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1

Your attempt in case 2 is wrong, because you have mistakenly assumed that $\det(a_{ij}+t)$ is a nonzero polynomial. Consider, e.g. $A=\pmatrix{1&1\\ 1&1}$. Then $(a_{ij}+t)=\pmatrix{1+t&1+t\\ 1+t&1+t}$ is always singular, whatever $t$ is. There is an easy fix to your proof, though: simply replace $(a_{ij}+t)$ by $A+tI$. When $t\ne0$ is ...


0

It depends how you want to go about it. With this I mean that you need to structure the way you want to visualize things for example eigen vectors a good way to visualize them is by doing 3D or 2D orthogonal transformations a very cool application with great visual implications is Principal components analysis for example. If you are familiar with ...


0

Maybe she would benefit talking a look in the local library, especially in the department of mathematics's one. In my case I was quite happy to find "the linear algebra a beginning graduate student ought to know" by Golan and "problems and theorems in linear algebra" by Prasolov.


1

You need an operation called vectorization (${\rm vec}\,\left[\begin{matrix}1&2\\3&4\end{matrix}\right] = \left[\begin{matrix}1\\3\\2\\4\end{matrix}\right]$); then you can use $({\rm vec}\,A^\top)^\top$ to convert it into a row vector.


1

Another way to the formula: I think that you have almost finished the job. I consider your formula with the $v_k$. Replace them by $w_1,..,w_n$ any vectors in $E=\mathbb{R}^n$. Your formula give then an application $f(w_1,..w_n)$ from $E^n$ to $\mathbb{R}$, and you have see that this is $a{\rm det}(w_1,...,w_n)$ for a constant $a$, depending on the $x_k$, ...


1

An inner product is a lot less unique than one might first suspect. As you have essentially noted: Theorem: Let $V$ be a finite-dimensional linear space over the field $\mathcal{F}$ of real or complex numbers with basis $B=\{ b_1,b_2,\cdots,b_n\}$. Then there exists an inner product $(\cdot,\cdot)_{B}$ on $V$ with respect to which $B$ is an orthonormal ...


1

The 4th row of your matrix is missing a minus sign, i.e. $$ \begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix} $$ $$ $$ But instead of forming an ...


0

As described, your observed values in terms of 'stars' are clearly ordinal. That means both the 'mode' and the 'median' are legitimate ways to express a 'typical' value. The 'mode' is the most frequent value, '1 star' in your case. The 'median' is found by sorting the values in order and picking the middle one of the sorted values. (You can think of ...


0

Let E = number of Estrada supporters Let A = number of Arrayo supporters Then E + A = 8600 Estrada's majority -- we don't know what it is yet -- is E - A In the second scenario, 1/3 of Estrada's stay home, so he gets 2/3 E votes. Likewise, 1/2 of Arrayo's fans are at home watching Dr. Who, so he's left with 1/2 A votes. The new spread, still a ...


2

Here's a hint: Let $(\lambda,x)$ be an eigenpair of $A$ and consider $$ \exp(A)x = \sum_{n=0}^\infty \frac{A^n x}{n!} = \sum_{n=0}^\infty \frac{\lambda^n x}{n!} $$


2

What you're missing is the 'cross term' in the dot product. The dot product is defined as a distributive operation, so if we call $$[0, 1] = \vec{e}_1 , \ [1/2, 1/2] = \vec{e}_2,$$ then your vectors can be written $$V_1 = \vec{e}_1, \ V_2 = - \vec{e}_1 + 2\vec{e}_2$$ and so the dot product, $$V_1 \cdot V_2 = \vec{e}_1 \cdot (-\vec{e}_1 + 2 \vec{e}_2)= ...


0

1420264 $\begin{matrix} \frac1{frank}&+&\frac1{bob}&+&\frac1{diane}&=&\frac1{12}\\ \frac1{frank}&+&\frac1{bob}&&&=&\frac1{18}\\ \frac1{frank}&&&+&\frac1{diane}&=&\frac1{16} \end{matrix}$ $\begin{matrix} frank&=&28.8\\ bob&=&48\\ diane&=&36 \end{matrix}$


2

The inner product does not in fact depend on a particular coordinate system. The error you are making is in the statement: "Then, relative to the new basis, V1 has coordinates [1,0] and V2 has coordinates [2,−1]. So the inner product V1⋅V2=1∗2+0∗−1=1 " Summing up the product of coordinates of two vectors gives the inner product only if the coordinates are ...


