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## New answers tagged linear-algebra

0

I'll do the first one for you. The second is done in exactly the same way. Here I'll use your two axioms, plus one more: $k\langle x,y\rangle = \langle kx, y\rangle = \langle x, ky\rangle$ where $k\in \Bbb R$ and $x,y\in V$. \require{enclose}\begin{align}\langle u-5v,3u-2v\rangle &= \langle u-5v, 3u\rangle + \langle u-5v, -2v\rangle \\ &= ... 0 I assume your definition of the convex hull of A, is the definition you gave for C. If this is true, then the convex hull is unique and so saying "a" convex hull makes no sense. A better way to argue is to assume B\supseteq A is a convex set and show C\subseteq B. 0 No, your argument is circular. The confusion seems to stem from talking about "a" convex hull. But the convex hull is unique. In a successful proof, you'll use the assumption that B is a convex set. What does this assumption mean, by definition? How does it relate to C? 0 The matrix can be written as A+bI, where A is the matrix with all rows equal to a_1,a_2,\dots,a_n. The determinant in question is (-1)^n\chi_A(-b), where \chi_A is the characteristic polynomial of A. Since A has rank 1, we have \chi_A(x)=x^n-tr(A)x^{n-1} (see this for instance). Finally, the determinant in question is ... 0 For n \ge 2 using Laplace expansion on the last row gives \begin{align} f_n &= \begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-3} \\ & & & c_{n-3} & a_{n-2} & b_{n-2} \\ & & & & c_{n-2} & a_{n-1} & b_{n-1} \\ & ... 0 Write the matrix as A+bI. Here, all the rows of A are the same, and so A is rank 1, and therefore the kernel is of dimension n-1 and there is only one non-trivial eigenvalue, \operatorname{tr}(A). Therefore the characteristic polynomial of A is p_t(A)=\det(tI-A)=t^{n-1}(t-\operatorname{tr}(A)). It is now straightforward to calculate ... 0 Set A=\sum\limits_{i=1}^na_i. By multilinearity, \begin{align*} &\begin{vmatrix} a_1+b &a_2&\dots&a_n\\ a_1&b+a_2 &\dots&a_n \\ \vdots&\vdots&&\vdots\\ a_1 a_2&\dots &a_n+b \end{vmatrix}= \begin{vmatrix} A+b &a_2&\dots&a_n\\ A+b&b+a_2 &\dots&a_n \\ ... 2 In either case, we have \phi(x + y) = \phi(x) + \phi(y), but the difference is under what conditions we can pull scalars out; for what scalars k we are guaranteed to have \phi(kx) = k\phi(x). For a \Bbb C-linear map, this happens whenever k \in \Bbb C, and for an \Bbb R-linear map, this happens whenever k \in \Bbb R. Source: Remmert's Theory ... 5 For example, \mathbb C is both a real vector space and a complex vector space. The "complex conjugate" map z \mapsto \overline{z} is real-linear but not complex-linear. explanation Of course it is additive: \overline{z+w} = \overline{z}+\overline{w} $$and real scalars are OK$$ \overline{rz} = r\overline{z}\qquad\text{if $r$ is real} $$But ... 1 The recurrence is obtained by developing the determinant along the last column (or, equivalently, along the last row). 1 A linear map is a map between vector spaces (over the same field) that preserves the vector operations of scalar multiplication and vector sum. The names r-linear and c-linear are just a reminder of what field we are looking at. 0 The structure of this matrix aloud to write this equation which does not generally hold$$det(A+Ib)=b^{n-1}(tr(A)+b)$$I guess that's related with the fact that$$A^2=tr(A)A$$but just now I don't see how... If someone can see just edit in the comments please 0 Yes. (There is a minimum of 30 characters) 1 if f is a polynomial then you have:$$f(x)=a_nx^n+...+a_1x+a_0$$Then you have$$f(P^{-1}AP)=a_n(P^{-1}AP)^n+...+a_1(P^{-1}AP)+a_0I$$which is$$f(P^{-1}AP)=a_n(P^{-1}APP^{-1}AP...P^{-1}AP)+...+a_1(P^{-1}AP)+a_0P^{-1}IP$$or$$f(P^{-1}AP)=P^{-1}a_nA^nP+...+P^{-1}a_1AP+a_0P^{-1}IP$$which finally gives ... 1 The classic way of solving such a question (for polynomials) is to group a, b and c by their coefficients.$$T(a,b,c) = (a−b−2c)x^2 + (2b+c)x + (a+b−c)= a(x^2+1) +b(-x^2+2x+1) +c(-2x^2+x-1)$$Cool! We managed to find an expression of any polynomial belonging to the image or range of T in terms of three polynomials. So all that is left is to trim ... 1 Hint: Properties 1 and 2 will hold regardless of your choice of t. On the other hand, positivity will hold if and only if for any non-zero p \in P_2, we have \langle p,p \rangle > 0. For which t will this be the case? 1 Hint: In vector notation:$$ T \begin{bmatrix} a\\b\\c \end{bmatrix}= \begin{bmatrix} p_1\\p_2\\p_3 \end{bmatrix}= \begin{bmatrix} a-b-2c\\ 2b-c\\a+b-c \end{bmatrix}=(a-b-2c)x^2+(2b-c)x+(a+b-c)1 $$Where \{x^2,x,1\} is the canonycal basis of P_2. Now note that p_3=p_1+p_2 so two linearly independent vectors span the range of T. 0 Use induction over the dimension of the matrix. In the induction step evaluate the determinant. You will get subdeterminants inside that are of the same form 1 The change of basis doesn't preserve the metric so there is no reason for TST^{-1} to be an isometry. TST^{-1} is an isometry for the metric d \circ (T^{-1},T^{-1}) where d is the classical metric. In general, TST^{-1} is an isometry iff T is a scalar mutiple of an isometry, or equivalently, iff T preserves the equality of distances. (in the ... 3 If all matrix entries are integers and the solution over \Bbb R is also integer than it is clear that the equations that express the fact that this is a solution remain true when reducing modulo some p. In a suitable sense this is even true for rational solutions (with denominator not a multiple of p). However, there may be additional solutions if ... 0 To me it is the same linear map. It's just a change of the representation base for the endomorphism. So according to the original base it is still the orthogonal or unitary endomorphism. For the definition of an isometry by a matrix the rows/cols have to be orthonormal or more general a endomorphism is orthogonal or unitary if, and only if its matrix ... 0 An easy way to get an arrangement by hand is to group the marbles into two groups of twelve, say the yellow+green and everything else. We can make an arrangement of the two groups that meets your requirement a by doing a checkerboard. There are twelve light squares and twelve dark squares. Now distribute the yellow+green any way you like on the (say) ... 4 Yes, it is a type of Circulant Matrix. 3 Suppose e_k are orthonormal, then \|f-\sum_k \alpha_k e_k \|^2 = \|f\|^2 -2 \sum_k \alpha_k \langle f, e_k \rangle + \sum_k \alpha_k^2. To find the best approximation, we minimise the above over the \alpha_k. This is a convex quadratic separable problem, and the solution is given by just differentiating with respect to the \alpha_k. This gives ... 3 Counterexample: (using your S)$$ S = \pmatrix{0&-1\\1&0}, \quad T = \pmatrix{2&0\\0&1} \implies\\ TST^{-1} = \pmatrix{0&-2\\1/2 & 0} $$1 Since you seem to be unsure whether your matrix is 2x2 or 3x3, here's the answer for the 2x2 case, for matrix A = \pmatrix{a&b\\c&d}: Find the determinant det(A) using Cramer's rule. In the 2x2 case, this is ad-bc. A is invertible iff det(A) \neq 0. The inverse of A is given by \frac1{det(A)} \pmatrix{d&-b\\-c&a}. 1 Since you're stating "along the 3rd row", I'll assume you want to use the method using the "submatrices" : So we end up with :$$(-1)^0\cdot3\cdot \det\left(\begin{bmatrix}1+i&2\\0&1-i\end{bmatrix}\right) + (-1)^1\cdot 4i \cdot \det\left(\begin{bmatrix}0&2\\-2i&1-i\end{bmatrix}\right) + (-1)^2\cdot 0 \cdot ...

