# Tag Info

4

Partial answer: Your complete answer can be given by combining these two links link 1 link 2 and possible my partial answer If $A, B \in \mathbb{R}^{n\times n}$ are positive definite then $x^t A x > 0$, $x^t B x > 0$, $\forall x \neq 0$. This implies that $x^t (A + B)x = x^t A x+ x^t B x > 0$. Hence sum of two positive definite matrices ...

4

No. It would only prove the converse. $$\underbrace{P \implies Q}_{\text {implication}} \quad\not\equiv \underbrace{Q \implies P}_{\text{converse of implication}}$$ If you need to prove $P \implies Q$, you can prove its equivalent: $$\underbrace{\lnot Q \implies \lnot P}_{\text{contrapositive of implication}}$$

3

Error 1: $a-b=8$ is not the same as $a = \frac{8}{-b}$ but $a = 8 + b$ Error 2: $b+c=1$ is not the same as $b = \frac{1}{c}$ but $b = 1-c$ Error 3: $3d+c=7$ ... same as errors 1 and 2 Error 4: $2a-4d=6$ Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error

3

Here's a straight forward solution. Every linear transformation $T\colon \mathbb R^2 \longrightarrow \mathbb R^2$ can be represented as a $2\times 2$ matrix, so you want to find $a_1$, $a_2$, $a_3$ and $a_4$ such that \[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix} \: ...

3

Let $$A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$ Then $Ae_4 = -e_4$. On the other side, the given three conditions and orthogonality of $A$ imply $Ae_4 \in \{\pm e_4\}$, as $Ae_4 \cdot Ae_i = e_4 \cdot e_i = 0$ for $i \in \{1,2,3\}$ hence $Ae_4 \cdot e_j ... 2 If you are in$\mathbb{R}^n$of in any vector space then you can always compare two vectors. They are just two elements of a set. Equality perfectly make sense. Perhaps the following description can help you. If you rewrite this equation, you will get $$(\lambda -\alpha)a=(\beta -\mu)b$$ which is same as $$a=(\beta -\mu)/(\lambda -\alpha)b.$$ That is a is a ... 2 We have $$\dot \psi u= A \psi u$$ Suppose that$\psi(t_0)\neq0$for some$t_0$. Then by continuity, in some neighbourhood $$A u = \frac{\dot\psi}\psi u$$ (and hence$\dot \psi/\psi$is an eigenvalue of$A$assuming$u\neq0$.) But the left-hand side is a constant; therefore so is the other, and the result follows. 2 There are 4 different types of (non-trivial) isometries of the plane. They are rotations, reflections, translations and glide translations. Hint: The non-trivial isometries of the plane are classified according to the existence of fixed points, and if they preserve orientation. Hint: Both$T_1 (T_2)$and$T_2 (T_1)$have/don't have fixed points, and ... 2 To help on this point of logic, let me try something I suggest to my students. You take the sentence If it's raining, then the ground is wet. as a reasonable one. And the contrapositive If the ground is dry, then it is not raining. seems equivalent and totally reasonable, as well. (So these are logically equivalent statements.) However, the ... 2 First, symmetric matrix and unitary matrix are two unrelated concepts. It is easy to find symmetric matrices that are not unitary, and vice versa. For an example where$T+S$is not normal, we can take$T=\begin{bmatrix} 0 & i \\ -i & 0\end{bmatrix}$, and$S=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$with ... 2 The set of matrices with rank exactly$k$is the open subset of those of rank at most$k$defined by the non-vanishing of at least one$k$by$k$minor---thus, it is not naturally realized as a closed subvariety of the space of all matrices. You can use the usual trick to identify this with an affine variety if you wish: the point is that the set of non-zero ... 2 Your attempt is correct, but note that your understanding of$W_1 + W_2$might be wrong. By definition,$W_1 + W_2 = \{\vec{w}_1 + \vec{w}_2 | \vec{w}_i \in W_i, i=1, 2\}.$However, This doesn't mean that the bases have to be added together. In other words, you can simply take$\{a_1,a_2,b_1,b_2\}$as a spanning set for the space$W_1 + W_2$and then toss ... 1 Assuming you mean that$N^\prime(x) = G(x)N(x)$, it follows by linearity of the operator$A \mapsto A^\ast$that $$(N^\ast)^\prime(x) = (N^\prime(x))^\ast = (G(x)N(x))^\ast = N(x)^\ast G(x)^\ast.$$ Hence, by the noncommutative Leibniz rule,$$(N(x)^\ast\sigma N(x))^\prime = (N(x))^\ast \sigma N(x) + N(x)^\ast \sigma^\prime N(x) + N(x)^\ast \sigma ... 1 I'll answer this for$n\times n$matrices, which gives only slightly more of a challenge. A diagonalisable matrix$A$is completely determined by its eigenspaces, which form a direct sum giving the whole space, and of cuorse$A$determines those subspaces as well. Over$\Bbb F_2$only two eigenspaces are possible, for eigenvalues$0$and$1$. The matrix$A$... 1 How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. That is, rewrite the system as follows:$ ...

1

Edit: Whether $\psi$ is surjective depends on the definition of its codomain. For example, if $n=2$ and $k=1$, the image of $\psi$ is a curve on $\mathbb{R}^2$. So, if the codomain of $\psi$ is $\mathbb{R}^2$, $\psi$ is certainly not surjective. At any rate, $\psi$ may not be injective even if the origin is not a global minimum. Consider $a=(1,0,0)^T$ and ...

1

I believe what the book is showing is that if $a$ and $b$ are not parallel vectors (read: linearly independent), and say $w$ is a third vector such that $w=\alpha a+\beta b$ for some scalars $\alpha$ and $\beta$, then that expression of $w$ is unique. If you thought that you could be clever and write $w=\lambda a+\mu b$, because of the fact that $a$ and ...

Only top voted, non community-wiki answers of a minimum length are eligible