# Tag Info

1

Compute the eigenvalues of $A$ and $B$. These are of course $1, 1, 3$. Compute eigenvectors of $A$. These are $$\begin{bmatrix}-2\\1\\0\end{bmatrix}, \begin{bmatrix}0\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix},$$ where the first two are relative to $1$, and the third to $3$. Simlarly for $B$ one has $$\begin{bmatrix}0\\1\\0\end{bmatrix}, ... 0 Let a_{ij} be the entries of the matrix A. Our basis is given by$$ v_1 = \pmatrix{1&0\\0&0}, \quad v_2 = \pmatrix{0&1\\0&0}, \quad v_3 = \pmatrix{0&0\\1&0}, \quad v_4 = \pmatrix{0&0\\0&1} $$We note that$$ T(v_1) = a_{11}v_1 + a_{21} v_3\\ T(v_2) = a_{11}v_2 + a_{21} v_4\\ T(v_3) = a_{12}v_1 + a_{22} v_3\\ T(v_4) = ...

3

If the two matrix $A$ and $B$ are similar then they are similar to the same Jordan matrix $J$; so find the change matrices $S$ and $T$ such that $$A=SJS^{-1}\quad;\quad B=TJT^{-1}$$ so $$A=ST^{-1}BTS^{-1}$$ and then we have $P=TS^{-1}$.

3

Suppose, you have $$A=S^{-1}DS$$ and $$B=T^{-1}DT$$ Then we have $$D=TBT^{-1}$$ So, we get $$A=S^{-1}TBT^{-1}S$$ So, with $$P:=T^{-1}S$$ you have the required transformation.

1

Let $x$ an eigenvector of $B^TB$ associated to the eigenvalue $\lambda$, which is real since $B^TB$ is symmetric, then $$\lambda ||x||^2=\langle B^TB x,x\rangle=\langle Bx,Bx\rangle=||Bx||^2\implies \lambda\ge0$$

2

They are always non-negative. Suppose $\lambda$ is an eigenvalue of $B^TB$ corresponding to an unit eigenvector $v$, then $\langle v, B^T B v \rangle = \lambda = \langle Bv, B v \rangle = \|Bv\|^2$. Hence $\lambda \ge 0$.

-1

Yes. $$\sum_{i=1}^n \sum_{j=1}^n a_i a_j 2^{x_ix_j}=\left(\sum_{i=1}^n a_i2^{x_i}\right)^2.$$

0

Attempt: W.L.O.G $\det B=0$ $\operatorname{rank}[(A+A^T)(B+B^T)]=\operatorname{rank}[AB+AB^T+A^T(B+B^T)]=\operatorname{rank}[AB^T+A^T(B+B^T)] \le \operatorname{rank}(AB^T)+\operatorname{rank}[A^T(B+B^T)] \le \operatorname{rank}B^T+\operatorname{rank}A^T=\operatorname{rank}A+\operatorname{rank}B \le (2k+1)+\operatorname{rank}AB=2k+1$ So that, if equality ...

0

I assume that $B^T$ has more columns than rows, and that $m=2n$. Then let $C_{n \times n}$ be a rotation-matrix (=orthogonal) and $D_{n \times n}=I_{n \times n}$ Then let $B^T= 0.5 \cdot [C \quad D]$ being a blockmatrix, then $B^T \cdot B = I_{n \times n}$

2

$$(A+B)^{-1}=[B(B^{-1}A+I)]^{-1}=[B(B^{-1}+A^{-1})A]^{-1}$$ Now use that $(XY)^{-1}=X^{-1}Y^{-1}$ to get $$[B(B^{-1}+A^{-1})A]^{-1}=A^{-1}(B^{-1}+A^{-1})^{-1}B^{-1}$$ Notice that there is a missing $-1$ in your formula. To get the other equation start with $$(A+B)^{-1}=[A(I+A^{-1}B)]^{-1}=[A(B^{-1}+A^{-1})B]^{-1}$$ and then ...

