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1

In other words, you multiply each row and column of the square matrix $A$ by the same $\omega_q$. Therefore the determinant changes by $\prod \omega_q^2 = 1$.


1

Row sum is multiplication from the right with the column vector $\bf 1$ consisting of ones. That row sum is 1 basically means that $\bf A1 = 1$, i.e. that the vector $\bf 1$ is an eigenvector with eigenvalue $1$. By multiplying with $\bf A$ to the left you will see that ${\bf A}^2 \bf1 = A1$, and you see that by systematically multiplying with $\bf A$ to the ...


1

Assume $P^k$ has all row sums $1$ for some $k \geq 1$. Then the $i$-th row sum of $P^{k+1} = P^k \cdot P$ is $$\sum_{j = 1}^n \sum_{m=1}^n (P^k)_{im} P_{mj} = \sum_{m=1}^n (P^k)_{im} \sum_{j=1}^n P_{mj} = 1 \cdot 1 = 1$$ By induction $P^k$ has all row sums $1$ for all $k \geq 1$.


0

Recall that if $f(a)=0$ then there is a $q(x)$ that $f(x)=(x-a)q(x)$. in your question $-5^3+5^2+16.5+20=0$.


3

This is not true even for positive-definite real matrices. If $\lambda$ is the leading eigenvalue with eigenvector $v$ (and assuming that this eigenvalue is simple) then the derivative of $\lambda$ with respect to the $(i,j)$th entry of $M$ is given by $$v^T(e_{i} \otimes e_{j}) v = v_iv_j.$$ Of course to keep $M$ symmetric you need to vary both the $(i,j)$ ...


3

Note that $x^3-x^2-16x-20=(x-5)(x^2+4x+4)=(x-5)(x+2)^2$.


0

To continue the answer that vadim123 provided: Division of nonzero integers is not closed. Here is how we know. Suppose that a,b are each nonzero integers. Is it guaranteed that a/b is an integer? Well, to prove that the division of nonzero integers is not closed, you only need to enumerate one example that demonstrates an integer a/b not being in the set ...


1

If for any row $$|a_{ii}|>\sum_{j=1,j\ne i}^n |a_{ij}|$$ Then the matrix is called diagonally dominated and can be proved to be invertible. Moreover, if $a_{ii}>0$, then it can be shown to be positive definite, i.e. all eigenvalues are positive.


1

Partial answer: Take the matrix $\left( \begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right)$, it's determinant is $0$. Take the identity matrix $\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$, it's determinant is $1$. Take the matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$, it's determinant ...


3

Hint: If all steps were correct, then the answer would be yes. You already know ($A=4I, B=I$) that the answer is no. Therefore, there is a mistake in one of the two "they are similar" statements. The second statement looks pretty rock solid. The matrix $$A^{1/2}BA^{-1/2}$$ is clearly similar to $B$. Therefore, the mistake must be...


1

The matrix $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has two one-dimensional eigen spaces, one generated by the vector $[1,1]^T$ and one by $[1,-1]^T$. On the other hand, $A^HA=I$, so every vector in $\mathbb C^2$ is an eigenvector of $I$. Conclusion: No. The set of eigenvectors of $A^HA$ is not the same as the set of eigenvectors of $A$.


0

Since $\rho(A)<1$, there is an induced norm $N(.)$ s.t. $N(A)<1$. Let $\mu=N(A)$; thus $N(A^k)\leq \mu^k$. Since the norms are equivalent, there is a fixed $C$ s.t. $||.||\leq CN(.)$. Finally $||A^k||\leq CN(A^k)\leq C\mu^k$.


0

OK. So the whole thing was very simple. First note that both $\tilde{H}$ and $\tilde{L}$ commutes with $J$. Now $\tilde{H} + \imath J \tilde{L}$ \ge 0$ is equivalent to \begin{equation} \begin{bmatrix} \tilde{H} & -J \tilde{L} \\ J \tilde{L} & \tilde{H} \end{bmatrix} \ge 0. \end{equation} Which is possible if and only if $\tilde{H} \geq - J ...


