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0

I do not know what you did for $B\cdot A$, but it was no matrix multiplication: $$ (BA)_{ik}=\sum_{j=1}^n b_{ij} a_{jk} $$ but $B$'s column index $j$ would range from $1$ to $3$ while $A$'s row index $j$ can only take the value $1$, so this will not work as common ranges are needed.


0

It sounds like the grader wants you to use the transpose of each matrix. In general, $$(AB)^T=B^TA^T$$


0

In general, matrix multiplication is not commutative, so $$A\cdot B \neq B \cdot A$$ this means that you cannot arbitrarily choose one product over another. You must choose the right product given the context of the situation, even if both products give the same numbers. In your specific situation, it looks like you are computing the transpose of the ...


0

You are actually allowed column operations in this situation, because $\mathbb{Z}^{3}$ admits an automorphism permuting the three copies of $\mathbb{Z}$; thus $\mathbb{Z}^{3}/<(13,9,2), (29,21,5), (2,2,2)> \simeq \mathbb{Z}^{3}/<(9,13,2),(21,29,5), (2,2,2)>$ and so on. You normally don't want to use column operations on a matrix, because it is ...


2

Hint (For (b).) Suppose that there are (at least) two nonpositive eigenvalues, and let $\Bbb W$ denote the span of their eigenspaces. What can we say about $v^T A v$ for $v \in \Bbb W$?


7

I guess it's easier to go for a reduction formula. I proceed along the generalization mentioned in comment: Call the determinant $M = M_n(x_1,x_2,\cdots, x_n)$ $$\displaystyle \begin{align}M &= \left|\begin{matrix}\dfrac{1}{x_1+x_1} & \cdots & \dfrac{1}{x_1+x_n}\\ \dfrac{1}{x_2+x_1} & \cdots & \dfrac{1}{x_2+x_n}\\ \cdot & \cdot ...


6

We consider a general case. Let $$ |A|=\left|\begin{array}{}{\dfrac1{a_1+b_1}\cdots\dfrac1{a_1+b_n}\\ \vdots\hspace{20 mm} \vdots \\\dfrac1{a_n+b_1}\cdots\dfrac1{a_n+b_n}} \end{array}\right| $$ In your case, just set $a_i=i,b_j=j$. By multiplying $i^{th}$ row with $\prod\limits_{j=1}^{n}(a_i+b_j)$ each, we have $$ ...


1

The most general name for it is a change of basis matrix. Since $A$ is diagonalizable, its eigenvectors span its domain, so every vector in its domain can be written as a linear combination of eigenvectors. Thus, the action of $A$ can be described by its action on each of the eigenvectors (which is multiplication by the eigenvalues). So if we choose to ...


0

For the transition matrix from $\alpha$ to $\beta$: note that $$ \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix} = 1 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 1 \cdot \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + (-1) \cdot \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0 & 0 \\ 0 & 1 ...


0

I believe the lower bound on the eigenvalues is 0. The eigenvalues of $\Gamma=\left[\begin{array}{cc} I & B \\ B^{*} & I \end{array} \right]$ are $1+\beta$ where $\beta$ is an eigenvalue of $B$. Every eigenvalue has a multiplicity of 2 (or, rather, twice the multiplicity of the corresponding eigenvalue of $B$). The eigenvectors are ...


0

Consider sequence $A067689$ in the website oeis.org . This sequence consists of the reciprocals of the functions of $n$ that you seek. These achieve astronomical values very quickly. For example, for $n=9$, the function yields a value of just over 50 quattourdecillion. Why not be satisfied with entering the first few values in a table and referring to them ...


0

Unless the smallest eigenvalue of $A$ is $\ge $ largest eigenvalue of $B$ you can do the same trick with diagonal matrices that you did above. Hence the condition is : $\lambda_n(A) \ge \lambda_1(B)$.


0

I assume A is quadratic(otherwise the term in your question does not hold) If you know that: any symmetric matrix $A$ can be decomposed to $A=LU$ $\nabla Ax=A$ [and $\nabla x^TA = \nabla(A^Tx)^T=(\nabla A^Tx)^T=(A^T)^T=A$] The product rule holds: $\nabla uv=(\nabla u)v+u\nabla v$ You can write: $$\nabla x^TAx \\ =\nabla x^TLUx \\ =(\nabla x^TL) Ux + ...


0

My guess for the meaning of the question: Assume $w_1=0.4, w_2=0.5, w_3=0.1$ and matrix $X$ is $$\begin{pmatrix} 7 &4 &9 \\6 &4 &12 \\3 &2 &17 \end{pmatrix}$$ then compute $$ \sum_{i=1}^3\sum_{j=1}^3 w_i w_j X_{ij} . $$ Is that what you mean? If so, in this group, you should show your work before expecting us to help you.


1

It can be easily proved with the generalisation of Laplace's formula to the expansion of a determinant along a group of columns – here along the $n_1$-first columns. For any minor made up of elements in the first $n_1$ columns and in a subset $H$ of $n_1$ rows, one associates the complementary minor, made up of elements in the remaining $n_2$ columns and ...


