# Tag Info

-1

A $4\times 3$ matrix will have a nullity (dimension of its nullspace) + rank(dimension of its column space AND also the dimension of its row space (they both have the same dimension)) = to $3$ (the dimension of its domain). A line is $1$ dimensional, so if the nullspace has dimension $1$ (and is thus a line) the row and column spaces will each have dimension ...

4

Hint: by the rank-nullity theorem, the number of columns of the matrix equals the sum of the dimension of the nullspace and the dimension of the column space. What does this say about the first question? For the second question, play around with some simple $2 \times 4$ matrices.

0

I suggest that you find a basis for $\psi(\mathbb{R}^4) = T$. This could be done by simply calculating $\psi(e_1), \dots \psi (e_4)$ and then removing one by one those of them which breaks the linear independence. When you have the basis of T, it's time to calculate $T \cap W$. One way to do this is by solving several systems of equations. Namely you can ...

0

The simplest thing I can think of is to take the QR decomposition of $A=BC$, then $\det(M)$ is simply the square of the product of the diagonal elements of $R$.

2

Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix: \begin{align*} |M| &= -\begin{vmatrix} \sin x & \cos x & \sin 2x & \cos 2x \\ \cos x & ...

1

Hint: Add $4$ times the first row to the third and $4$ times the second row to the fourth. That should get you started.

1

There is no simple formula for the derivative of the exponential of a matrix valued function. However, there is a well known integral representation of the derivative. $$\frac{d}{dt} e^{X(t)} = \int_0^1 e^{\lambda X(t)} \left(\frac{dX(t)}{dt}\right) e^{(1-\lambda)X(t)} d\lambda$$ Look at the wiki page for matrix exponential, there are some references for ...

2

A standard application in undergraduate physics is to find the principal axes of inertia of a solid. For any solid object, we can construct an inertia matrix $I_{ij}$, which is a 3x3 symmetric matrix describing how the object's mass is distributed in space, with respect to a specific Cartesian reference frame $\{x_1,x_2,x_3\}$. Its diagonal elements ...

1

If $X(t)X'(t)=X'(t)X(t)$, for all $t$, then $$\frac{d}{dt}\mathrm{e}^{X(t)}=\mathrm{e}^{X(t)}\frac{dX(t)}{dt}.$$ If not, then the above does not hold.

0

Every symmtric matrix is always diagonalizable and eigenvalues of symmtric matrix are always real so 5 is most correct.

0

I'm using matrix operations when work with neural networks. It helps to represent weights and inputs of neural network as matrix which also give me the way to parallel computations at different processors or cores of processors. Before usage of matrix operations I didn't know how to spread calculations at neural networks in parallel mode

2

This is the transpose of the dual unit $\epsilon$. So if you represent your duals by $\begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix}$ it will not correspond to any "real" number.

1

I think, that the claim is not true. If you really allow all $n\times n$-matrices with integer coefficients, then it is obvious that the coefficients can be arbitrary large odd integers. Because you are mentioning that $\det(M)=\pm 1$, I suppose that you mean $M\in GL_n(\mathbb{Z})$. But then let $M$ be $-I_n$. Then $-1=m<n$. Also, for $n\ge 2$ it is ...

1

Hint: $$\det\begin{pmatrix} \varepsilon a_{1,1} + 1 &\varepsilon a_{1,2} & \cdots & \varepsilon a_{1,n} \\ \varepsilon a_{2,1} &\varepsilon a_{2,2}+ 1 & \cdots & \varepsilon a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ \varepsilon a_{n,1} & \varepsilon a_{n,2} & \cdots & \varepsilon ... 0 The matrix represents a linear application \Bbb R^6\to\Bbb R^3. Since the rank is 3, the dimension of the image is 3; then, the application is surjective, and the systen Ax=b has always solution. Now, the dimension of the kernel is 6-3=3, so the dimension of a space of solutions is also 3. 1 Since the square of your matrix is 0, I would say that your matrix is \epsilon. Note that the representation given in the Wikipedia article you linked to is not the only representation using 2 \times 2 real matrices. 0 To check if it is a metric, first you should define what ||A|| means; it is of course a matrix norm, but which one? Unfortunately not all norms work; for example, to satisfy the property d(A, B ) = 0 \iff A = B we need ||I|| = 1 but this is not true for all the matrix norms. Anyhow you're book seems to assume that your matrix already satisfy the ... 1 You could use the identity that as t\rightarrow 0, (see Equation 4, here)$$\det(I+tv)=1+t\cdot tr(v)+O(t^2)\sim1+t\cdot tr(v)$$Thus we have$$\lim_{t\rightarrow 0}\frac{\det(I+tv)-1}{t}\sim\frac{1+t\cdot tr(v)-1}{t}=tr(v)$$The above identity is obtained using the derivative of the determinant and a Taylor expansion, where (from here) ... 1 Ax=\lambda x\Rightarrow (pId+qA)x=px+qAx=px+q\lambda x=(p+q\lambda)x -2 Ok, just try it with a diagonal matrix:$$ A+Ix = diag(a_{11}+x,\ldots, a_{nn}+x)$$then$$ \det(A+Ix)=\prod_{i=1}^n (a_{ii}+x)$$Then the derivative is for example if n=2$$ a_{22}+a_{11}$$Then you can generalize for other matrices. 0 Convince yourself that if \;\{w_1,..,w_n\}\; is any orthonormal basis, then for any \;v\in V\; we have the easy expression$$v=\sum_{k=1}^n \langle\,v\,,\,w_k\,\rangle\,w_k$$Thus, in our case we have that$$\forall\;1\le i\le n\;,\;\;v_i=\sum_{k=1}^n\langle\,v_i\,,\,u_k\,\rangle\,u_k\implies A=\left(\langle\,v_i\,,\,u_j\rangle\right)_{1\le ...

