# Tag Info

4

The characteristic polynomial of a $2 \times 2$ matrix can be written as: $$p(\lambda) = \lambda^2 - \textrm{tr}(A)\lambda + \textrm{det}(A)$$ (Check here). If a matrix $A$ is symmetric then it is diagonalizable such that: A = Q \Lambda Q^T \quad \textrm{ where } \quad \Lambda= \textrm{diag}(\lambda_1,\lambda_2) ...

3

The answer is positive, assume $A \in \mathbb{R}^{n \times n}$, then for any $x \in \mathbb{R}^n$, take $(x^T, x^T)^T$ to test the diagonal matrix: $$\begin{pmatrix} x^T & x^T \end{pmatrix} \begin{pmatrix}A & 0 \\ 0 & A \end{pmatrix} \begin{pmatrix}x \\ x\end{pmatrix} = 2x^TAx \geq 0$$ implies that $x^TAx \geq 0$, i.e., $A$ is positive ...

3

Hint What can you say about the traces of the given matrices? (Alternatively, for three of the choices, one can find a suitable matrix $B$ for which the equation holds for all $A$.)

3

That's wrong. For $x = (0,1)$ and $y = (1,0)$, $$x^t y = \begin{pmatrix} 0\\ 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$$ is not diagonizable.

3

The operator $2 \, I$ is always self-adjoint. Hence, $2 \, I = I$ or $2 \, I = 0$. This yields $I = 0$, hence, $\mathcal{H} = \{0\}$. Edit: Since every orthogonal projection onto a subspace is self-adjoint, it is quite easy to reconstruct $\mathcal{H}$ from its self-adjoint operators.

2

So $\forall\ \textbf{x}\in\mathbb{R}^{2n}$ we have $\textbf{x}^T\begin{pmatrix}A & 0\\ 0 & A\end{pmatrix}\textbf{x}\geq0$. Let $\textbf{y}\in\mathbb{R}^n$ arbitrary, we want to show $\textbf{y}^TA\textbf{y}\geq0$. To this end, let $\textbf{0}\in\mathbb{R}^n$ and define $\textbf{x}=\left(\begin{array}{c}\textbf{y}\\ \textbf{0}\end{array}\right)$. Then ...

2

Yes : Take the companion matrix of the coefficients of $p$.

2

Well I think we could proceed like this: $$(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=(I-\lambda^{n}P^{n})$$ Which if $\lambda<1$ gives you $$(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=I$$ from there you see that since $P^n=...=P^2=P$ then: ...