1

Suppose $\left\Vert A\left(x^{\star}\right)\right\Vert=K<1$. For all $\epsilon>0$, there exists $r>0$ s.t. $0<\left\Vert x^{\star}-x\right\Vert<r$ implies $$\left|\left\Vert A\left(x\right)\right\Vert-\left\Vert A\left(x^{\star}\right)\right\Vert \right|\leq\left\Vert A\left(x\right)-A\left(x^{\star}\right)\right\Vert<\epsilon.$$ If ...


1

Ok, after some fun roundabouts and a nice long learning experience, I have the following. it's long, but it answers things. The algebra I was attempting to perform was giving erroneous results because $e^{Ax}=(e^A)^x$ is only strictly true with matrix $A$ when $x$ is an integer. Otherwise calculations of $(e^A)^x$ will involve solving a set of ...


1

I assume you're talking about real or complex matrices. Hint: What is the characteristic polynomial of $A$? Apply the Cayley Hamilton theorem. If "eigenvalue" in this context specifically means "real eigenvalue", then I suppose you'll need a counterexample. Note that the matrix $$ \pmatrix{0&4\\-1&0} $$ is not skew-symmetric, but has purely ...


4

a is false. It means $2A=0$, and as I suppose you do not work in a field of characteristic $2$, this means $A=0$. Why should it be true? Here is a $5\times5$ (Jordan) matrix with $0$ as sole eigenvalue, that satisfies neither a, nor b nor c: $$A= \begin{bmatrix} 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0 ...


0

The conjecture appears true: For any $a_1,\ldots,a_n\in\mathbb Z$, the determinant of a matrix with them in its first row can be any multiple of their $\gcd$. It clearly suffices to handle the case in which their $\gcd$ is $1$. I found a recursive procedure that seems to work. For $n=2$ we take $$\begin{pmatrix}a_1&a_2\\-b_2&b_1\end{pmatrix}$$ ...


1

Answer a) is correct. Note that the characteristic polynomial $\chi_A$ of $A$ is of degree at most $2$, hence $f(x)=\chi_A(x)g(x)+c_1x+c_0$ for some poylnomial $g$ and constants $c_0,c_1$ (division with remainder). Now plug in $A$ for $x$ and use $\chi_A(A)=0$. b) is false because $f$ can be arbitrary, even constant nonzero c) and d) are false because $f$ ...


1

You're on the right way with the Cayley-Hamilton theorem. Recall that the characteristic polynomial of $A$ is of degree $2$ and use euclidean division on $f$ and the characteristic polynomial.


1

Let $\chi_A(x)$ be the characteristic polynomial of of $A$ (it has degree 2). Then by the division algorithm $f(x)=q(x)\chi_A(x)+r(x)$ where $\deg r<2$. Now, $f(A)=q(A)\chi_A(A)+r(A)=r(A)$ since $\chi_A(A)=0$. Since $\deg r<2$, $r(x)=c_0+c_1x$ for some $c_0,c_1\in\mathbb{C}$.


0

No, $AD$ is not equal to $$\sum_{i=1}^n a_id_i$$ For every $i$, the value $a_id_i$ is a vector (of dimension $M\times 1$). Therefore, the sum of such vectors is also a $M\times 1$ vector. On the other hand, since $A$ is a $M\times n$ matrix and $D$ is a $n\times n$ matrix, the matrix $AD$ is also a $M\times n$ matrix.


1

Not quite. Notice that you are multiplying an $M\times n$ matrix by an $n\times n$ matrix, so you should get back an $M\times n$ matrix, but $\sum_i^n \mathbf{a}_id_i$ is an $M\times 1$ matrix, i.e. a vector.


1

In your second manipulation step an error occured. If you subtract $[2 ,-1, 2| -1]$ by $[2, \frac{4}{3}, \frac{2}{3}| 0]$ you obtain $[0, -\frac{7}{3}, \frac{4}{3}|-1]$. Now add the second row to the third row.


1

No. Consider $v_A=(1,0,0)^T,\ v_B\in\mathbb R^2$ and $$ C=\pmatrix{ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&1\\ 0&0&0&1&1&0\\ 0&0&0&1&1&0\\ 0&0&1&0&0&1}. $$


3

The equation $A^2 -4A +4I = 0$ can be written as $(A -2I)^2 =0$ and hence $A-2I$ must be a nilpotent matrix of order $2$. Therefore the only eigenvalue of $A-2I$ is $0$ and so the only eigenvalue of $A$ is $2$. Now it is easy to see that $A$ and $A-2I$ have the same eigenvectors and therefore the dimension of their respective eigenspaces must be the same. ...


