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0

Let $y$ be in the range of $U$. Then there is $x \in V$ such that $y = Ux$. Note that $$Ty = TUx = UTx \in \operatorname{range}(U).$$ So $T(\operatorname{range}(U)) \subseteq \operatorname{range}(U)$.


0

Suppose that $U(T(v))=T(U(v))$ for all $v\in V$. Now suppose that $w=U(v)$ for some $v$. Then $T(w)=T(U(v))=U(T(v))$. Thus $T(w)$ is also in the range of $U$.


2

You can show that there is no matrix $X = \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$ such that $X^2 = \left( \begin{array}{ccc} 0 & 1 \\ 0 & 0 \end{array} \right)$.


7

No: the matrix $E=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is such that $E^2=0$. If there were a matrix $X$ such that $X^2=E$, it would be nilpotent. However, by Hamilton-Cayley, the index of nilpotency (the smallest $k$ such that $X^k=0$) is at most $2$ in a space of dimension $2$, so that we would have $X^2=0=E$.


1

We consider the COMPLEX case. Proposition. if $A,B$ are invertible and symmetric complex matrices , then they are congruent Proof. We use a pretty result due to Horn and Sergeichuk. Let $U,V\in GL_n(\mathbb{C})$. Then, they are congruent iff $U^{-T}U$ and $V^{-T}V$ are similar. That's all folk.


13

No. Let $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Note that $A^2 = 0$. Thus, if $B^2 = A$, then $B^4 = 0$. But since $B$ is a 2 by 2 matrix, if $B^4 = 0$, then $B^2 = 0$. Therefore $A$ has no "square root."


3

No it's not true but such a matrix exists if $A$ is hermitian. You can show this by diagonalization.


1

Over $\mathbb{R}$ $B$ is congruent to $C$ and as you said above $A$ is not congruent to $B$ and $C$ since the signature of $A$ is $(1,1,0)$ ad the signature of $B,C$ is $(2,0,0)$. Over $\mathbb{C}$ the situation is quite different. Infact $A$ is equivalent to $B$ and $C$; clearly $B$ is still equivalent to $C$ since real coefficient are also complex ...


0

There is an easy example of a matrix with this property: the identity matrix $$\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$$ You can check that scalar multiplication fails by using any scalar that is not $\pm 1$ and the identity matrix. (Where do the scalars come from?) But I want to address another part of your question: how to come up with other ...


1

Try the identity matrix: $I^2=I$, so $I\in U$. What can you say about $2I$?


0

For example, to check whether it is close under scalar multiplication, you should check whether the matrix $cA$ also satisfies the definition of $U$, i.e., whether $$(cA)^2=cA$$ given $c$ a scalar. Matrix operation gives $(cA)^2=c^2A^2=c^2A$ since $A^2=A$. Does this satisfy the definition of $U$? Similarly, to check whether it is close under addition, you ...


0

define $y = U^T x$, it is the orthogonal projection of $x$ onto the vectors $\{u_1, \cdots u_n\}$ the rows of $U$. thus : $$||y||^2 = ||U^T x ||^2 = x^T U U^T x = x^T I x = x^T x = ||x||^2$$ by definition of $||.||^2$ and that $U$ is orthogonal. this is one of the definitions of an orthonormal basis : the projection of a vector onto the basis keeps the ...


1

Assume that we are given a space curve $$\gamma:\quad s\mapsto{\bf x}(s)\qquad(0\leq s\leq L)$$ parametrized by arc length, and assume that its curvature $\kappa(s):=|\ddot{\bf x}(s)|$ is nonzero for all $s$. The three vectors $$\dot{\bf x}(s),\quad {\bf n}(s):={\ddot{\bf x}(s)\over\kappa(s)},\quad {\bf b}(s):=\dot{\bf x}(s)\times {\bf n}(s)$$ then form an ...


1

The coefficient at place $(i,j)$ in $U^TU$ is $u_i^Tu_j$ which is precisely the scalar product of $u_i$ by $u_j$. So the set is orthogonal and its elements have norm $1$; along with the fact that this set is a basis because $U$ is invertible this suffices to prove the assertion. Note that this is an orthonormal basis of $\mathbb{R}^n$ under the standard ...


0

In general if you have $n$ vectors linearly independent in $\mathbb{R}^n$ they form a basis of $\mathbb{R}^n$. You know that $UU^T=I$ this implies $|u_j|=1$ for every $1\leq j\leq n$ and $(u_i,u_j)=0$ if $i\neq j$. It follows that $(u_1,...,u_n)$ is an orthonormal basis.