1

Can you show that the constant zero function belongs to $W$? Can you show that, if $f,g\in W$, then $f+g\in W$? Can you show that, if $f\in W$ and $a\in\mathbb{R}$, then $af\in W$? Can you do the same verifications for the set of continuous functions? Hint for 2. Set $h=f+g$. Then $$h(\tfrac{1}{2})-3h(\tfrac{\pi}{4})= f(\tfrac{1}{2})-3f(\tfrac{\pi}{4})+ ... 1 if |\lambda| \lt 1 then we may define$$ Q = \sum_{n=1}^{\infty} (\lambda P)^n = (\sum_{n=1}^{\infty} \lambda^n) P \\ =\lambda(1-\lambda)^{-1}P $$and$$ (I-\lambda P)(I+Q) = I $$it can be seen by calculation that the restriction on \lambda is not required, as long as \lambda \ne 1 1 For \lambda=0 the assertion is obvious. Else we can write$$I-\lambda P=\lambda(\lambda^{-1} I -P)$$Note that the only eigenvaules of P are 0 and 1, hence the latter is invertible whenver \lambda \neq 0,1. 1 Let's assume that \exists\lambda\neq 1,\, I-\lambda P non inversible. This means that \exists x\in\mathbb{K}^n,\, x\neq 0\land x=\lambda Px. And this leads immediately by multiplying by P to the left to Px=\lambda Px i.e (1-\lambda)Px=0 and because \lambda\neq 1 we have Px=0 and therefore x=\lambda Px=0 a contradiction and we have proven ... 1 What you looking for is a Thomas algorithm which is a simplified form of a Gaussian Elimination or if you want LU decomposition. Matlab looks fast because Matlab identifying such special cases and call in such case to a very effective solver for banded-matrices. This solver is little bit more general then Thomas algorithm. The very rough idea is that you ... 1 The characteristic polynomial of your matrix is $$p(x) = {x}^{3}-t{x}^{2}- \left( 2\,{t}^{2}+3\,t \right) x-3\,{t}^{2}$$ Which has roots: $$\begin{array}[ccc] \\ x_1 = -t,& x_2 = t+\sqrt{t^2+3t},& x_3 = t-\sqrt{t^2+3t} \end{array}$$ The values for which x_2,x_3 exists are such that t^2 + ... 1 Hint$$\left|\begin{pmatrix} 2t & 1 & 0 \\ 3t & 0 & 0 \\ t & 0 & -t \end{pmatrix}-\lambda I\right|= \left|\begin{pmatrix} 2t-\lambda & 1 & 0 \\ 3t & -\lambda & 0 \\ t & 0 & -t-\lambda \end{pmatrix}\right| =-(t+\lambda)(\lambda(\lambda-2t)-3t)=0 $$The roots are \lambda =-t, \lambda=t\pm\sqrt{t^2+3 t} ... 1 Note that -x^3+6x^2+9x-14=-(x-1)(x+2)(x-7), so we may assume that M is the diagonal matrix with entries 1,-2,7 on the diagonal. Then it's easy to see that the characteristic polynomial of M^{-3} is given by (x-1)(x+(1/2)^3)(x-(1/7)^3). 1 As abiessu says, you have$$y=16+2x\\\frac 14y-\frac 12x=2$$You can rewrite these to$$y-2x =16\\\ y- 2x=8$$and as you say these cannot give a solution simultaneously. 1 I think you are confusing a function and its derivative. You say, I want it to expand quickly to start with with the expansion slowing over time. i.e, an inverse square. So if r(t) is the radius of the sphere, this sentence means you want r'(t) = \frac{1}{t^2}. But the way you programmed it is if r(t)=\frac{1}{t^2}+\text{initial radius}. So, ... 1 Well, think about all these definitions as about rules written in blood and agreed by the community. This is slightly extreme and even somewhat wrong analogy, but the answer is in the "mood" of the question. With theorems it is slightly more complicated, they are result of many years of research and work of many scientists (well this is also true for some ... 1 You can eliminate answers A and D by noting that the zero matrix B always satisfies that identity. 1 Just to expand on the comment of @Myself above... Adjoining algebraics as matrix maths Sometimes in mathematics or computation you can get away with adjoining an algebraic number \alpha to some simpler ring R of numbers like the integers or rationals \mathbb Q, and these characterize all of your solutions. If this number obeys the algebraic equation ... 1 Note that A = \operatorname{Re}A +i\operatorname{Im}A. Also$$z(x) = \begin{bmatrix}\operatorname{Re}x\\\operatorname{Im}x\end{bmatrix}\qquad z^{-1}\left(\begin{bmatrix}v_1\\v_2\end{bmatrix}\right) = v_1 + iv_2$$So, since K = z\circ A\circ z^{-1},$$\begin{align}K\begin{bmatrix}v_1\\v_2\end{bmatrix} &= zAz^{-1}\begin{bmatrix}v_1\\v_2\end{bmatrix}\\ ...

1

As stated, the result is false. For example, let $x=\begin{pmatrix}1&-1 \end{pmatrix}$ and $y=\begin{pmatrix}1&1\end{pmatrix}$ then $$x^Ty=\begin{pmatrix}1&1\\-1&-1 \end{pmatrix}.$$ That matrix has $0$ as only eigenvalue and it is clearly not diagonalisable. Actually, the statement should be Let $x,y$ be two non zero row vectors, ...

1

Hint. (1) To show that $\def\M{\mathcal M_{m,n}}\M$ is open, note that if $M \in \M$, then $M$ has a $d \times d$-submatrix (where $d := \min\{m,n\}$) $A_d(M)$ with nonzero determinant and the map $\mathcal M \to \mathbf R$, which maps $M \mapsto A_d(M)$ is continuous. (2) By (1), $\M$ is an open subset of $\mathbf R^{mn}$ and hence carries naturally a ...

1

I think the result is this one $$A^x=\left(\begin{array}{c} \begin{array}{ccccc} 3x+1 & 3x\\ 3x & -3x+1\\ \end{array}\end{array}\right)$$ Now I have to exit, but as soon as I'm back I will explain it if needed...

1

A linear complex structure on $\mathbb{R}^{2n}$ is the structure of a complex vector space on it compatible with its real vector space structure. $J$ is multiplication by $i$. Since there is only one $n$-dimensional complex vector space up to isomorphism, any two such complex structures give rise to two complex vector spaces $V, V'$ such that there must be ...

1

$$x+y+4z=6\\ 2x+4y-4z=16$$ $$5y=22-3x\\ 10z=4-x$$ $$7x+10y+10z=7x+44-6x+4-x=48+0\cdot x=60$$ $$0\cdot x=12$$ $$\therefore \text{No solution}$$

1

These topological arguments involve the same basic idea: it's often easy to prove things for a subset of matrices which are dense in the space of all matrices. Any "continuous" fact (e.g. the assertion that two continuous functions are equal) can be proven for all matrices by proving it for this dense subset. For example, if $L$ is diagonalizable with ...

1

This is a good approach, but you need to show that any subgroup of $\mathbb{Z}^r$ is finitely generated. This is not extremely hard, but it's not trivial either. Here are two approaches. Perhaps a more direct way to approach this problem is to use the structure theorem for finitely generated abelian groups.

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