6

The matrix $A = \begin{bmatrix}i&1\\1&-i\end{bmatrix}$ is (complex) symmetric but has Jordan form $A = VJV^{-1}$ where $J = \begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $V = \begin{bmatrix}i&1\\1&0\end{bmatrix}$. So, not every (complex) symmetric matrix is diagonalizable. The rotation matrix $R = \begin{bmatrix}\cos\theta & ... 0 A spanning set in$W$is a collection of elements in$W$which spans$W$, i.e. a family$f_1, \dots, f_n \in W$such that each$x \in W$can be written as$x = a_1 f_1 + \dots a_n f_n$for real coefficients$a_i$. The spanning set is minimal if you can't find a spanning family with fewer vectors. The number of vectors in a minimal spanning set of$W$is the ... 16 This is a simpler example, but maybe that'll be good for undergraduate students: Linear algebra is a central tool in 3-d graphics. If you want to use a computer to represent, say, a spaceship spinning around in space, then you take the initial vertices of the space ship in$\mathbb{R}^3$and hit them by some rotation matrix every$.02$seconds or so. Then ... 4 One of the examples my students have absolutely loved in the past is Hill cipher. It is a "real" application although outdated but the students do love playing with it. Using some sort of a encoding, convert a message to a bunch of numbers, and then a key matrix is chosen. The message can be organized into an array and multiplied by the key for encryption. ... 2 I like the Google Page Rank and Adjacency Matrix points. Linear Algebra is a deep subject that is readily connected to computer science, graph theory, and combinatorics in unexpected ways. The traditional connection is with numerical analysis. However, Linear Algebra is closely related to graph theory. There is a field known as algebraic graph theory which ... 3 I worked as a software engineer for 27 years for a large Defense corporation. They used Finite Element software tools to model spacecraft designs for stress tests, amount of construction material required, simulated launch testing, etc. Finite Element theory is based on a matrix of vectors that describe the connections and forces on elements of a structure. ... 5 Stoichiometry (not that our students would ever stoop to linear algebra for it) is a very elementary place it shows up. Quantum mechanics is an advanced place. Linear programming is ubiquitous in business applications, as is game theory in economics and political science, and a lot of game theory is based on linear algebra and Markov processes. Least squares ... 6 Multiplication of a graph's adjacency matrix can be used to calculate the number of walks of length$n$from one vertex to another. In particular: Proposition. For any graph formed of vertices connected by edges, the number of possible walks of length$n$from vertex$V_i$to vertex$V_j$is given by the$i,j^\text{th}$entry of$A^n$, where$A$is the ... 2 Using Vandermonde matrices, one can show that for any$k$and any$n$, there exists$k$points in general position in$\Bbb R^n$. Indeed, given$k$(assuming$k>n$), pick$k$distinct real numbers and consider${\bf v}_i=(r_i,r_i^2,\ldots,r_i^{n})$for$i=1,\ldots,k$and use Vandermonde's determinant to prove the claim. 7 The restricted isometry property (RIP) of matrices is something not too hard for undergraduates to understand: it means that a (rectangular) matrix$A$satisfies $$(1-\delta) \|x\| \le \|Ax\|\le (1+\delta)\|x\| \tag{1}$$ for all vectors$x$with at most$s$nonzero components. The constant$\delta$should be small, and of course independent of$x$. The ... 2 If we only allow$(\pm 1, 0)$or$(0, \pm 1)$as steps, $$A = \pmatrix{1&0&0\\0&0&0\\0&0&1}$$ suffices as a counter-example (no permutation will move the two ones into the same row or column). Allowing diagonal steps$(\pm 1, \pm 1)$should work make the conjecture true, but I don't know how to prove it. (thinking about this at the ... 2 Anything to do with scheduling and maximising linear systems: an airline scheduling planes and pilots to minimise the time airplanes are stationary and pilots are just sitting around waiting is an example. Linear optimisation saves millions if not billions of dollars each year by allowing companies to allocate resources optimally, and it's basically an ... 9 We can also use Linear Algebra to solve Ordinary Differential Equations An ODE is of the form $$\underline{u}'(t) = A \underline{u}(t) + \underline{b}(t)$$ with$A \in \mathbb{C}^{n \times n}$and$\underline{b}(t) \in \mathbb{C}^{n \times 1}$. If we have an initial condition $$\underline{u}(t_0) = \underline{u_0}$$ this is an initial value problem. ... 