# Tag Info

0

There's two cases: if the matrix is diagonalizable hence the dimension of every eigenspace associated to an eigenvalue $\lambda$ is equal to the multiplicity $\lambda$ and in your given example there's a basis $(e_1)$ for the first eigenspace and a basis $(e_2,e_3)$ for the second eigenspace and the matrix is diagonal relative to the basis $(e_1,e_2,e_3)$ ...

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The eigenspace can mean the span of the matrix's eigenvectors.

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We can rephrase $Bu = v$ as $$U^T A U u = v \implies A(Uu) = A(Uv)$$ So, we have $A(Uu) = A(Uv)$ and $Ax = Ay$. So, $Uu$ and $Uv$ satisfy the same equation as $x$ and $y$. The most concise way we can phrase this is to say that $x - y$ and $U(u-v)$ are both in the nullspace of $A$. We don't know anything outside of this from the information provided.

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If $U$ is an orthogonal matrix, $U{-1} = U^T$. So if $B = U^T A U$, then $U B U^T = U U^T A U U^T = A$. On the other hand, $U^T B U = (U^T)^2 A U^2$, which is not the same as $A$ unless $U^2$ happens to commute with $A$.

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$$2x+5y+7z=86\\$$ $$3x+y+5z=60\\$$ $$x+4y+3z=54$$ $\\$ $$x=\frac{86-5y-7z}{2}\\$$ $$3(86-5y-7z)+2(y+5z)=2\times 60\\$$ $$86-5y-7z+2(4y+3z)=2\times54$$ $\\$ $$x=\frac{86-5y-7z}{2}\\$$ $$13y+11z=138\\$$ $$3y-z=22\\$$ $\\$ $$x=\frac{86-5y-7z}{2}\\$$ $$y=\frac{138-11z}{13}\\$$ $$3(138-11z)-13z=13\times22\\$$ $\\$ $$x=\frac{290}{23}\\$$ $$y=\frac{190}{23}\\$$ ...

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Let $A=Q_1R_1=Q_2R_2$ then we have $Q_2^{-1}Q_1=R_2 R_1^{-1}$ but $Q_2^{-1}Q_1$ is unitary matrix so it's diagonalizable and their eigenvalues belong to the unit circle of the complex plane and it's a triangular matrix with positive diagonal elements (eigenvalues) so these eigenvalues are equal to $1$ and then this matrix is similar to $I_n$ hence it's equal ...

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This is Joel's comment, expanded into an answer. I'm not looking to steal his points, so if he posts an answer, I'll delete mine. $$\|QA\|_2=\sup_{x\neq 0}\frac{\|QAx\|_2}{\|x\|_2}=\sup_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}=\|Ax\|_2.$$ Now, this second step obviously requires $\|QAx\|_2=\|Ax\|_2$. But this is easier to prove because we're now working with a ...

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In order to go the other direction, you can simply note that $$A = Q^*(QA)$$ where $Q^*$ is unitary. From there, apply your inequality to find $$\|Q^*(QA)\|_2 \leq \|Q^*\|_2\|QA\|_2 = \|QA\|_2$$

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Let's find the intersection points of the plane and the line. Let $M(x,y,z)$ a point in the plane and in the line. So $31x+3y+18z=62$ and $\exists t\in\mathbb{R},\,\left \{ \begin{array}{c @{=} c} x=2-3t \\ y=1+t \\ z=1+5t \end{array} \right.$ So $31(2-3t)+3(1+t)+18(1+5t)=62$ which means that $83=62$ which is a contradiction. So the ...

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The entries of the covariance matrix of $\vec{Y}$ are $$C^Y_{ij}=Cov(Y_i, Y_j)$$ for $i=1,2$. Moreover the matrix is symmetric since $$C^Y_{ij}=Cov(Y_i, Y_j)=Cov(Y_j,Y_i)=C^Y_{ji}$$ Thus ...

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The line $X(t) = (2-3t,1+t,1+5t)$ must satisfy the plane equation for all values of $t$. We have: $$31(2-3t) + 3(1+t) + 18(1+5t) = 83 \neq 62,$$ so the line isn't in the plane. The normal vector to the plane is $(31,3,18)$. So, take a point in the line, say, $(2,1,1)$. Consider the line $Y(\lambda) = (2,1,1) + \lambda(31,3,18)$. Let's find the point where ...