0

Yes, the simplest example is $U=0$. Just take $v\in V$, $v\ne0$ and consider $m=v, n=-v$.

2

Yes, consider $V=k^2$ as a $k$-vector space for a field $k$ and consider the subspace $k \times \{0\}$. Then $(0,1)$ and $(0,-1)$ are not in that subspace but their sum is.

3

Take $A$ to be an invertible matrix .

1

A very important identity that you may have seen in your course so far is $$\text{Im}(M)^{\perp}=\text{Ker}\left(M^{T}\right).$$ This is true because if $\vec{y}\in\text{Im}(A)^\perp$ then by definition it satisfies $\langle\vec{y}\cdot M\vec{x}\rangle=0$ for all $\vec{x}$ and so $$\langle\vec{y}\cdot ... 1 Since A is symmetric, there exists an orthonormal basis v_1, \ldots, v_n of eigenvectors of A. Write Av_i = \lambda_i v_i. Since A is positive definite, \lambda_i > 0. Show that$$ \{ x \in \mathbb{R}^n \, | \, ||x||_A = 1 \} = \{ a_1 v_1 + \ldots + a_n v_n \, | \, \lambda_1 a_1^2 + \ldots + \lambda_n a_n^2 = 1 \} $$which means that \{ x ... 0 In the example you have A=PLU, i.e. PA=LU so PA is not tridiagonal anymore. However, PA is still banded matrix and can be effectively solved. The idea is very similar to the tridiagonal case. This is still an LU based algorithm, but concentrated to run inside the band. The efficiency is therefore depends on the width of the band, i.e. a distance ... 0 This looks good to me, and I think you can exploit the first few lines of your argument, which proves (if we replace l by arbitrary 0 \leq k \leq n) that the group GL(W) acts transitively on the space Gr(k, W) of k-dimensional subspaces of W, and which in particular says that for any g \in GL(W) the map \Phi_g : Gr(k, W) \to Gr(k, W) ... 1 Hint: If B=P^{-1}AP, what is B(P^{-1}A^{-1}P)? 0 If B = P^{-1}AP and x is an eigenvector of A with eigenvalue \lambda, then$$B(P^{-1}x) = P^{-1}A(PP^{-1}x) = P^{-1}Ax = \lambda P^{-1}x.$$1 A, B similair \implies there exists a matrix P such that B = P^{-1}A P Now, using the fact that det(P^{-1}) = \frac{1}{det(P)} and det(AB) = det(A)det(B), det(B) = det( P^{-1}A P)= det(P^{-1})det(A)det(P) = det(A) Since A is invertible then det(A) \neq 0. Thus B is invertible. 1 The fact that P^{-1}, A, and P are invertible means their product P^{-1}AP is invertible. Why is this true? What is a candidate for the inverse of this matrix? 0 Given u^TBv = u^TCv \forall u,v\in U\subset\mathbb{R}^3 Consider A=(B-C), let v\in U, then$$\|Av\|_2^2 = (Av,Av)=v^tA^TAv=0$$we got that \|Av\|_2=0,\forall v\in U, therefore A=0 and hence B=C. This could only be true for an open subset (of \mathbb{R}^3 in this case), because otherwise U could be a discrete set, e.g. subset of the ... 0 Clearly B_{il} isn't a_{il} but a_{jk}. Hence$$ e_{ij}Ae_{kl}=B=a_{jk}e_{il}, $$i.e. e_{ij}Ae_{kl} is the matrix whose entry at position (il) is a_{jk} and whose other entries are zero. 2 OK, Let us say that the eigenvalue of A is \lambda and then call the corresponding eigenvector x_{\lambda} and hence$$Ax_{\lambda}=\lambda x_{\lambda}$$and observe that$$x_\lambda ^TA{x_\lambda } = \lambda x_\lambda ^T{x_\lambda } = \lambda (1) = \lambda $$so you should find the maximum eigenvalue of A. So the maximum value of the quadratic ... 2 If x\neq 0 is an eigenvector of A with unit length and associated eigenvalue \lambda, then x^TAx=x^T\lambda x=\lambda x^Tx = \lambda. So the maximum value of x^T Ax where the maximum is taken over all x that are the unit eigenvectors of A is simply the value of the largest eigenvalue of A. As pointed out by @Ant, the quantity x^TAx arise ... 3 This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient 1$$x^TAx=\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$2 I think the reason is that the quadratic form associated with any matrix A is equal to the quadratic form associated to its symmetric part. To prove this, first consider that any matrix can be written as the sum of symmetric and anti-symmetric matrices as follows$${\bf{M}} = {1 \over 2}\left( {{{\bf{M}}^T} + {\bf{M}}} \right) + {1 \over 2}\left( ...

1

Maybe we need to take it step by step $-2Z^T(y-Z\beta)+2\lambda \beta = 0$ $-2Z^Ty + 2Z^TZ\beta + 2\lambda\beta = 0$ $2Z^TZ\beta + 2\lambda\beta = 2Z^Ty$ Now when you factor the left hand side and know that $\lambda$ is a scalar and $\beta$ is a vector, the only matrix which can give element wise multiplication like that is $\lambda I$.

0

The cardinality of $P$, i.e. $|P|=2^n$ is a number of possible different combination of a binary (components $0$'s or $1$'s) vector of a length $n$. $W$ has at most $m$ basis vectors from $P$ (it is not stated that it generated by $m$ linearly independent elements from $P$, so it could be less) therefore its cardinality cannot exceed $2^m$. Since the ...

0

For a vector norm $\|\cdot\|_\star$ (e.g. $\star=1,2,\infty$) one define an induced matrix norm by $$\|A\|_\star=\max_{\|x\|_\star\ne 0}\frac{\|A x\|_\star}{\|x\|_\star}$$ for any finit dimentional matrix $A$. it can be proved from that the following is equivalent definition $$\|A\|_\star=\max_{\|x\|_\star=1}{\|A x\|_\star}$$ Let $x^*$ be a vector such that ...

2

If I understand correctly, you want to find $x$ such that $x^HAx = B$, where $B \geq 0$ is a real number. Let $v_1, \ldots, v_4$ be an orthonormal basis of eigenvectors of $A$ with $Av_i = \lambda_i v_i$. Write $x = a_1 v_1 + \ldots + a_4 v_4$. Then you have $$x^HAx = \left< Ax, x \right> = \sum_{i=1}^4 \lambda_i |a_i|^2 = B.$$ This gives you all ...

1

Using $1=pr+qs$ one can easily deduce $\operatorname{ker}(p(A)) \subset \operatorname{Im}(q(A))$, thus we get $$n = \operatorname{rank}(p(A)) + \dim \operatorname{ker}(p(A)) \leq \operatorname{rank}(p(A))+\operatorname{rank}(q(A)).$$ This proof gives you for free: Equality holds iff $\operatorname{ker}(p(A)) \supset \operatorname{Im}(q(A))$ iff ...

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