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4

(I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.) Hint What are the possible eigenvalues of $M$? What can one say about the eigenvalues of real $2 \times 2$ matrices...?


3

One way to prove this is through eigenvalues. Since $A$ is positive definite, it is a symmetric matrix. $$A \text{ is positive definite } \iff \text{ all eigenvalues are positive. }$$ It is known that for any matrix $M$: $$\lambda \text{ eigenvalue of matrix } M \implies \lambda^k \text{ eigenvalue of matrix } M^k,\quad k=1,2,\ldots$$ Also, you can use ...


3

by row reducing $A-3I,$ using my ti-83, i get $$\pmatrix{1&-2&3&2\\3&2&1&-4\\-1&6&-7&-7\\-2&0&-2&1}\to\pmatrix{1&0&1&-0.5\\0&1&-1&-1.25\\0&0&0&0\\0&0&0&0} $$ we can see that two linearly independent vectors $a, b$ such that $(A-3I)a = 0, (A-3I)b = 0$ where ...


3

I assume you are working with $n \times n$ matrices over the complex numbers. Let $A^\ast$ denote the Hermitian adjoint of $A$, i.e. the complex conjugate of the transpose. Then, by definition, $A$ is Hermitian if and only if $A = A^\ast$. Now suppose that $A$ is an arbitrary complex $n \times n$ matrix. Set $B = \frac{1}{2}(A + A^\ast)$ and $C = ...


2

$$A-I=\begin{bmatrix} 3 & -2 & 3\\ 0 & -2 & 3\\ -1 & 2 & -3 \end{bmatrix}$$ And you can see the two last columns are proportionate so the matrix is not invertible and $\lambda=1$ is an eigenvalue My computation of $\det(A-\lambda I)=-\lambda^3+\lambda^2+13\lambda-13$


2

Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$. Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and ...


2

no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$ pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = ...


1

Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that ...


1

The general matrix is given by the sum between the identity matrix and a circulant matrix, hence its characteristic polynomial over $\mathbb{Q}$ is given by: $$ p(\lambda)=(\lambda-1)^n-1.$$ Over $\mathbb{F}_2$ such a matrix cannot be invertible since the sum of the elements in every row/column is zero, hence $(1,1,\ldots,1)$ is an eigenvector associated ...


1

Try using the definition of semi-positive ($v^*Av\geq0$), the fact that all $x_i\in[0,1]$ (you need both bounds!) and Young's inequality ($2ab<a^2+b^2$) for $n$ terms. A 0-eigenvector (which is a slick way to say solution to the homogeneous system) is perhaps written in front of you?


1

$$\begin{pmatrix} 1 & -2 & 3 & 2\\ 3 & 2 & 1 & -4\\ -1 & 6 & -7 & -7\\ -2 & 0 & -2 & 1 \end{pmatrix}$$ is a zero divisor since : column$(3)=$column$(1)-$column$(2)$. So the matrix: $$B=\begin{pmatrix} 0 & 0 & -a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 ...


1

Your matrix is the sum between an identity matrix and a circulant matrix, so the characteristic polynomial is given by: $$ p(\lambda)=(1-\lambda)^n-(-1)^n \tag{1}$$ and the determinant is given by $(-1)^n p(0)$, so it is $2$ if $n$ is odd and zero otherwise.


1

We have $$\begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = 0\cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ So the column vector is indeed an eigenvector of the matrix, with eigenvalue zero. I'm not sure what you ...


1

For the eigenvalue $\lambda=0$, that means that for the corresponding eigenvector, $x$, you have $Ax = \lambda x$ which in this case corresponds to $Ax=0$. To solve for $x$, you can set up and row reduce the augmented matrix: $\left[\begin{array}{ccc|c} 0 & -1 & -1 & 0\\ 0 & -1 & -3 & 0\\ 0 & 0 & -2 & ...


1

(1) Consider the map $$ \rho : S_n \rightarrow M_n(\mathbb{R}^n) $$ where $S_n$ is symmetric group. So if $A$ is a permutation, then it is an image of $\rho$. Let $\rho(t)=A$. And note that if $t=t_1\cdots t_m$ where $t_i$ is a single permutation, then $$ A_i:=\rho(t_i),\ A=A_1\cdots A_m$$ To show that $ADA$ where $D$ is diagonal, we suffice to show that ...



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