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Well, I showed you how to try to attack these problems a couple of days ago, and you can use exactly the same strategy here. A problem on positive semi-definite quadratic forms/matrices Once again using MATLAB Toolbox YALMIP to compute a sum-of-squares certificate. Ordering of the variables is not required, but the equality appears to be necessary to ...

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Hint: whether it's 3 or 4 will affect your answer, the choice of R vs. C will not. Note that the degree of a factor in the characteristic polynomial is the multiplicity of the corresponding eigenvalue. That is, the degree is the sum of the sizes of the blocks associated with that eigenvalue. The degree of a factor in the minimal polynomial is the size of ...

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In the unlikely case by semisimple you mean "has zero radical" (this is sometimes called "Jacobson semisimple" ) then yes, it's always true that $\mathrm{rad}(V/\mathrm{rad}(V))=\{0\}$. But if you mean "is a direct sum of simple modules" then (as PavelC has already pointed out) there are many counterexamples. The polynomial ring is nice, but I'll try to ...

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With this particularly special matrix, you can observe that if you look at a particular row $r$, then it is the average of row $r - i$ and row $r + i$ (provided $r - i \geq 1$ and $r + i \leq 5$, so that the rows we're talking about are actually there). More generally, though not necessary, is that row $a$ + row $d$ = row $b$ + row $c$ for $1 \leq a \leq b ... 2 For non-finite-dimensional algebras certainly not. Take$A=k[X]$and$V=k[X],$the regular module. Then$\mathrm{rad} V=0,$basically since$k[X]$is a PID, but$V$is not semisimple (because, for example, all simple$A$-modules are easily seen to be finitely dimensional. However,$A$is not, and if$A$was semisimple, it would be a sum of finitely many ... 0 Hint: $${ 3 \choose 1}+{ 3 \choose 2}+{ 3 \choose 3}$$ 2 The shortcut is a method to (more or less) quickly solve full rank$(n-1)\times n$linear systems (i.e., all$n-1$lines are linearly independent). It is IMO awkwardly presented by writing a the determinant of a matrix in which some entries are "unit vectors" (which makes no sense, matrix entries must be scalars). The proper way to present is it is as ... 0 We have that $$\dim_{\Bbb F}\Bbb F[x]=\infty\;,\;\;\dim_{\Bbb F}N=n+1\implies \dim_{\Bbb F}\Bbb F[x]/N=\infty$$ For$\;\Bbb F[x]/E\;$, we have that $$x+E\;,\;\;(x+x^3)+E\;,\ldots,(x+x^{2n-1})+E\;,\ldots$$ are all linearly independent in the quotient space, since $$\sum_{k=0}^na_k\left((x+x^{2k-1})+E\right)=\overline 0 (=E)\;,\;\;a_k\in\Bbb F\iff$$ ... 0 Your method$(3)$has nothing to do with eigenvectors per se. Instead it is a quick method for solving homogeneous systems of rank$2$in three variables; but it can only be used in this case, namely rank$2$and dimension$3$. Given such a system $$a_{i1}x_1 +a_{i2} x_2+a_{i3}x_3=0\qquad(1\leq i\leq m)\ ,\tag{1}$$ where usually$m=2$or$m=3$, it is easy ... 0 The point of the theorem is the existence of the inner product. It is true (and easy to show) that if you are given an inner product$\langle -, -\rangle$then you can always define a norm via$||x||^2 = \langle x, x\rangle$, and this norm satisfies the parallelogram law and the polarisation identity. But that isn't the situation the theorem is considering. ... 1 Can you find the co-ordinate vectors of all$x\in\mathbb{R}^n$with respect to any basis? For example, the co-ordinate vector of$(1,1,1)^t$with respect to the standard basis is$(1,1,1)^t$, but with respect to the basis$(1,-1,0)^t,(0,1,-1)^t,(-1,0,1)^t$it's something else. If you know how to solve these problems in$\mathbb{R}^n$, then the principle is ... 1 The question can be made a lot clearer by given $$v = at^3 + bt^2 + ct + d$$ find the representation $$v = x_1v_1 + x_2v_2 + x_3v_3 + x_4v_4$$ See that $$v = x_1(1) + x_2(1+t) + x_3(t+t^2) + x_4(t^2+t^3) = x_4 t^3 + (x_3+x_4) t^2 + (x_2+x_3) t + (x_1+x_2)$$ so you need to solve ... 