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Suppose $\;0\neq w\in\Bbb F\;$ is such that $$\forall\,v\in V\;,\;\;0=w\cdot v\stackrel{\text{mult. by}\;w^{-1}}\implies 0=w^{-1}wv=1\cdot v=v$$ where we used associativity, inverse and etc. from the axioms. The claim now follows if there exists $\;0\neq v\in V\;\ldots$
Every linear combination of $EV_{1}=\pmatrix{1\\0\\0}$ and $EV_3=\pmatrix{0\\1\\0}$ is a eigenvector with eigenvalue $1$. $EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right)\}$ is the same as $EV_{1,3} = span\{\left( ... 1 Update: I have undeleted my answer because I think it is fixed now. You got $$V_{\lambda_2} = \left(\begin{array}{ccc} 0 \\ 1 \\ 1 \end{array} \right)$$ correct but then copied it down wrongly.(I think..) Then you correctly wrote down the case$\lambda_1$. From $$\left(\begin{array}{ccc } 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 ... 1 This is a very common mistake. You have no equation telling you about x_2. Lots of people assume this means x_2 must be zero. On the contrary, since you have no information about x_2, it could be anything! So the solution is$$x_1=0\ ,\quad x_2=t\ ,\quad x_3=0$$and the nullspace is$$\left\{t\pmatrix{0\cr1\cr0\cr}\ :\ t\in{\Bbb R}\right\}\ ,$$... 1 It is not always true. For example$$ A=\left(\begin{matrix} 3 & 1\\ 1 &3\end{matrix}\right)\quad\text{and}\quad B=\left(\begin{matrix} 1 & 0\\ 0 &3\end{matrix}\right). $$Then$$ AB=\left(\begin{matrix} 3 & 3\\ 1 &9\end{matrix}\right), $$which is non-symmetric. Positive operators NEED to be symmetric. 1 The equation ax + by = n has integer solutions in x, y if and only if the GCD of a and b divides n. Hence part (b) and part (c) does indeed have integer solutions, while part (a) indeed has none. Make use of Bezout's Identity (I'll leave the workings to you, so as to let you get a feel for this) to derive that the integer solutions for (b) is$$x ... 1 Denote$Q(v) = Q(v_1,\dots,v_n) = \frac1{2}v^T H v + q^T v + c$where$v = (v_1,\dots,v_n)$. Fix some$v$for which you want to calculate$Hv$Function$f(x) = Q(x,v_2,\dots,v_n)$is quadratic function and$f'(x) = \frac{\partial Q}{\partial v_1}(x,v_2,\dots,v_n) = (Hv+q)_1$, by$(Hv+q)_1$I mean first component of$Hv+q$. Now we use fact, that central ... 1 Since S is 5-D subspace of$ R^6$,so S consists of 5 Linearly Independent Six dimensional vectors.Now if you add one more Six dimensional vector((not 5-dimensional) This is where u went wrong,) then you will have 6 Linearly indep. vectors which will span six dimensional space i.e$R^6$For part (B)converse is true 1 You seem to think that if$v$is a vector in a$5$dimensional vector space, then$v$has$5$components, i.e., that it looks something like$v=(x_1,\cdots ,x_5)$. This is not the case. In fact, vectors in general need not look anything like a list of numbers. For instance, the space of all continuous functions$f:\mathbb R \to \mathbb R$is a linear space. ... 1 If$\alpha v = 0$for$\alpha \in F$,$0 \ne v \in V$, then we must have$\alpha = 0$, for if$\alpha \ne 0$, then$v = \alpha^{-1} \alpha v = \alpha^{-1} 0 = 0$, a contradiction. So we must then have$\alpha = 0$, which shows$\alpha$is unique. In this sense scalar-vector multiplication has the "no zero divisors" property;$\alpha v = 0$if and only if ... 1 Usually the way to approach matrix polynomial equations (of one variable) is to figure out what the minimal and/or characteristic polynomial of the solution matrices would be. Note that any conjugate of a solution to$p(A)=0$is also a solution, so if there is one solution there will be infinitely many (except in cases where the only solutions are scalars). ... 1 There are many solutions of$A^4=I$in$GL(2,\Bbb R)$, because of conjugation. However it is fairly easy to see that one can nevertheless not embed$Q_8$in$GL(2,\Bbb R)$. Assuming one has such an embedding, then since$Q_8$is finite one can find a positive definite bilinear form that is invariant under all elements of$Q_8$(this a standard averaging ... 1 Suppose that the quaternion group can be embedded into$M_2(\mathbb{R})$. Then it is isomorphic to a finite subgroup of$GL_2(\mathbb{R})$. Since all finite subgroups of$GL_2(\mathbb{R}$have a faithful real character of degree$2$, this would apply to the quaternion group, too. But this is not the case, see here, a contradiction. Actually, the question has ... 1 Rewriting$t$as you have is the correct first step. From there recall that as$\mathbb Z_2$is a field everything you learned in linear algebra still holds. You can see that the rank of$t$is$1$and the image is spanned by the vector $$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$ You know that the kernel has dimension$2$so you just need to find two ... 1 The matrix exponential is given by: $$\tag 1 e^{At} = \sum_{k=0}^{n-1} \alpha_k A^k$$ where the$\alpha_i$'s are determined from the set of equations given by the eigenvalues of A, as: $$\tag 2 e^{\lambda_i t} = \sum_{k=0}^{n-1} \alpha_k \lambda_i^k$$ We are given:$\$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & ...