Assignment-daixieTM为您提供哥伦比亚大学Columbia University Math UN2010 Linear algebra线性代数代写代考和辅导服务！
Instructions:
This course provides a thorough introduction to linear algebra, including the study of matrices, vector spaces, linear transformations, eigenvalues and eigenvectors, canonical forms, and their various applications. Students will gain a deep understanding of these topics through rigorous mathematical proofs, which will be introduced as a central component of the course.
Throughout the course, students will learn how to think critically about linear algebra concepts and apply them to solve real-world problems. They will also learn how to communicate their ideas effectively and write mathematical proofs, which are essential skills for further study in mathematics and other quantitative fields.
Upon completion of this course, students will have a solid foundation in linear algebra and will be well-prepared for advanced courses in mathematics, engineering, physics, computer science, and other related fields. They will also have gained valuable skills in critical thinking, problem-solving, and mathematical communication that can be applied in a wide range of settings.
If A is an n × n matrix and it satisfies the equation A3 − 4A2 + 3A − 5In = 0, then A is nonsingular
We can rewrite the equation as:
A(A2−4A+3I)=5I
If we assume that $A$ is singular, then $0$ is an eigenvalue of $A$. Thus, $\det(A) = 0$, and $\det(A^2-4A+3I)=0$. But $\det(A^2-4A+3I)$ is just the product of the eigenvalues of $A^2-4A+3I$, which are $1$ and $3$. This means that $\det(A^2-4A+3I)\neq 0$ and hence $A$ cannot be singular. Therefore, $A$ must be nonsingular.
The equation of the plane which passes through the points $(0,-1,-1),(5,0,1)$, and $(4,-1,0)$ is $a x+b y+c z=1$ where $a=$ ,$b=$ , and $c=$
One possible method to find the equation of the plane is to use the cross product to find the normal vector to the plane, and then use one of the given points to write the equation of the plane in the form a x+b y+c z=d or that lie in the plane by taking the differences of the points:
$$
\vec{v}_1=\left(\begin{array}{l}
5 \
0 \
1
\end{array}\right)-\left(\begin{array}{c}
0 \
-1 \
-1
\end{array}\right)=\left(\begin{array}{l}
5 \
1 \
2
\end{array}\right)
$$
and
$$
\vec{v}_2=\left(\begin{array}{c}
4 \
-1 \
0
\end{array}\right)-\left(\begin{array}{c}
0 \
-1 \
-1
\end{array}\right)=\left(\begin{array}{l}
4 \
0 \
1
\end{array}\right)
$$
Then, we take the cross product of these vectors to find the normal vector to the plane:
$$
\vec{n}=\vec{v}_1 \times \vec{v}_2=\left(\begin{array}{l}
5 \
1 \
2
\end{array}\right) \times\left(\begin{array}{l}
4 \
0 \
1
\end{array}\right)=\left(\begin{array}{c}
1 \
7 \
-4
\end{array}\right)
$$
Therefore, the equation of the plane is of the form \$ a x+b y+c z=d \$, where \$(a, b, c) \$ is the normal vector we just found. To find \$ d \$, we can use one of the given points, say \$(0,-1,-1) \$ :
$$
a(0)+b(-1)+c(-1)=d \quad \Rightarrow \quad d=-b-c .
$$
Substituting the components of the normal vector, we get:
$$
d=-b-c=-7 a+4 b=-a-4 c .
$$
Solving for a, b, c , we find a=1 / 3 , b=-2 / 3 , and c=-1 / 3 . Therefore, the equation of the plane is:
$$
\frac{1}{3} x-\frac{2}{3} y-\frac{1}{3} z=1
$$
or equivalently,
$$
x-2 y-z=3 \text {. }
$$
Find an orthogonal matrix $Q$ (that is, a matrix such that $Q^t=Q^{-1}$ ) which diagonalizes $T$. What is the associated diagonal form $\Lambda$ of $T$ ?
To diagonalize a matrix $T$, we need to find a matrix $Q$ that satisfies $Q^{-1}TQ=\Lambda$, where $\Lambda$ is the diagonal matrix containing the eigenvalues of $T$.
The orthogonal matrix $Q$ can be obtained by finding the orthonormal eigenvectors of $T$. Let $\lambda_1,\lambda_2,\dots,\lambda_n$ be the eigenvalues of $T$ and let $v_1,v_2,\dots,v_n$ be the corresponding eigenvectors. Then, we can construct the matrix $Q$ by taking the eigenvectors as its columns and normalizing them to have unit length, i.e.,
$$Q=\begin{bmatrix} \frac{v_1}{|v_1|} & \frac{v_2}{|v_2|} & \cdots & \frac{v_n}{|v_n|} \end{bmatrix}$$
where $|v_i|$ is the Euclidean norm of $v_i$.
Note that since the eigenvectors are orthonormal, $Q$ is an orthogonal matrix, i.e., $Q^tQ=QQ^t=I$. Also, since $T$ is diagonalized by $Q$, we have $Q^{-1}TQ=\Lambda$, where $\Lambda$ is the diagonal matrix with entries $\lambda_1,\lambda_2,\dots,\lambda_n$.
Therefore, the diagonal form of $T$ is $\Lambda$, and the orthogonal matrix that diagonalizes $T$ is $Q$.