2

Let $A\in M_n(\mathbb{R})$. $A$ has $n$ distinct real eigenvalues iff there are $\lambda\in\mathbb{R}$ and $P,Q$ real symmetric $>0$ s.t. $(A+\lambda I)^2=PQ$ AND the dimension of the commutant of $A$ is $n$.


0

HINT: You have $\|\frac{1}{2}x + \frac{1}{2}y\| = 1$. Let's show for instance that $\|\frac{1}{3}x + \frac{2}{3}y\| = 1$. Note that $$\left\|\frac{1}{3}x + \frac{2}{3}y\right\| \le \frac{1}{3}\|x\| + \frac{2}{3}\|y\|=1$$ If we had $\|\frac{1}{3}x + \frac{2}{3}y\| <1$ then $$\left\|\frac{1}{2}x + \frac{1}{2}y\,\right\| = \left\|\frac{1}{4} x + ...


3

You're thinking too Euclidean! It is entirely possible in an arbitrary normed linear space to have points other than antipodal points that satisfy this property. As an example, consider the points $(1, 0)$ and $(0, 1)$ in the unit sphere under the $1$-norm on $\mathbb{R}^2$. Good question though!


1

Well, there's actually a step in between in the Matrix Cookbook, which makes this a lot clearer: $$ \frac{\partial\left(\mathbf X^{-1}\right)_{kl}}{\partial X_{ij}}=-\left(\mathbf X^{-1}\right)_{ki}\left(\mathbf X^{-1}\right)_{jl} $$ This you can get from the relationship that you quoted by setting $x=X_{ij}$: \begin{align} ...


2

If $A$ and $B$ are square matrices in $\mathbb R^{n\times n}$ such that $AB=I$, then we can prove that $BA=I$ too. One way to see this is to consider the $n$ column vectors $B\mathbf e_1, B\mathbf e_2, \ldots, B\mathbf e_n$, where $e_i$s are the standard basis for $\mathbb R^n$. The $B\mathbf e_i$s must be linearly independent (because if we have a linear ...


2

Forget about linearity for the moment. If $X$ and $Y$ are sets and $f : X \rightarrow Y$ is some function that is injective, then there exists a function $g : f(X)\rightarrow X$ such that $$ g(f(x))=x,\;\;\; x\in X. $$ Even though $f$ may not be surjective, you can apply $f$ to both sides of the above in order to obtain: $$ ...


5

Commutativity is part of the definition of the inverse, but it is justified by the following fact on monoids: If an element $a$ in a monoid $M$ has a right inverse $b$ and a left inverse $c$: $ab=e$, $ca=e$ (the neutral element in $M$), then $b=c$ — in other words, $a$ has an inverse. This results very simply from the associativity of the monoid law: $$b= ...


2

If at least one of $a_1,\dotsc,a_n$ is nonzero and $a_{n+1}$ is fixed, then the set $$ \{(x_1,\dotsc,x_n)\in\Bbb R^n:a_1x_1+\dotsb+a_nx_n+a_{n+1}=0\} $$ is called an affine hyperplane of dimension $n$. More generally, if $A$ is an $m\times n$ matrix of rank $k$ and $\vec a_0\in\Bbb R^m$, then the set $$ \{\vec x\in\Bbb R^n:A\vec x+\vec a_o=\vec 0\} $$ is an ...


0

We are given an $m\times n$ matrix $A$ with $\DeclareMathOperator{rank}{rank}\rank(A)=n$. Your $T$ is a linear transformation $\Bbb R^m\to\Bbb R^k$ for some $k$. Every such transformation is of the form $T(\vec x)=C\vec x$ for some $k\times m$ matrix $C$. Thus your question is whether the columns of $B=CA$ are linearly independent. Equivalently, your ...


1

Sometimes I think the art of teaching is 90% mind reading. The fact is, I'm not well versed in these subtle and powerful techniques to read your mind over the internet, so unfortunately, I can't tell you with any certainty exactly what you've gotten wrong, or what you're missing. You seem to have slightly misread, or maybe misunderstood, or perhaps you've ...


0

I have mixed feelings when I found that this unanswered thread is among the first few results returned by Google when searching for "colleague matrix". Here's my attempt at making this search result more useful than it currently is. The term "colleague matrix" was coined by I. J. Good in this paper; it is one of a family of so-called "comrade matrices" that ...