3

You can construct matrices with this property as follows: assume $m\geq n$ and endow $\Bbb R^m$ with the standard euclidean inner product. Pick vectors ${\bf v}_1,...,{\bf v}_n$ such that ${\bf v_i}\cdot{\bf v_i}=\delta_i^j$. Then the matrix $B$ whose rows are the vectors ${\bf v}_1,...,{\bf v}_n$ (in any coordinate system) has the desired property.

2

Here's an example: $B = \begin{pmatrix}1 \\ 0 \end{pmatrix}$ then $B^T\cdot B=1$. For something less simple, you can consider $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$.

3

Hints : The diagonal elements have to be $0$ If the lower left triangle is given, the upper right triangle is determined by $a_{ij}=-a_{ji}$

0

on the other hand, I can't find any pair of a,b such that...Why is that? You cannot find such a pair because your assumption but from the text it seems that the matrix of the form ..is the most general form of O(2) is not true: As you noted every rotation $$R=\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right]$$ has determinant ...

1

HINT: $a^2+b^2=1$ and $T(e_1) =(a,b)^t$, give you two solutions not one: $T_1(e_2) =(-b,a)^t$ and $T_2(e_2) =(b,-a)^t$.

0

0

If you want to form the $(m_1+m_2) \times (n_1+n_2)$ block matrix \begin{equation*} M = \begin{bmatrix} A & B\\ C & D \end{bmatrix} \end{equation*} where $A$ is $m_1 \times n_1$, $B$ is $m_1 \times n_2$, $C$ is $m_2 \times n_1$ and $D$ is $m_2 \times n_2$, you just need to type M = [A, B; C, D];

1

Hint. The key is to show that the rank of at least one of the two matrices is at most $k$. Once this is shown, use the facts that $\det(XY)=\det(X)\det(Y)$ and $\operatorname{rank}(C+D)\le\operatorname{rank}(C)+\operatorname{rank}(D)$ to complete the proof.

2

After an orthogonal change of basis $A^TA$ becomes a diagonal matrix $D$ with nonnegative entries $d_1,\dots,d_n$. The question is then whether $$f(\lambda) = \det(\lambda I+D) = \prod_{i=0}^n(\lambda+d_i)$$ increases or decreases as $\lambda$ increases. For $\lambda\geq0$, we see that each term in the product is positive and increases with $\lambda$ so ...