0

They did not derive the basis of the image from the kernel. The simple rule is that the columns which contain the pivots (or the leading $1$'s in each row) in the rref form are the first and the second columns. The rule for finding the basis of the column space is simply that you should choose the corresponding columns from the original matrix. In other ...


0

In the reduced echelon form of the matrix there are leading $1$'s in the first and second column. This tells you that the first and second columns of the original matrix give a basis for the image of that matrix.


0

It's saying that if x and y (or their derivatives) are multiplied together in any way, it's not considered a linear differential equation because it's not solvable in the usual ways that linear ODE's are. This relates to normal linear equations in that if you have an equation where x and y are multiplied or otherwise modify each other in a way that prevents ...


0

(1)For any $x$ with $||x||\leq 1$ there exists $y$ with $||y||=1$ and $|T(x)|\leq |T(y)|$. PROOF.If $x=0$ let $y$ be any vector of norm $1$ because $T(0)=0$. If $ 0<||x||\leq 1 $ let y=x/||x|| (Because $||x||\leq 1 \implies |T(y)|=|T(x)|/||x|| \geq |T(x)|$.) .....................(2) Therefore $$||T||= \sup \{|T(x)| : ||x||\leq 1\} \leq \sup \{|T(y)| : ...


5

The desired result drops right out if you apply the fundamental theorem of calculus to the function $$ g(t) = f\bigl(x + t(y - x)\bigr), $$ since $$ g'(t) = \langle \nabla f\bigl(x + t(y - x)\bigr), y - x\rangle $$ by the chain rule.


2

As it happens, the identity you give: $$\sqrt{a_i^2}+\sqrt{b_i^2}\le\sqrt{(a_i+b_i)^2}$$ is false. Take for example $a_i=1$ and $b_i=-1$ to see why it is false.


0

You put them as the rows of the matrix in order to determine whether or not they are all linearly independent. Try it --- when you put the vectors as the rows of a matrix and transform it into reduced echelon form, (2, 2, 2) will cancel because it is the combination of u1 + u2.


0

It is mainly for a technical reason: I guess you know Gauß's row reduction for solving linear equations. In a system of linear equations , it lets you detyermine the number of linearly independent equations. For the span of a number of vectors, you should perform an analogous column reduction. It is simpler to transpose the matrix and perform row reduction. ...


1

Hint: the constant is the total distance covered by all the vehicles. It will take Diane $36$ days, Bob $48$ days and Frank ${144\over 5}=28,8$ days.


1

$A$ is primitive because each of the $k$ states is path connected to the others. Hence the eigenvalue $1$ is dominant and simple (Perron-Frobenius theorem). It follows that if $v$ is the left eigenvector of $A$ corresponding to the eigenvalue $1$, and it is scaled to become a probability vector (so that the sum of its entries is $1$), then every row of ...


1

Judging from the additional context provided in a comment$^1$, the notation means pointwise multiplication, that is $$(fg)(x) := f(x) \cdot g(x)\\ (fg)(-1) = f(-1)\cdot g(-1) = -4\cdot 24 = -96$$ Similarly, you can define other pointwise operations as $$\begin{align*} (f/g)(x) &:= \frac{f(x)}{g(x)}\\ (f+g)(x) &:= f(x) + g(x)\\ (f-g)(x) &:= f(x) - ...


0

HINT Let $E$ denote the number of voters from Estrada and $A$ - from Arrayo. Then, the total is $A+E = 8600$. Translate the second sentence into an equation and solve them together.


0

Since $W\in S_m$, $DT_W:V\in S_m\rightarrow 3tr(W^2V)$. We may assume that $W^2=diag((\lambda_i)_i)$ where $\lambda_i\geq 0$. We consider the Frobenius inner product (cf. joriki post) $<A,B>=tr(A^TB)$. We seek $\sup_{||V||=1,V\in S_m}tr(W^2V)$. One has $tr(W^2V)=<W^2,V>=\sum_i\lambda_i v_{i,i}=\leq ||W^2||||V||=||W^2||=\sqrt{\sum_i\lambda_i^2}$. ...