1

You're applying Laplace's formula as if it applied to blocks, but it applies to elements. (This is also reflected in the notation; usually lowercase letters are used for matrix entries and uppercase letters for matrices.) If Laplace's formula held for blocks (with the determinant of the block where you're using the block itself), then what you're trying to ...


1

You can see that the sup norm of matrix A, i.e., $\max\{a_{ij}\}$, is 3/4. Then $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a contraction. Since $R^n$ is a complete metric space there is a unique $y\in \mathbb{R}^n$ such that $Ay=y$ and we have $y=\lim_{n\to \infty}A^nz$, for every $z\in \mathbb{R}^n$, in particular for the given $x$. So, find y such that ...


1

We can simplify GFR's approach and go further: let $X$ be the matrix $$ \left( \begin{matrix} 0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0 \end{matrix} \right) $$ Then $$M = I + X + X^2 + X^3 = \sum_{k=0}^{\infty} X^k = (I - X)^{-1}$$ Consequently, $$M^n = (I-X)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} (-X)^k = ...


1

Go with Gilbert Strang's Linear algebra. He also has video lectures here. If you have never encountered matrices before, or want to revise, you should start with this book by Eves


0

If $X$ is invertible, then $I-O=(X-OX)X^{-1}$ and: $$\|I-O\| \leq \|X-OX\| \|X^{-1}\| \leq \epsilon\|X^{-1}\|$$ If $X$ is not invertible, take $\|X^{-1}\|=\infty$. I don't think it's possible to get a bound in terms of $\|X\|$. Consider the $2\times 2$ case with $O=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$ and $X=\left[ ...


0

The book "Linear Algebra" by Hoffmann and Kunze is a really good book. However, you can try "Linear Statistical Inference" by C.R. Rao as well. It is a pretty short encyclopaedia on a lot of topics in matrix algebra, although it is difficult to read and has some really hard problems. In fact, you can go through "Matrix Algebra from a Statistician's ...


0

Linear Algebra by Hoffman Kunze is a great book for beginners. Then if you want to learn more about algebra I would suggest : 1)Artin's Algebra 2)Abstract Algebra by Dummit and Foote


0

If the points are nearly coplanar (co-hyperplanar?), then the problem is ill-conditioned. No algorithm you use will give you a "robust" answer in floating-point arithmetic. You can try this for yourself with nearly colinear points in the plane. That said, there are approaches that are better than others. For example, QR factorization using Householder ...


3

Since $ (AB)^T=B^TA^T$ and $(A^T)^T=A$, for all matrices $A,B$ we have: $$ Y^T=\left(Axx^TA^T\right)^T=\left(x^TA^T\right)^T\left(Ax\right)^T=Axx^TA^T=Y $$


1

If you only consider the states $\{Q_k\}$ then with respect to those alone $Q_k$ is not a DPMC. If it were, then we could determine the distribution of the next state $Q_k$ solely in terms of the current state $Q_{k-1}$. The state $Q_k$ is determined by $$Q_k = (Q_{k-1} - 1) + V_k$$ At step $k$, $G_1^m, \cdots, G_{k-1}^m$ have already been chosen and we ...


1

I wasn't able to figure out an exact formula for eigenvalues and eigenvectors for a pentadiagonal Toeplitz matrix. Knowing them for a tridiagonal Toeplitz, though, is very helpful. (Those can be found, for example, in this paper on the sensitivity of the spectrum of a tridiagonal Toeplitz matrix.) As you've already discovered, the eigenvalues are ...


2

If you write a matrix so: $$M=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ you can't say, in general, that $$\det M=\det A\det D-\det B\det C$$ Just note that even $A$ and $D$ are square, $B$ and $C$ needn't be. But if $B$ or $C$ are zero, then we have $$\det M=\det A\det D$$ To show it, note that if you consider the entries of $M$ lying in an ...


8

Maybe not exactly what you are asking, but notice that your matrix is lower triangular, and can be written as $M=I+N$, with $I$ the identity and \begin{equation} N=\begin{pmatrix} 0&0&0&0\\ 1&0&0&0\\1&1&0&0\\1&1&1&0 \end{pmatrix}.\end{equation} N is nihilpotent, $N^4=0$ and (obviously) commutes with $I$, hence ...


0

You can use the $2 \times 2$ definition of the determinant and reduce all larger matrices into smaller ones repeatedly


4

These are diagonals of Pascal's Triangle.


1

Laplace formulas, Gauss elimination, product of the eigenvalues are the first which come to (my) mind. Surely there are others...


1

The question is a bit unclear but perhaps this might help. You are given A(AA*) = (AA*)A , so by associativity : A(AA*-A∗A) = 0 and the definition of normal is AA*-A*A = 0, so we have to get from the matrix expression above to the expression AA * -A*A = 0 Let {D0,D1,D2....Dn} be the n-vectors which span the eigenspace of AA* , then we have : ...