0

I would do it one transformation at a time and then form the product of the matrices. So, to begin with, we have $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$ and by implication $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} ... 1 Hint: What happens when you expand and simplify the polynomial product$$(1 - x)(1 + x + \cdots + x^n)?$$Can you produce a polynomial analogue of the resulting equation? 0 It has failed because: \dim (A+(-B)) \ge \dim A 0 It says you get B from A (or vice versa) by adding a matrix of rank s. What else do you want it to say? EDIT: The column space of A has dimension K = \text{rank}(A). The column space of B could be anything from K - s to K. 1 The Gauss' map is \Omega\to\Bbb R^3 given by p\longmapsto \dfrac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}, where \partial_1=\frac{\partial f}{\partial v} and \partial_2=\frac{\partial f}{\partial w}. We can see this map as a vector field N:\Sigma\to\Bbb R^3, i.e. an assignation  \left(\begin{array}{c} x\\ y\\ z ... 1 I'm going to give a complete proof of this result, which has a surprisingly conceptual basis, using the quaternions. I believe it can possibly be simplified further. Let \mathbb{H} be the division ring of the quaternions. It contains \mathbb{C} as a subring. View \mathbb{H} as a left-\mathbb{H}-vector space over itself. An arbitrary endomorphism of ... 0 This won't quite work: e.g. you might have A = I, in which case your T could be any invertible matrix, and there's no reason to think T B T^{-1} will be in Jordan normal form. What you want to do is choose T on each eigenspace of A to put into Jordan form the restriction of B to that eigenspace. 1 Assume that W_c is the matrix with orthonormal columns forming a basis of V_c and let W_c^\perp be such that W:=[W_c,W_c^\perp] is unitary (W_c contains a basis of the orthogonal complement of V_c). We have AW_c=cW_c, so$$ A[W_c,W_c^\perp]=[W_c,W_c^\perp]\begin{bmatrix}cI & B_{12} \\ 0 & B_{22}\end{bmatrix}=:WB, $$that is, A is ... 3 A simple and classic result (prove it): if two endomorphisms commute then the eigenspace of one is invariant by the other. Now since A and A^* commute then V_c is invariant by A^* hence for all w\in V_c we have A^*w\in V_c so for v\in (V_c)^\perp:$$\langle Av,w \rangle =\langle v,A^*w\rangle=0$$and the result follows 0 The proof for the Wikipedia formula was given in this thread, which unfortunately still leads to a system for which an inverse needs to be taken in order to solve the determinant. At least I was able to prove that in this general case it's impossible: Inserting my matrix into the proof for \begin{bmatrix}A&B\\B &A \end{bmatrix} Gives ... 8 Let AB\in F^{n,n}, F any field, n odd. Assume they satisfy A^2 = AB+BA. From A^2=AB+BA it follows$$ AB-BA= A^2-2BA= (A-2B)A. $$Then it holds$$ \det(AB-BA) = \det((A-2B)A) = \det(A(A-2B)). $$Now we have$$ A(A-2B) = A^2-2AB = BA-AB. $$This proves$$ \det(AB-BA) = \det(BA-AB) = \det( -(AB-BA)) = (-1)^n \det(AB-BA), $$and since n is odd, ... 3 This was easier than I originally thought. Note that$$ AB - BA = AB + BA - 2BA = A^2 - 2BA = (A-2B)A\\ -(AB - BA) = AB + BA - 2AB= A^2 - 2AB = A(A - 2B) $$and that we have both \det(AB - BA) = \det(-(AB - BA)) and \det(AB - BA) = -\det(-(AB - BA)). My original approach: note that AB - BA has trace zero. Then, we note that$$ AB - BA = AB + BA - ...

0

What did you try? You could use the well-known formula $$\DeclareMathOperator{\Var}{Var} \Var(AX) = A \Var(X) A^T$$ You can find such formulas here: https://en.wikipedia.org/wiki/Multivariate_random_variable

2

A partial answer. I usually prove that $A=(I-Q)(I+Q)^{-1}$ is skew-Hermitian by showing that $A^\ast=-A$ rather than $A+A^\ast=0$. The merit of doing this is that we can factor out $Q^\ast$ and make the proof shorter: $$A^\ast = \left[ (I-Q)(I+Q)^{-1} \right]^\ast \color{red}{=} (I+Q^\ast)^{-1}(I-Q^\ast) \color{blue}{=} ... 2 According to your link a submatrix is just a block of another matrix. So this is just a connected 'rectangle' of numbers of the original matrix, that means (as you already assumed) that you cannot skip rows or columns. But if you consider following link, you can skip rows or columns, that means you can construct any submatrix by deleting whole columns and ... 5 Hint$$\dfrac{1}{2}[(a+b)^2+(b+c)^2+(a+c)^2]=a^2+b^2+c^2+ab+ac+bc\le 0$$so we have$$a=-b,b=-c,c=-a\Longrightarrow a=b=c=0$$0 After two well chosen row operations you can get$$ \left( \begin {array}{cccccc} a&1&a&0&0&0\\ -ab&0&1- ab&b&0&0\\ 0&0&c&1-cd&0&-cd\\ 0 &0&0&d&1&d\end {array} \right) $$Since we cannot have both -ab and 1-ab as zero, the rank is at least 3. The rank can only be 3 if rows 2 ... 0 You're given the matrix in row echelon form!$$A= \begin{pmatrix} a & 1 & a & 0 & 0 & 0 \\ 0 & b & 1 & b & 0 & 0 \\ 0 & 0 & c & 1 & c & 0 \\ 0 & 0 & 0 & d & 1 & d \\ \end{pmatrix} $$Now, there's no way row 1 and row \;2\; are linearly dependent , as it'd have to be ... 0 we can show that rank(A) > 2 by leaving out the first column. second row is replaced by second row minus b times the first row. we have two cases: ab = 1 and ab \neq 1. let us take the easier case ab \neq 1. now you have pivots 1, 1-ab, 1 in rows 1, 2,  and 3. so rank(A) \ge 3 the second case ab = 1, now we have pivots 1 on rows ... 0 Looking at these matrices as maps, we have that \;B:\Bbb F^m\to\Bbb F^n\; , and thus \;\dim\text{Im}\,B\le n\; , so \;\dim\text{Im}\,A(B)\;\le n . Or shortly: linear maps (or matrices) cannot increase the dimension of the domain. 2 Notice that$$E_{ij}E_{kl}=\delta_{jk}E_{il}$$so if a matrix$$A=\sum_{1\le k,l\le n}a_{kl}E_{kl}$$commutes with the all the matrices then it commutes with E_{ij} hence we get$$AE_{ij}=E_{ij}A\iff \sum_{k=1}^n a_{ki}E_{kj}=\sum_{l=1}^n a_{jl} E_{il}$$so we see that$$a_{ii}=a_{jj}=:\lambda\;\forall i,j\quad \text{and} \quad a_{ki}=0\;\forall k\ne ...

