(a^2 + ac + c^2)] = \]
\[a^2 + ac +c^2 = (a + \frac{1}{2}c)^2 + \frac{3}{4}c^2\]
Tā kā visiem \(a, b, c \in R \ z^*M z \geq 0,\) tad šī Hermita matrica ir pozitīvi semi - definēta. 3) \[z^T \cdot M \cdot z = \left[\begin{matrix} a +2b & 2a +b & 3c\end{matrix} \right] \cdot \left[\begin{matrix}a \\ b \\ c\end{matrix}\right] = \left[a^2 +2ab + 2ab +b^2 +3c^2 \right] = \left[a^2 + (2ab)^2 + 3c^2\right]\]
Šī matrica arī ir pozitīvi semi - definēta. 4) \[A^*A = \left(\begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & 1 \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & 1 \\ \frac{1}{\sqrt{18}}& 0 & -\frac{2}{\sqrt{6}}\end{matrix}\right) \cdot \left(\begin{matrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{18}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 1 & 1 & -\frac{2}{\sqrt{6}}\end{matrix}\right) \]
AA <- matrix(c(1/sqrt(3), 1/sqrt(3), 1/sqrt(18), 1/sqrt(2), -1/sqrt(2), 0, 1, 1, -2/sqrt(6)),
ncol = 3, nrow = 3)
A <- t(AA)
AA %*% A## [,1] [,2] [,3]
## [1,] 1.8333333 0.8333333 -0.6804138
## [2,] 0.8333333 1.8333333 -0.6804138
## [3,] -0.6804138 -0.6804138 0.7222222A %*% AA## [,1] [,2] [,3]
## [1,] 7.222222e-01 -1.779619e-17 0.9622504
## [2,] -1.779619e-17 1.000000e+00 0.0000000
## [3,] 9.622504e-01 0.000000e+00 2.6666667Šī matrica nav normāla.
Matricu klase, kas ir vienlaicīgi Ermita un unitāras raksturjoums ir, ka pašas matricas un to inversās matricas ir vienādas. Šīs klases matricu sasaiste ar ortogonāliem projektoriem ir tāda, ka šīm matricām ortogonālie projektori ir Ermita matricas ar eigenvērtībām 1 vai 0.
\[A = \left(\begin{matrix} 1 & \frac{1}{\sqrt{2}} & 0 \\ 0 &-\frac{1}{\sqrt{2}} & 1\end{matrix}\right)\]
\[\begin{cases} |\phi_1⟩ = (1/\sqrt(2))(|0⟩ - |1⟩) & & \text{ar varbūtību } \frac{3}{4};\\ |\phi_2⟩ = (1/\sqrt{2})(|0⟩ + |1⟩) & & \text{ar varbūtību } \frac{3}{4}; \end{cases} \]
\(x_{12} = z_1 + iz_2, \ \ z_1, z_2 \in R\) \(x_{21} = z_1 - iz_2\) \(y_{12} = z_3 + iz_4, \ \ z_3, z_4 \in R\) \(x_{21} = z_3 - iz_4\) \[A = \left(\begin{matrix} a & z_1 + iz_2 \\ z_1 - iz_2 & b \end{matrix}\right) \ \ B = \left(\begin{matrix} c & z_3 + iz_4\\ z_3 - iz_4& d \end{matrix}\right)\]
\[AB = \left(\begin{matrix}ac+ (z_1 + iz_2)(z_3 - iz_4) & a(z_3 + iz_4) \\ c(z_1 - iz_2) + b(z_3 - iz_4) & (z1 - iz_2)(z_3 + iz_4) + bd \end{matrix}\right)\] \[tr(AB) = ac + bd + z_1z_3 -iz_1z_4 + iz_2z_3 +z_2z_4 + z_1z_3 +iz_1z_4 -iz_2z_3 + z_2z_4 = ac + bc + z^2_1z^2_3 + z^2_1z^2_4 +z^2_2z^2_3 + z^2_2z^2_4\] Visi saskaitāmie ir reāli skaitļi => \(tr(AB)\in R\) Tā kā \(A\) un \(B\) ir ermita matricas, tad \(x_{ij} = y_{ji}.\) No tā izriet, ka \[tr(AB) = x_{11}y_{11} +x_{22}y_{22} + x_{33}y_{33} + x_{12}^2 + x_{13}^2 + x_{21}^2 + x_{23}^2 +x_{31}^2+x_{32}^2\]
Pozitīvi pus-definētai matricai \(A\) eksistē matrica C, tāda, ka: \[A = C^TC\] Tāpat arī \(B=D^TD.\) Tad \[AB = (C^TC)(D^TD) = C^TCD^TD.\]
Izmanto trases ciklisko īpašību: \(tr(C^TCD^TD) = tr(DC^TCD^T).\) Apzīmē \(M = CD^T.\) Tad \[DC^TCD^T = (CD^T)^T (CD^T) = M^TM.\] Līdz ar to \(tr(AB) = tr(M^TM).\) Tagad, ja M ir jebkura matrica, tad \(tr(M^TM) = \sum_{ij} M^2_{ij} \geq 0,\) jo tā ir visu elementu kvadrātu summa. Tātad $tr(AB) = tr(M^TM) . $ Tādejādi abos veidos iegūstam, ka, ja \(A\) un \(B\) ir pozitīvi semi - definētas matricas, tad \(tr(AB) \geq 0.\) 5. uzd.
data = data.frame(x1 = c(1/2, 1/3, 1/4, 1/7), x2 = c(1/3, 1/4, 1/5, 1/8), x3 = c(1/4, 1/5, 1/6, 1/9),
x4 = c(1/7, 1/8, 1/9, 1/12))
E1 = as.matrix(data)
I = diag(x = 1, ncol = 4, nrow = 4)
E2 = I - E1
E2## x1 x2 x3 x4
## [1,] 0.5000000 -0.3333333 -0.2500000 -0.1428571
## [2,] -0.3333333 0.7500000 -0.2000000 -0.1250000
## [3,] -0.2500000 -0.2000000 0.8333333 -0.1111111
## [4,] -0.1428571 -0.1250000 -0.1111111 0.9166667