1

The minimum polynomial is $$ m(\lambda)=\lambda^2-4\lambda+4=(\lambda-2)^2 $$ So the only eigenvalue of $A$ is $2$. Since the $m(\lambda)$ is not of linear factor, $A$ is not diagonizable. The Jordan block has maximum size of $2$. An example of $A$ is $$ A=\pmatrix{2 & 1 & 0 \\0 & 2&0 \\ 0 &0 &2} $$


4

"Hint": Consider the matrix $$A:=\begin{pmatrix}2&1\\0&2\end{pmatrix}.$$


1

Here is a sketch of these proofs. The details of this are described in great detail in Michael Artin's Algebra - a book well-worth reading. 1a) For the $2\times 2$ case: If $A$ is orthogonal, then multiply by a suitable reflection to ensure that $\det(A) = 1$. Then note that $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ where the vector ...


1

This is a terrific question. I will try to answer all of your concerns. In general, you should be aware of a few things. First, when we say things like $g_{ij}$, $i$ and $j$ are just placeholders for integers. So it doesn't matter what letters are in the subscript, they mean the same thing. Second, not all tensors can be represented as matrices. Only tensors ...


1

This works for all fields (with characteristic 0 or finite characteristic). Let $E_{ij}$ denotes the matrix with a 1 at the $(i,j)$-th entry and zero elsewhere. Consider the set consisting of $I$, $I+E_{ij}$ for any $i\ne j$, $I+E_{ii}+E_{i1}+E_{1i}$ for every $i\ne 1$. From (1) and (2), one can obtain any $E_{ij}$ with $i\ne j$. Therefore, one can ...


1

Such a basis can be constructed by combining three types of matrices: The $n^2 - n$ matrices that have a $1$ on the diagonal, a $1$ on a single off-diagonal element and are $0$ elsewhere. The identity matrix. The matrices $\mathrm{diag}(2, 1, \ldots, 1), \ldots, \mathrm{diag}(1, \ldots, 1, 2, 1)$. More generally, one can construct many bases that only ...


1

As a remark, if the Jordan form $J$ is known to be a $2\times2$ Jordan block with eigenvalue $\lambda$, you don't really need to calculate $S$ in order to find $e^M$. More specifically, suppose $M=SJS^{-1}$ where $$ J=\pmatrix{\lambda&1\\ 0&\lambda}=\lambda I+N. $$ Then \begin{align} e^J &= I + (\lambda I+N) + \frac{(\lambda I+N)^2}{2!} + ...


0

Take $A=I$ over any field that has characteristic 2 (i.e. any field with $1=-1$).


1

You can take any real-valued square orthogonal matrix $A$ (e.g. reflection, rotation, any distance preserving linear transformation) and for this matrix $A$ you will have by definition $A^TA = I$. Then multiplying $A$ by $i = \sqrt{-1}$ will give you what you want. Note also the equivalent definition of a real orthogonal matrix which is perhaps more ...


1

$$\begin{pmatrix} a & c \\ b & d\end{pmatrix} \times \begin{pmatrix} a & b \\ c & d\end{pmatrix}= \begin{pmatrix} a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \end{pmatrix}= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$ Taking $b=c=0$ implies $a=d=i$, where $i^2 = -1$.


1

That is correct. If $y$ is an arbitrary vector, let $x=Hy$, so that $$y^{*}H^{*}MHy=\left(Hy\right)^{*}M\left(Hy\right)=x^{*}Mx\geq0$$ by the PSD property of $M$.


2

The notation $\Sigma^{1/2}$ means the square root of the matrix $\Sigma$, which exists because your covariance matrix $\Sigma$ is symmetric positive definite, and thus $\Sigma = MDM^{-1}$ where $D$ is a diagonal matrix with all positive diagonal entries, so then $\Sigma^{1/2}$ is obtained by replacing each diagonal entry in $D$ with the square root of the ...