1

Hint: $Tr(A) = A_{11} + A_{22} = 4$ (sum of eigenvalues) Another Hint: $Det(A) = A_{11} * A_{22} - A_{12} * A_{21}= 13$ (product of eigenvalues)


1

Hint: If $a$ and $b$ are real numbers, what are the eigenvalues of $\begin{bmatrix}a & -b \\ b & a\end{bmatrix}$? Alternatively, notice that $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ (the matrix which represents a rotation by $180^{\circ}$) has eigenvalues $\pm i$. Can you manipulate this matrix into one which has eigenvalues $2 \pm 3i$? ...


1

Hint: start by finding the characteristic polynomial of $A$, and try working from there.


1

There is no need to take normal forms, you just have to check that we have positive definiteness by its definition. Let $M=A^T A+I$. Obviously $M=M^T$ and we have: $$\forall v\neq 0,\qquad \langle v,Mv\rangle = \|v\|^2 +\|A v\|^2\color{red}{>}0 $$ as well as: $$ \inf_{v:\|v\|=1}\langle v,Mv\rangle \geq 1.$$


1

I'm on the right track now, I did as I've suggest earlier : Should I apply the object's transform to the target point and to a given vector between the camera eye point and target point, then I'll get my new target and my new camera eye point (which is the end of my new vector)? Will that be enough information to setup the camera pose properly? but ...


0

Take $p\ge q$ and $rank(A)=p$, i.e., $A$ has full rank. Then $AA^T$ is invertible and positive definit. Let $\lambda_1$ be the smallest eigenvalue of $B$. Let $\sigma$ be the smallest eigenvalue of $AA^T$, i.e., which is the square of the smallest singular value of $A$. Now take $x\in \mathbb R^q$. Then $$ x^TXx = x^TABA^Tx \ge \lambda_1 \|A^Tx\|^2 = ...


5

Here's a useful fact: if $ x $ is an eigenvector of $ M $ with eigenvalue $\lambda$, then $ x $ is an eigenvector of $ M + c I $ with eigenvalue $\lambda + c $. Proof: $ (M + c I) x = Mx + cx = (\lambda + c) x. $ (Similarly, if $ x $ is an eigenvector of $ M + c I $ with eigenvalue $\lambda $, then $ x $ is an eigenvector of $ M $ with eigenvalue $\lambda ...


5

Well, since $A^T A$ is positive definite, it is diagonalisable with positive real eigenvalues. That is, if $A$ is $n \times n$, then for some $n \times n$ invertible matrix $P$, we have: $$PA^TAP^{-1} = \left(\begin{matrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 ...


0

I suspect your issue is that you're really thinking of rotating both the object and camera about their respective centers - this is not applying the same rotation, hence transformation, to each. If you do apply the same rotation (in the same plane, about the same point) and translation to each, this is an affine transformation, which preserves points, ...


0

Do you want an invertible matrix $A$ over $\mathbb{Z}/2\mathbb{Z}$ or over $\mathbb{R}$ ? In the first case, the Blue's method seems to be good. Yet, in the second case, it is not, because $\det(A)=\pm 1$ that is far from random.


0

Let $f:A\in M_n\rightarrow A\circ B\in M_n$. $f$ is linear ! Then $Df_A:H\in M_n \rightarrow H\circ B$.


4

The matrix $T$ does not need to be orthogonal unless $a=1$. Something else can be though if $a>0$. You can write $T^\dagger A T=A$. If $A$ is positive definite (if $a>0$), it has a unique positive definite square root $A^{1/2}$ (in this case, a diagonal matrix with square roots of the corresponding diagonal entries on the diagonal). Multiplying by ...


1

For notational convenience, define $$\eqalign{ M &= \sigma^2 I + \lambda^{-1}K_{\theta}^{-1} \cr dM &= -\lambda^{-2}K_{\theta}^{-1}d\lambda \cr }$$ Then write the function in terms of $M$ and the Frobenius product, and take its differential $$\eqalign{ f &= y:M^{-1}y \cr &= yy^T:M^{-1} \cr\cr df &= yy^T:d(M^{-1}) \cr &= ...


1

No, it does not imply. It may well happen that $AY + YA^T$ is sign indefinite. To find an example choose $A$ Hurwitz (but not diagonal) and pick $Y=I$. Then it is not difficult to find a $1 \times n$ matrix $C$ that results in $C (A + A^T) = 0$.


1

Note: This is a slightly clumsy but systematic approach. On the plus side, this allow you solving similar equations of the form $$\begin{cases} x^2 - Ayz &= D\\ y^2 - Bxz &= E\\ z^2 - Cxy &= F \end{cases}$$ without knowing how to complete the squares. On the minus side, you need to factor a quartic polynomial in the middle of the process. ...


2

Yes, it is possible. Furthermore, there does exist some algorithms to do this, such as the fraction-free Gaussian elimination, see, e.g., E H Bareiss. Sylvester's identity and multistep integer-preserving Gaussian elimination. Math. Comput., 22(103):565-578, 1968.


0

Yes, it is possible. For example, to eliminate that $3$ in the lower left, instead of subtracting $\frac32$ times the first row, you can first multiply the third row by $2$, then subtract $3$ times the first row. More generally, instead of using $b - \frac{b}{a}\cdot a = 0$, you can multiply through by $a$ and use $a\cdot b - b\cdot a = 0$ for your ...



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