1

Set $A_t = 1 + t(A - 1)$. Since $A$ is unipotent, $\log A_t = a_n t^n + \cdots + a_0$ is a polynomial in $t$. The coefficients $a_n$ commute, since they're polynomials in $A - 1$. Furthermore, since $A$ is unipotent, it follows that the $a_n$ are nilpotent. Hence $$\exp(\log A_t) = (\exp t^n a_n)\cdots (\exp t a_1) (\exp a_0)$$ is a polynomial in $t$. The ...


2

In the hint you have proved that for each element $ij$ of the matrix, the function $t\mapsto(\exp(\log(A(t))))_{ij}$ is a polynomial in $t$, and that there is an $\epsilon>0$ such that this polynomial agrees with $\delta_{ij}+t(a_{ij}-\delta_{ij})$ everywhere on $[0,\epsilon]$. (Here $\delta_{ij}$ is the Kronecker delta, that is, the elements of the ...


0

Essentially what you are saying is that you have NOT learned anything from the answers that you have been given to other, similar questions. So is there really any point in continuing to answer your questions?


0

Each of the equations represent a line passing through the origin $(0, 0)$ hence let's compare the slopes for respective cases as follows For infinitely many solutions the lines must have equal slope, hence $$-(\lambda-3)=\frac{1}{-(\lambda-3 )}$$ $$(\lambda-3)^2=1$$ $$\lambda-3=\pm 1$$ $$\implies \lambda= 4, 2 $$ For exactly one solution, there are ...


0

To find the maximum, you can use the fact that any vector $v$ can be written as $v=a_1\phi_1+a_2\phi_2+a_3\phi_3$ for some $a_1,a_2,a_3\in \mathbb{R}$, which can be rewritten as $$v=\Phi.\begin{pmatrix} a_1\\ a_2 \\ a_3\end{pmatrix}.$$ According to question $(2)$, you then have $\|v\|^2=a_1^2+a_2^2+a_3^2$. On the other hand, since the $\phi_i$ are ...


0

Why not to use simplest method. Each of equations represent a straight line hence we have $$4x+ky=6\implies \frac{x}{\frac{3}{2}}+\frac{y}{\frac{6}{k}}=1 $$ $$kx+y=-3\implies \frac{x}{\frac{-3}{k}}+\frac{y}{-3}=1 $$ For having infinite many solutions, both the lines must intersect each other at infinite no. of points i.e. they must coincide with each other ...


1

For infinitely many solutions, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\frac{4}{k} = \frac{k}{1} = \frac{6}{-3}$ From first and second ratios, we get $k = 2, -2$ and comparing with the third ratio eliminates $k = 2$ and gives $k = -2$.


3

Suppose that $T$ has a proper invariant subspace $U \subsetneq V$. An element of $V/U$ is an equivalence class $[x]=\{ x + m : m \in U\}$. Note that $V/U$ is a non-trivial vector space (contains something other than the $[0]$ vector) because $U \ne V$. $T$ induces a linear map $\dot{T} : V/U \rightarrow V/U$ defined by $$ \dot{T}[x] = ...


0

The result of addition or substraction of these two equation has to be zero on both sides. Then the two equations are linear dependent. And you have infinitely many solutions. Therefore you have to add the double of the second equation to the first equation. The RHS gets 0. $2(kx+y)=2(-3) \Rightarrow 2kx+2y=-6$ (second equation) $4x+ky=6$ (first equation) ...


0

Hint: instead of solving $$\begin{pmatrix} 4 & k\\ k & 1 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 6\\ -3 \end{pmatrix}$$ you can instead look at $$\varphi_k:\mathbb{R}^2\rightarrow\mathbb R^2,~\begin{pmatrix} x \\ y \end{pmatrix} \mapsto A\cdot\begin{pmatrix} x \\ y \end{pmatrix}$$ with $A=\begin{pmatrix} 4 & k \\ k ...


1

Assume $A$ is real symmetric and $x^TAx>0$ for every $x\in \mathbb R^n$. Let $z\in \mathbb C^n$. We can decompose $z=x+iy$ for some $x,y\in \mathbb R^n$. Testing for positive-definiteness: $z^*Az=(x^T-iy^T)A(x+iy)=x^TAx+y^TAy+i(x^TAy-y^TAx).$ Since $A$ is symmetric, $x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx$. Thus, $z^*Az=x^TAx+y^TAy>0$, by assumption.