2

The generalized eigenvector you want is any solution to the equation $$ (M - 6I)x = v_1 $$ We can solve this equation by row-reduction: $$ \pmatrix{2&-1&1\\4&-2&2} \leadsto \pmatrix{2&-1&1\\0&0&0} $$ So, we can take for instance the generalized eigenvector $\pmatrix{0\\-1}$. More generally, any vector of the form ...


2

Note that this just reduces to the case where $J = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$, nothing interesting is happening in the higher dimensions.


1

If $U \subset W$ (or vice versa) then $U \cup W = W $ so if W is a subspace so is the union. Otherwise there is a ∊ U , a ∉ W and b ∊ W, b ∉ U So, a, b ∊ U ∪ W But a + b ∉ U otherwise with (-a) ∊ U then (-a) + a + b ∊ U ⇒ b ∊ U And similarly a + b ∉ W Therefore a + b ∉ U ∪ W, i.e. there are vectors a, b in U ∪ W whose linear combination is not in U ∪ ...


1

Well, there is explanation in the book at the same page: Note that since each element of $\Gamma(s)$ (the adjoint matrix of $sI-A$) is obtained from the determinant of $(n-1)\times(n-1)$ matrix, each element $\Gamma_{ij}(s)$ is a polynomial in $s$ of degree no greater than $s^{n-1}$ And since we can group terms with the same degree and put the ...


0

I haven't encountered such truncating function, but you're basically getting a submatrix of a matrix. So let me say the following. In Horn & Johnson's Matrix Analysis, they introduced the notation: Given $A\in M_{n}$, if $\alpha,\beta\subseteq\{1,\ldots,n\}$, then $A[\alpha, \beta]$ denotes the submatrix of entries that lie in the rows of $A$ indexed by ...


2

Let $$ C=\pmatrix{I&0\\0&D}, $$ where the identity $I$ and an arbitrary symmetric $D$ have the same dimension greater than one. For $v_A=[1,0]^T$ and any nonzero $v_B$ (of the same dimension as $I$ and $D$), $v=v_A\otimes v_B$ is an eigenvector of $C$. The matrix $C$ does not need to have a Kronecker product form since $D$ is arbitrary symmetric.


2

This is in general false. Consider $v_A = v_B = (1, 1)^t$ and $C = \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1\end{pmatrix}$. Then $v_A \otimes v_B = (1, 1, 1, 1)^t$ and $C (v_A \otimes v_B) = 10 (v_A \otimes v_B)$, but clearly $C$ is not of the form $A \otimes I_2$. In ...


3

This is a wrong question because the limit of $HAH^{-1}$ in general may not exist. The right questions should be: provided that $\lim_{H\to 0}HAH^{-1}$ exists, is it necessarily similar to $A$? The answer, unfortunately, is no. Here is a counterexample: $$ HAH^{-1} = \pmatrix{x^2&0\\ 0&x} \pmatrix{1&1\\ 0&1} \pmatrix{\frac1{x^2}&0\\ ...


1

In terms of the $n\times n$ blocks of $B$ and $C$, this is $\sum_{ij}\operatorname{tr}XB_{ij}XC_{ji}$, so the derivative with respect to $X$ is $$\sum_{ij}\left(B_{ij}XC_{ji}+C_{ij}XB_{ji}\right)^\top\;.$$


1

https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem We can use the Cayley Hamilton Theorem to do this. $M^{-1}=\frac{adj(M)}{det(M)}$. So,$adj(M)=M^{-1}det(M)$. If $M=sI-A$, then $adj(sI-A)=(sI-A)^{-1}det(sI-A)$. But $det(sI-A)$ is the characteristic polynomial of $A$.


1

The assumed equality isn't true. The matrices {$L, S, X, Z$} must all be the same dimensions for the math (addition, subtraction, scalar product, etc) to make sense. Yet, if you generate 4 random $3 \times 5$ matrices, and evaluate the above expression with random values of $r$, you will find that the RHS does not equal the LHS. Try it again with random ...


1

Note that the left side of $$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$ can be written as $$ \operatorname{tr}(Z^T(X-L-S)) + \frac { r }{ 2 } \operatorname{tr}( (X-L-S)^T(X-L-S) ).$$ And this is the same as $$ ...



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