0 From your first set of equations I find$v_A=50-v_B$. From your second set I find$v_A=v_B+10k$(by simple substitution). From these two I find$v_B=25-5k$and$k<5$. Try working from there. 4 It looks like you are computing the cross product of two rows of the matrix from the second method. This will give a vector that is orthogonal to those two rows. An eigenvector will be orthogonal to all rows, since it is in the null space of the matrix. So, if the two chosen vectors span the row-space, the third method gives an eigenvector. 0 Adding to mnz's answer you need to solve a set of linear equations $$P_{\infty}^T T = P_{\infty}^T$$ for the entries of$T$. The problem will most likely be underdetermined, there can be many solutions as pointed out in above comment. At the same time you have to satisfy constraints imposed by the structure of your graph and by the requirements ... 0 Here we have a solution for$n=2$. I don't know how to do for$n>2$. I hope some of these ideas can help. Let us prove this result by induction on$m$. First for$m=3$. Suppose$A_1+A_2+A_3=Id_{2\times 2}$. We may assume that each$A_i$has a zero eingenvalue, otherwise one of these$A_i$is invertible. Now, if each$A_i$has eigenvalues$0$and$1$... 1 Here are some ideas. For the first: To show that this is not a linear subspace, consider$b_4-b_3$. Is it contained in$V_1 \cup V_2$? Remember that$V_1$consists of all linear combinations of certain vectors and the same with$V_2$. This should help you with this question. For the third: Closed under addition is immediate, since if you have two vector of ... 0 The plane has normal vector$n=(-3, - 2 , 2)$. You first find the projection of$u$on$n$, then use$u$minus it to get the direction of the projection line on the plane. 0 Maybe a picture helps (from Wikipedia): If$a=(1,2,x)$,$b=(-2,1,0)$,$c=(1,1,3)$, then this is your parallelpiped. You can convince yourself, that its volume is given by the determinant: $$\left|\det\left(\begin{matrix}1 & -2 & 1\\2 & 1 & 1\\x & 0 & 3\end{matrix}\right)\right|$$ 0 Inner product tells you how much of one vector is pointing in the direction of another one. If e is a unit vector then$<f, e>$is the component of f in the direction of e and the vector component of f in the direction e is$<f, e>e$. The vectors f and e are orthogonal when$<f, e> = 0$, in which case f has zero component in the direction ... 1 Three points define a flat parallelogram, the fourth point will make sure how that parallelogram gets extruded into the third dimension. For the volume you need to derive three vectors, by determing the difference vectors $$u = \vec{PQ} = Q - P$$ between two points each. Just use the point at 0 as second point, so it gets easy and take the absolute ... 1 If$u,v,w$are three vectors in$\mathbb{R}^3$that don't lie on the same plane,$|<u,v\times w>|=|\det(u,v,w)|$gives you the volumn of the parallelepiped. 1 The inner product is not necessarily the best way to think of it; I would argue that orthogonality is the more useful concept. It's nice to know when things are perpendicular, and the inner product provides us with a way of determining that in contexts more general than, say,$\mathbb{R}^N$. Why is it useful? Well, it's nice to be able to decompose vector ... 1 Inner product is a generalization for dot product on a space$X$. And with this generalization, we can define the angle between two elements in$X$and we can talk about perpendicular, projection of one element on the other element's direction, minimum distance and so on. For example, we can define an inner product on$C[0,1]$by$<f,g>=\int_{0}^1 ...

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The essence of the Gauss-Jordan elimination is performing certain updates on the rows of the matrix. For an $m\times n$ matrix, for each row $i=1,\ldots,m$ we 1) scale the row to normalize the leading row entry, 2) update the rows $j=i+1,\ldots,m$. This gives in total $O(m^2)$ row operations on $n$-vectors and hence the complexity is roughly $O(m^2n)$. Of ...

1

$-3$ Is an eigenvalue, as $\text{rank }(A+3I) < 3$. The sum of th eigenvalues is $\text{tr }A = -7$, and their product is $\det A = -6$. So if $a_{1,2}$ are the other eigenvalues, $a_1 a_2 = 2$ (so they has the same sign) and $a_1 + a_2 = -4$ so they are $<0$. So all the eigenvalues are $<0$.

