Assignment-daixieTM为您提供哥伦比亚大学Columbia University MATH UN2015 Linear Algebra & Probability线性代数代写代考和辅导服务！
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That sounds like a very interesting course that combines linear algebra and probability/statistics, which are two important fields in mathematics with many real-world applications. The course seems to cover the essential topics in linear algebra, including systems of linear equations, matrices, determinants, vector spaces, bases, dimension, eigenvalues, and eigenvectors. These topics are fundamental to many areas of mathematics, including calculus, differential equations, and more advanced topics in linear algebra.
The probability theory part of the course also covers important topics, such as conditional probability, random variables, probability distributions, and limit theorems. These topics are essential for understanding probability and statistics, which are used in many fields, including economics, finance, engineering, and the social sciences. The course also covers more advanced topics, such as Markov chains, curve fitting, regression, and pattern analysis, which are important for data analysis and machine learning.
Overall, this course seems like an excellent introduction to the applications of linear algebra to probability and statistics, with many real-world examples and applications to the life sciences, chemistry, and environmental sciences. Students with an interest in these fields should find this course both interesting and useful, and it should provide a solid foundation for more advanced courses in probability, statistics, and data analysis.
Write $T$ as a linear combination of the projections found in (e).
Answer: $[T]=$ $E_1$___+$ $E_2$___+$ $E_3$___.
Without context, it is difficult to determine what the matrices $E_1$, $E_2$, and $E_3$ represent. However, assuming that they are projection matrices onto some subspaces $V_1$, $V_2$, and $V_3$, respectively, we can write $T$ as a linear combination of these projections as follows:
$$T = TE_1 + TE_2 + TE_3$$
where $TE_i$ is the projection of $T$ onto the subspace $V_i$. To see why this is true, note that each projection matrix $E_i$ satisfies $E_i^2=E_i$ and $V_i=\text{range}(E_i)$. Thus, $TE_i$ is the projection of $T$ onto the subspace $V_i$, which is given by $TE_i=E_iT$.
Substituting this expression for each $TE_i$ in the above equation, we get:
$$T = E_1 T + E_2 T + E_3 T = (E_1 + E_2 + E_3)T$$
Therefore, we can express $T$ as a linear combination of the projection matrices $E_1$, $E_2$, and $E_3$ as:
$$T = E_1 + E_2 + E_3$$
Note that this assumes that the projections $E_1$, $E_2$, and $E_3$ are orthogonal to each other. If they are not, we would need to orthogonalize them first before writing $T$ as a linear combination of projections.
Let $N$ be a normal operator on a finite dimensional complex inner product space $\mathrm{V}$. Show that $|N \mathbf{x}|=\left|N^* \mathbf{x}\right|$ for all $\mathbf{x} \in V$.
Since $N$ is a normal operator, it commutes with its adjoint $N^$. That is, $N N^=N^*N$. Therefore, for any vector $\mathbf{x}\in V$, we have:
\begin{align*} |N\mathbf{x}|^2 &= \langle N\mathbf{x}, N\mathbf{x}\rangle \ &= \langle NN^\mathbf{x}, \mathbf{x}\rangle && \text{since } N \text{ is normal} \ &= \langle N^N\mathbf{x}, \mathbf{x}\rangle && \text{since } N \text{ commutes with } N^ \ &= \langle N^\mathbf{x}, N^\mathbf{x}\rangle && \text{since } N \text{ is normal} \ &= |N^\mathbf{x}|^2. \end{align*}
Therefore, $|N\mathbf{x}|=|N^*\mathbf{x}|$ for all $\mathbf{x} \in V$.
Determine the number of solutions of the following system
\left{\begin{array}{l}
x+2 y-3 z=4 \
4 x+y+2 z=6 \
x+2 y+\left(a^2-19\right) z=a,
\end{array}\right.
depending on the parameter $a \in \mathbb{R}$.
Solution. The matrix associated to the system is
$$
\left[\begin{array}{ccc|c}
1 & 2 & -3 & 4 \
4 & 1 & 2 & 6 \
1 & 2 & a^2-19 & a
\end{array}\right]
$$
We get the row echelon form of the matrix subtracting the first row from the third, and then subtracting four times the first row from the second:
$$
\left[\begin{array}{ccc|c}
1 & 2 & -3 & 4 \
0 & -7 & 14 & -10 \
0 & 0 & a^2-16 & a-4
\end{array}\right] .
$$
Notice that the roots of $a^2-16=(a+4)(a-4)$ are $\pm 4$.
- If $a \neq \pm 4$, the term $a^2-16$ is not zero and we have a unique solution.
- If $a=-4$, the third equation of the system in echelon form becomes $0 z=-8$ and there are no solutions.
- If $a=4$, the third equation of the system in echelon form becomes $0 z=0$, which is satisfied by any value of $z$. Therefore the system has infinitely many solutions.