1

Intuitively, $K(x,y)$ is a "continuous" matrix $K$ acting on a "continuous" row vector vector $f$: $$ \sum_{x}f(x)K(x,y) = \int f(x)K(x,y)dx $$ The associated quadratic form would be $$ \sum_{x}f(x)K(x,y)g(y)^{\star} = \int\int K(x,y)f(x)\overline{g(y)}dxdy $$ This would be selfadjoint if $K(x,y)=\overline{K(y,x)}$, analogous ...


1

If $A$ is simetric and positive you have $(Ax,x)>0$ and if $x$ is an eigenvector you have $(\lambda x,x)>0$. So $\lambda >0$.


1

I am rewriting the problem as $$\left. y = \dfrac{x}{C_0+C_1 x+C_2 x^2} \right\}\;(C_0+C_1 x+C_2 x^2)y = x$$ with the constants $C_0=C$, $C_1=-C D$ and $C_2=C D S^2$ and $x=M$, $y=R$ You want to find the coefficients $C_0$, $C_1$, and $C_2$ given 3 pairs of $(x_i,y_i)$ This can be treated as a 3×3 linear system when the three points are used in $C_0 y_i + ...


2

If one wants to pick the maximum of its eigenvalues, will the value be positive? Of course not. Consider $$ \begin{pmatrix} -1 & 0\\ 0 & -2 \end{pmatrix} $$ For an adjacency matrix, it's true that the largest eigenvalue is non-negative, but that's not trivial; see eg here or here or here. Update: as A.G. notes in a comment, for a symmetric ...


2

Yes, because- $x^TAx=\sum_{i=1}^{n}A_{ii}x_i^2\geq0, \forall x \in \mathbb{R^n}-{\theta_n}$.


0

As another way of looking at this, you may remember $\mathbb{R}$ linear maps only having linear factors of what you're putting in? This also applies in the complex setting, but we also can't have complex conjugation showing up (messes up pulling out constants). In your example, we easily see that (noting $x = (z + \bar z)/2$ and $ y = ( z - \bar z)/2i$)). $$ ...


1

For $\mathbf{C}$ as $\mathbf{R}^2$: $T(v_1 + v_2) = T(x_1 + x_2 + i(y_1 + y_2)) = ... = T(v_1) + T(v_2)$ (This is easy to check) Similarly, $T(\alpha v) = T(\alpha x + i \alpha y) = ... =\alpha T(v)$ is also easy to check. This proves the linearity of T when $\mathbf{C}$ is viewed as $\mathbf{R}^2$. Calculating Matrix: $T(\begin{bmatrix} 1\\ 0 ...


1

Hint : Let $A=\begin{vmatrix} 2m & 6 \\ 4 & -(1-m) \end{vmatrix}$ Then , the system has no solution if : $\det(A)=0$


1

Perhaps you want to prove that if $\lambda\in \mathbb{C}$ is an eigenvalue of $A$ with multiplicity $\alpha$, then the same is true for $1/\lambda$. Since $A$ satisfies $A^{-1}=\Omega^{-1}A^T\Omega$, $A^{-1}$ is similar to $A^T$; consequently $A^{-1}$ is similar to $A$ and we are done.


0

An $n \times n$ matrix $A$ (for $n \geq 2$) will satisfy $$ \det(p(A) - p(0)) = p(\det(A)) - p(0) $$ for all polynomials $p$ if and only if it is singular. To show that any such $A$ is singular, it suffices to consider the polynomial $p(x) = 2x$ (as you have done). We then need to show that the condition holds for any singular matrix. So, suppose that ...


5

If $\mathcal V$ is some field, then $\mathcal V^n$ is a vector field over $\mathcal V$ and it will always have dimension $n$, because $$\{(1,0,\dots,0), (0,1,\dots,0), \dots, (0,0\dots, 1)\}$$ is always a basis for $\mathcal V^n$. However, you are right that just because a vector space consists of $n$-tuples, it may not be a $n$-dimensional vector space. ...