0

Let $U=PSQ^\ast$ be a singular value decomposition. Since $U$ has orthonormal rows, $S=(I_r,0_{r\times(n-r)})$. Therefore $UDU^\ast$ is unitarily equivalent to the leading principal $r\times r$ submatrix of $Q^\ast DQ$. In other words, you are essentially asking about the relationship between the eigenvalues of an arbitrary normal matrix and the eigenvalues ...


8

Hint: $$(a-b)^2+(b-c)^2+(c-a)^2= 2(a+b+c)^2-6(ab+bc+ca)=0$$


1

For (1), I think it is easy to show that, if $v\in V$ is nonzero and $w\in V$, there exists $a\in E$ such that $va=w$. If $w\in \text{span}_k(v)$, then $a$ is just a scalar multiplication. If $w\notin\text{span}_k(v)$, then you can extend the $k$-linearly independent set $\{v,w\}$ to a basis of $V$, and then the rest should be trivial. Hence, $V$ is a ...


1

Probably unrelated (so I don't want an upvote, but I won't miss the opportunity to tell you this). This is the way you "transport" structures along sets with the same number of elements, provided one of them is endowed with some additional structure. And one of the byproducts of this procedure is that some collections you can consider in Mathematics are ...


1

For $\lambda\in [0,1) $, you are taking the positive (rank-one) operator $\lambda |\psi\rangle\langle\psi|$ and you are adding a positive multiple of the identity, so the resulting operator is invertible. Thus your dimension (the range) will be $2^N $.


0

To complement TrialAndError's answer, notice that $$m_T(T) = \begin{pmatrix} \sum_{i=0}^{k-1} a_i (T|_W)^i + T|_W^{k} & * \\ 0 & * \\ \end{pmatrix} = 0$$ Which implies that $m_T(T|_W) = 0$, hence $m_T$ belongs to the ideal of polynomials that annihilate $T|_W$, meaning that the unique monic generator of the ideal ($m_{T|_W}$) must divide it: ...


1

You can solve for the formula given the description. Projecting into the plane perpendicular to $b$ means you have $$v \cdot b = 0$$ That you're projecting $a$ in the direction given by $c$ means you have $$ v = a + t c $$ for some scalar $t$. To make them both true (so that you're projecting to the plane defined by $b$), you solve the system of ...


0

This is false even in one dimension. Let $V = \mathbb{C}$ as a complex vector space, $u = 1$, $v = -i$. Then $\langle u,v\rangle = 1\cdot\overline{-i} = i$, so $\text{Re}\langle u,v\rangle = 0$, yet $\langle u,v \rangle \neq 0$, i.e. $u\not\perp v$. The statement about the projection also fails for this choice of $u$ and $v$, as the projection would be zero, ...


1

Hint: to show it is not a subspace, you can show that it is not closed under scalar multiplication. To do this, you just need to find a vector $(x,y,z)$ with $x$ is an integer and multiply it by a real number so that the product's first coordinate is not an integer: For example, if your vector has the form $(1,y,z)$, can you think of a real number $r$ such ...


1

If $m(\lambda) = \lambda^{k}+a_{k-1}\lambda^{k-1}+\cdots a_1\lambda +a_0$ is the minimal polynomial for $T$, then $$ 0=m(T) = T^{k}+a_{k-1}T^{k-1}+\cdots + a_1 T + a_{0} I. $$ If $p$ is any other non-zero polynomial for which $p(T)=0$, then $m$ divides $p$. The restriction $T_W$ of $T$ to the invariant subspace $W$ also satisfies $$ ...


0

The set of polynomials in $\mathbb{F}[T]$ annihilating $V$ are closed under multiplication by $\mathbb{F}[T]$ and addition, and so form an ideal of the polynomial ring $\mathbb{F}[T]$. Since $\mathbb{F}$ is a field, $\mathbb{F}[T]$ is a principal ideal domain, and hence the ideal of polynomials in $T$ annihilating $V$ is generated by $T$'s minimal polynomial ...


0

The eigenvalues of $M$ are convex combinations of the eigenvalues (i.e. the diagonal entries) of $D$. To see this, note that $P=U^*U$ is a projection, since $P^2=U^*UU^*U=U^*U=P$. Concretely, it is the projection onto the subspace spanned by the rows of $U$. Since the eigenvalues of $AB$ are the same as the eigenvalues of $BA$, the eigenvalues of ...


3

The answer is no in general since Hermitian matrices have real eigenvalues. So for example, there is no Hermitian matrix whose characteristic polynomial is $X^2+1$. Even if you restrict to polynomials with real roots, I doubt you can find a simple formula : I think that if there is a rational formula that works for every polynomial with real roots, it ...



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