# Tag Info

7

notice that the expression for your vector does not depend on the magnitudes of $\vec b$ or $\vec c$ . It can be written in terms of unit vectors. $$\vec v = \vec a - \frac{\vec a \cdot \hat b}{\hat b \cdot \hat c}\hat c$$ imagine that $\hat b$ defines the normal vector to the ground and $\hat c$ defines the direction of the sun's rays , then $\vec v$ ...

7

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

5

In any function $x$ matches to only one $y$. That's literally the definition of a function

4

$C$ is a disk, so the closest point is on the line connecting $(x_1,x_2)$ and $y$. It is $$\frac{y-(x_1,x_2)}{\|y-(x_1,x_2)\|}+(x_1,x_2)$$

4

If you take the scalar product with $b$ you get zero, so the vector constructed here is perpendicular to $b$, so lies in a plane perpendicular to $b$ (defined by $b$) You obtain the vector by taking $a$ and adjusting it by a multiple of $c$, which might be what is meant by a projection along the vector $c$.

4

$\forall n, k \neq 0$ subtract $\frac{m}{n}$ to give $$A-\frac{m}{n}=\frac{n}{k}$$ Then re-arrange the left hand side so that we have an equivalent fraction $$\frac{An-m}{n}=\frac{n}{k}$$ Take the reciprocal $$\frac{n}{An-m}=\frac{k}{n}$$ From which $$n\left( \frac{n}{An-m}\right)=k$$ Or $$\frac{n^{2}}{An-m}=k$$ Writing in a way similar to the outset of the ...

3

By spectral theorem, there is invertible matrix $P$ such that $$P^{-1}(I+A)P=\pmatrix{1+\mu_1 \\ & \ddots & \\ & & 1+\mu_n}$$ where $\mu_i$ is eigenvalue of $A$ and $\mu_i\geqslant0$. So $$(I+A)^{-1}=P\pmatrix{\dfrac1{1+\mu_1} \\ & \ddots & \\ & & \dfrac1{1+\mu_n}}P^{-1}$$ And $$... 2 You don't define the notion "characterization of bases". At any rate such a characterization would be a definition, and as such needs no proof. Given that, a proof per se is not "useful" (apart from didactical purposes); it can only be correct, or wrong. It seems that your actual problem is: Why are the following statements useful when dealing with bases? ... 2 The idea is that a basis is sort of a "perfect" generating set. If a generating (spanning) set E is not linearly independent, it means we have some "redundancy" between the elements: we have more than we need to generate the space V. For example, if v_3 = v_1 + v_2, then we do not need all 3 of v_1,v_2 and v_3, we can do without v_3 if we have ... 2 If we have a n\times n tridiagonal Toeplitz matrix of the form:$$A = \begin{bmatrix} a & c & & & & \\ b & a & c &&\mathbf 0 \\ & b & a & c \\ &&\ddots&\ddots&\ddots& \\ &\mathbf 0&&&& \\ &&&&&&&\end{bmatrix},$$its eigenvalues are given ... 2 Ok, I think that this work. But it can be write better. The Frobenius automorphism of \mathbb{F}_{p^2} is an invertible \mathbb{F}_p-linear map. If we fix a basis of \mathbb{F}_{p^2} over \mathbb{F}_p, then \sigma is representable as a matrix of \mathrm{GL}_2(\mathbb{F}_p). We denote with S_F the associated matrix of \sigma. Fix an s \in ... 2 The block matrix (let us denote by M) can be expressed as the Kronecker product of matrices A and I (the fixed size identity matrix, of dimension n) as follows:-$$M=A\otimes I$$where A is the 3\times3 matrix:-$$A=\left[\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ ...

2

The vector is co-planar with $\underline {a}$ and $\underline {c}$ and is perpendicular to $\underline {b}$

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 $$... 1 We can use your observation: "T being over a finite dimensional complex vector space there is necessarily a complex eigenvalue" to solve the first part of the problem. Consider the operator T_{|U} : U \to U (T restricted to act on the proper subspace U).$$T_{|U}(u) = T(u) \,\,\, \forall u \in U$$Now that is a well defined linear operator that ... 1 Another route through: First multiply everything by kn to clear denominators$$Akn=km+n^2$$Next get all the terms in k on the same side of the equation and gather them together to identify the coefficient$$n^2=Akn-km=(An-m)k$$Finally divide through by the coefficient (on the assumption that this is non-zero i.e. An\neq m)$$k=\frac {n^2}{An-m}$$... 1$$A = \frac mn + \frac nk \quad \Rightarrow \quad \frac nk = A - \frac mn = \frac{An-m}{n} \quad \Rightarrow\quad \frac kn = \frac{n}{An-m} \quad \Rightarrow\quad k = \frac{n^2}{An-m}

1

Inverting a vector is not a valid function. That's where you're going wrong. Before that looked good. Done, even. If you want a more elementary-looking equation, you simply have to multiply out the dot product. Also, I think the notation is part of the confusion: using $x_1$, $x_2$, ..., $x_n$ for independent variable coordinates while using ...

1

If the inner product is $\langle X,Y\rangle = \operatorname{tr} (X^TY)= \operatorname{tr}(XY)$ ($X,Y$ are symmetric), then $v^TYv=\sum\limits_{i,j=1}^n Y_{ij}v_iv_j =\sum\limits_{i,j} Y_{ij}X_{ij} = \operatorname{tr}(XY)=\langle X,Y\rangle$, because you have that $X_{ij}=v_iv_j$.

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

This property of each x mapping to only one y is officially called "well-definedness", and every function (even linear maps defined by matrices) ought to satisfy it before any math can be done with the function. Every injective function is (implicitly) already well-defined.

1

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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