0

$A^n$ is singular for all $n\geq 1$ if and only if $A$ itself is singular. You have proven nicely that $A$ needs to be singular, so you can now safely assume that $A$ is a rank $1$ matrix (or rank $0$, but that's boring). We have two options. First off, if $A$ is diagonalizable: OK, so $A=PLP^{-1}$ where $L=\begin{bmatrix}\lambda & 0\\ 0 & ...


0

We have $\omega$ is a primitive $n$th root of unity, $\omega^{k}$ and $\omega^{n-k}$ are conjugates. This means that $(1/2)(\omega^{k}+\omega^{n-k}) = \mathrm{Re}(\omega^{k}) = \cos{\frac{2\pi k}{n}}$ (and so $\omega^{\frac{n-1}{2}} + \omega^{\frac{n+1}{2}} = 2 \cos{\frac{n-1}{n}}$). Also, $\omega$ is a root of $$z^n -1 = (z-1)(z^{n-1} + \ldots + z+1),$$ ...


1

Assume for simplicity (but without loss of generality) that you select last $n-k$ rows of $-I_n$ and first $k$ rows of $A$ for some $k$ between $0$ and $n$ to form your $C$. Partitioning $A$ in the form $$ A=\pmatrix{A_{11}&A_{12}\\A_{21}&A_{22}}, $$ where $A_{11}$ is $k\times k$, we have that $$ C=\pmatrix{0_{(n-k)\times ...


3

Case 0. If $A$ is square, then the answer is YES. Use determinants: suppose $LA = I$. Then $\mathrm{det}(L)\mathrm{det}(A) = 1$. So $\mathrm{det}(A)$ is non-zero. So $A$ has a two-sided inverse. Now use: Proposition. If an element $A$ of a monoid has a two-sided inverse, then every left inverse of $A$ is a two-sided inverse. So from $LA = I$ we deduce ...


4

Certainly not in general. Let'see this from the point of view of linear maps: $A$ is the matrix associated with a linear map $f\colon\mathbf R^m\to\mathbf R^n$, $L$ is associated with a linear mar $u\colon\mathbf R^n\to\mathbf R^m$. $LA=I_m$ means $\;u\circ f=\operatorname{id}_{\mathbf R^m}$, which implies $f$ is injective and $u$ surjective. On the other ...


3

It's not true for non-square matrices. Consider $$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & -1 \\ 0 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 ...


6

If $A$ is square and of full rank, and $L$ is its left inverse and $R$ is its right inverse, then from $$LA = I$$ We get (if we multiply both sides by $R$ from the right) $$LAR = IR\\ LI = IR\\ L = R$$ However, if $A$ is not square, then one of the two inverses does not exist and the other is not unique, so you cannot draw the same conclusion. For ...


1

What about a Method 3)? Your matrix has rank $1$ and its eigenvalues are $3,0,0$. Since it is a symmetric matrix, it is positive semi-definite.


1

The vector product of the two vectors given is $$\vec n=(1,-7,10).$$ All vectors in the spanned plane will be perpendicular to $\vec n$. So, the equation is $$\vec v\cdot \vec n =0.$$ Or, if $\vec v=(x,y,z)$ then the equation is $$x-7y+10z=0.$$


0

I feel stupid because I figured it out minutes after posting. It turns out that the matrix I was talking about is indeed symmetric under the stated assumptions. We have $$ (A^{1/2} B A^{-1/2})^\top = A^{-1/2} B^\top A^{1/2} = A^{-1/2} B^\top A A^{-1/2} = A^{-1/2} A B A^{-1/2} = A^{1/2} B A^{-1/2}. $$ Here we used $B^\top A = (AB)^\top = AB$, i.e., ...


1

One can do something similar to what you want in the realm of tensors. One can define a "tensor product" of second order tensors, $\odot$, such that for any second order tensors $\boldsymbol{S}$, $\boldsymbol{T}$, and $\boldsymbol{Z}$, $$\left[\boldsymbol{S}\odot\boldsymbol{T}\right]\boldsymbol{Z}=\boldsymbol{SZT}.$$ Note that ...


3

You have $$x=2-3t, y=-1-t, z=1-2t,$$ or $$\underline{r}=\left(\begin{matrix}2\\-1\\1\end{matrix}\right)+t\left(\begin{matrix}-3\\-1\\-2\end{matrix}\right)$$ which is the equation of a straight line. The equation of the plane you have found is one of infinitely many which contain this line



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