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You're on the right track, simultaneous row and column transformations are the way to go. So, as a first step, we will use the first row to eliminate the entry in the lower left corner: $$\begin{pmatrix} -3&0&-1 \\0&-3&0 \\ -1&0&-1 \end{pmatrix} \to \begin{pmatrix} -3&0&-1 \\0&-3&0 \\ ... 2 The vector perpendicular to your line will be just that, perpendicular to your line; nothing more is guaranteed. To be perpendicular to the plane, you must be perpendicular to all vectors (or lines) in the plane at once; there is only one direction (and its opposite) which does that. The vector you found perpendicular to (2,0,-3) could very well be inside ... 1 Let A=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. Then A^n \neq 0 for all n, A e_2 =e_1, but A^2 e_2 = 0. A is not diagonalizable (it is in Jordan form). Modulo permutations, this is the only Jordan form the matrix can have. A^3 =A^2 means any ... 2 The question asks you to show that A is nilpotent on some subspace. Now, from the assumptions you know that the minimal polynomial divides$$ t^3-t^2 = t^2(t-1). $$Hence 0 and 1 are the only possible eigenvalues. Moreover, the associated Jordan blocks have maximal dimension 2 (eigenvalue 0) and 1 (eigenvalue 1). It remains to determine size ... 0 Hint: You are looking for an element v \in ker(A^2) - ker(A) or you could also look for an w \in im(A) \cap ker(A) - 0. Think about why such an element would yield a solution to your problem. And A^{-1} is only defined for matrices with full rank. 1 1) Show that [LM : K] \le [L : K][M : K]. 2) WLOG let [L : K] = 2. Show that [LM : K]/[M : K] is either 1 or 2 (note: M \subseteq LM). 3) If [LM : K] = [M : K], show that L \subseteq M. Deduce that K = L \cap M = L, contradicting [L : K] = 2. 0 An inner product is a function of two vectors into \Bbb R that satisfies certain properties. You are asked to find a function f((a,b),(c,d)) that satisfies these. You can't have u_1=1 because u_1 and 1 are different kinds of things. You can have f(u_1,u_1)=1, which is what you want. The required linearity is a powerful constraint. Express ... 1 The plane is passing through two points and one line.... (x_1,y_1,z_1) = (4,2,−1) (x_2,y_2,z_2) = (1,1,1) (l,m,n) = (−2,0,3) Cartesian equation $$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 &y_2-y_1 &z_2-z_1\\ l & m & n \end{vmatrix} = 0$$ 3 It is clear that the matrix is symmetric. It suffices, therefore, to show that its eigenvalues are non-negative. Note that the eigenvalues of 11^T are 0 and n. Thus, the eigenvalues of H must be 1-\rho, and (1 - \rho) + \rho n = 1 + (n-1) \rho respectively. Note that both of these must be non-negative. 5 Using Cauchy-Shwartz's inequality we have:$$\begin{align*} \mathbf{x}^\intercal H \mathbf{x} &= (1-\rho)\sum_{i=1}^n x_i^2 + \rho\left(\sum_{i=1}^n x_i\right)^2\\ &\geq \frac{(1-\rho)}{n}\left(\sum_{i=1}^n x_i\right)^2 + \rho\left(\sum_{i=1}^n x_i\right)^2\\ &=\frac{1+(n-1)\rho}{n}\left( \sum^n_{i=1} x_i^2\right)^2\\ &\geq 0 \end{align*} ...

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An easier way is: You already have one vector and one point then: Let $L : B + t\overrightarrow{w}$ where $\overrightarrow{w} = (-2,0,3) \wedge B = (1,1,1), t \in \mathcal{R}$ Let $C = (4,2,-1)$ You look for a plane that contains this line and $C$, now: Let $\overrightarrow{z} = \overrightarrow{CB} = \overrightarrow{B-C}$ Let $H$ be the plane defined by ...

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One valid solution would be $(3,1,2)$. This is incorrect. Not all vector perpendicular to $(-2,0,3)$ gives you the normal vector of the plane. The correct way is pick a point on the line, for simplicity say $(1,1,1)$ then the normal vector of plane is perpendicular to $(-2,0,3)$ and $(4,2,-1)-(1,1,1)=(3,1,-2)$. Hence you need to calculate $n=(-2,0,3)\times ... 7 It is incorrect, the$0\$ matrix is symmetric but not invertable.

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