22 Another very useful application of Linear algebra is Image Compression (Using the SVD) Any real matrix$A$can be written as $$A = U \Sigma V^T = \sum_{i=1}^{\operatorname{rank}(A)} u_i \sigma_i v_i^T,$$ where$U$and$V$are orthogonal matrices and$\Sigma$is a diagonal matrix. Every greyscale image can be represented as a matrix of the intensity ... 32 I was a teaching assistant in Linear Algebra previous semester and I collected a few applications to present to my students. This is one of them: Google's PageRank algorithm This algorithm is the "heart" of the search engine and sorts documents of the world-wide-web by their "importance" in decreasing order. For the sake of simplicity, let us look at a ... 1 The matrix $$A = \frac{1}{\sqrt 3} \left( \begin{array}{ll} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \\ \end{array} \right)$$ is unitary. This is an instance of the more general Fourier Matrix, which answers your question for unitary matrices of arbitrary size. Note that these matrices are unitary and ... 0 There is a factorization for every square matrix$A$in$\mathbb{R}^{n \times n}$in the form$A=S_1S_2$where$S_1$and$S_2$are real symmetric matrices and one of them is necessarily nonsingular. Reference: Problem 12 of Chapter 9 of "Matrix Analysis for Scientists and Engineers", Alan Laub. 6 Using the identity $$\sin x=\cos\left(x-\frac\pi2\right)$$ we see that the given matrix is a matrix of rotation of angle$\frac{4\pi }{9}$hence the answer is$9$. 2 Here's a hint (assuming you are indeed missing a minus sign):$\sin \frac{\pi}{18} = \cos \frac{4\pi}{9}$. 1 It means that if you have two matrices:$\begin{bmatrix} a&b\\c&d \end{bmatrix}=\begin{bmatrix} h&i\\j&k \end{bmatrix}$then it must be the case that$a=h, b=i, c=j, d=k$. In your case, $$\begin{bmatrix}\mathbf{x}_1 + 3\mathbf{x}_3 & \mathbf{x}_2 + 3\mathbf{x}_4\\\mathbf{x}_1 + \mathbf{x}_3 & \mathbf{x}_2 + ... 2 A general linear algebra result says that for S\succeq 0, R\succeq 0 such that R-S\succeq 0, we may infer that |S|\leq |R|. Apply this with$$ S\equiv C^TC-C^TB(B^TB)^{-1}B^TC\succeq 0,\quad R\equiv C^TC\succeq0,\quad R-S\succeq0. $$0 To reinforce Laars Helenius's answer. Suppose you have the n \times n identity matrix:$$\begin{bmatrix} 1 & 0 & \cdots &0 \\ 0 & 1 &\cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1\end{bmatrix}$$If we want to find all permutation matrices we can think as following: The 1 in first ... 2 The square n\times n permutation matrices are in one-to-one correspondence with elements of the symmetric group on n elements. Since there are n! distinct permutations, then there are n! permutation matrices. 13 I'll assume your question is something like "Given A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 2 \\0 & 0 & 0\end{pmatrix}, what is e^{At}?" The definition of a matrix exponential is e^{tA} = \displaystyle\sum_{n = 0}^{\infty}\dfrac{t^n}{n!}A^n = I + tA + \dfrac{t^2}{2!}A^2 + \dfrac{t^3}{3!}A^3 + \cdots. In your title, you ... 1 Using the idea of Maestro Will: Necessarily \det(A-I) divides \det(B-I)\not= 0 or \det(B-I)=0. We calculate the Smith normal forms A-I=SDT,B-I=U\Delta V. Thus (U^{-1}XS)D(TYV^{-1})=\Delta. It suffices to solve the equation XDY=\Delta. where d_i divides d_{i+1} and \delta_i divides \delta_{i+1}. An instance without any solutions: ... 0 Note that "A, B hermitian and diagonal" is not enough for M to exist. In fact, M^{-1}AM=B implies that A and B have the same eigenvalues with the same multiplicities. In other words, the diagonal of B should be a permutation of the diagonal of A. So you can take M to be the permutation matrix that permutes the diagonal of A into the ... 0 The image of A consists of all columns of the form Ab, for whatever b you want. Since A^2b=A(Ab), you get that any vector in the image of A^2 is also in the image of A. No, it's not generally true for infinite dimensional spaces. For example, an infinite dimensional Hilbert space has a maximal orthogonal system such that the span is dense, so no ... 3 Hint If \lambda is an eigenvalue of A associated to the eigenvector v, then$$A^mv = A^{m-1}Av = A^{m-1}\lambda v = \ldots = \lambda^m v,$$i.e. \lambda^m is an eigenvalue of A^m. Now, can you find the eigenvalues of A? 1 Use the usual method$$\det (A - \lambda I) = 0$$that is$$\begin{vmatrix}2-\lambda & 0 & 0\\ 0 &2- \lambda & 4 \\ 0 &-1 &2 - \lambda \end{vmatrix} = (2 - \lambda)\begin{vmatrix}2- \lambda & 4 \\ -1 &2 - \lambda \end{vmatrix} = (2 - \lambda)\Bigg((2 - \lambda)^2 + 4\Bigg) = 0$\$ You can take it from here.

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