1

Hint: Compute $A E_{ij}$ and $E_{ij} A$. Force them to be equal.

2

If $A$ is orthogonal such that $\color{blue}{\text{$A+I$is invertible}}$, then $$X+X^T=(A+I)^{-1}(A-I)+(A-I)^T(A+I)^{-T} =(A+I)^{-1}[\underbrace{(A-I)(A+I)^T+(A+I)(A-I)^T}_{=0}](A+I)^{-T}=0.$$

2

Use the case for two matrices and use induction: $$(ABC)^T = C^T(AB)^T = C^TB^TA^T.$$ You can easily generalize this to arbitrarily large products and see that transposing a product inverts the order of the product. In your particular case you get that: $$A^TC^TB^T = (BCA)^T.$$

3

Absolutely no conditions (except for compatibility*, of course): $$A^T C^T B^T = A^T (C^T B^T) = A^T (BC)^T = ((BC)A)^T = (BCA)^T$$ By associativity of matrix multiplication and by $(AB)^T = B^T A^T$. *Matrices $A, B$ are said to be compatible if the product $AB$ exists, i.e. if the number of rows of $B$ is equal to the number of columns of $A$.

0

It's a trick. Since $X = IX$, then $aX = a(IX) = (aI)X$.

3

There are three kinds of 2-tensors, of type $(0,2)$, $(2,0)$ and $(1,1)$. Let's take the metric tensor $g$ of type $(0,2)$. If $p$ is a point in your manifold and $V:=T_p M$ the tangential space as a vectorspace, then the metric is a bilinear map $$g_p:V \times V \rightarrow \mathbb{R}$$ and "the matrix" of $g$ is a description of $g$ if you choose a ...

2

All reflections in the plane have matrices of the form $$\left( \begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & - \cos \alpha \end{array} \right)$$ or, for any $a^2 + b^2 = 1,$ $$\left( \begin{array}{cc} a & b \\ b & - a \end{array} \right)$$

1

Eigenvalues of $A\bar{A}$ have been thoroughly studied in D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13(1961), 694-704. In the paper, it is proved (in Lemma 5) that for any complex square matrix $A$, those eigenvalues of $A\bar{A}$ that are not real nonnegative must either be non-real and occur in ...

0

An $n\times n$ matrix with $n$ independent eigenvectors can be written as $A=P^{-1}DP$, where $D$ is the diagonal matrix $diag(\lambda_1\:\lambda_2\:...\lambda_n)$ and $P$ is the matrix $(\vec{v}_1\:|\:\vec{v}_2\:|...|\:\vec{v}_n)$ where $v_i$ is the corresponding eigenvector to $\lambda_i$. ...

2

Writing the matrix down in the basis defined by the eigenvectors is trivial. It's just $$M=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{array} \right).$$ Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely:  S = \left( \begin{array}{ccc} 1 & 1 & ...

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