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7

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, ...


4

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 $$a=-b,b=-c,c=-a\Longrightarrow a=b=c=0$$


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 $A$.


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 ...


3

Notice that $[A(\theta)]^n = A(n\theta)$, based on geometric interpretation of rotations. If you apply $A(\theta)$ twice, it is the same as rotating by $\theta$ twice, or $2\theta$. The same is true for an arbitrary number of rotations. That shows you have to write $A(\theta)^n$, namely as the same matrix with $\theta$ replaced by $n\theta$.


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 - ...


2

call the eigenvectors $u_1, u_2$ and $u_3$ the eigenvectors corresponding to the eigenvalues $1, -2, $ and $2.$ then $$A = 1\dfrac{u_1u_1^T}{u_1^Tu_1} - 2\dfrac{u_2u_2^T}{u_2^Tu_2} + 2\dfrac{u_3u_3^T}{u_3^Tu_3}$$ you can verify this by computing $Au_1, \cdots$. this expression for $A$ is called the spectral decomposition of a symmetric matrix.


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 & ...


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 ...


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

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 $a^2 + b^2 = 1,$ $$ \left( \begin{array}{cc} a & b \\ b & - a \end{array} \right) $$


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

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 ...


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: $$ A^TC^TB^T = (BCA)^T. $$


1

If terminology were consistent, the thing you ask for would be called an orthogonal transformation, and the things which are actually called orthogonal transformations would better be called orthonormal transformations. But since terminology is as inconsistent as it is, I believe the best you can do is stating that your transformation matrix is made from ...


1

Hint: start by adding row $1$ to row $2$, then add $-1$ times row $2$ to row $1$....


1

For completion, you can do this in-place (that is, no extra storage): As individual operations: Set row 1 to row 1 plus row 2. Set row 2 to row 2 subtracted from row 1. Set row 1 to row 2 subtracted from row 1.


1

The quotient space of $\mathbb{R}^3 / \text{Ker}(A)$ would be elements of the form $\text{Ker}(A)+x$. Thus, you need to find a basis for such elements. A short calculation reveals $\text{rref}(A) = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]$. It follows that $x \in \text{Ker}(A)$ has $x_1=-x_3$ ...


1

It is very easy to construct a counterexample. Take any 2x2 symmetric matrices and look at their product. Is it symmetric? Try. I tried with $$\begin{pmatrix} 1 & 2 \\ 2 & 0 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$ (just putting random numbers on the entries)


1

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


1

I got it! Because (in the second line), it poses the condition of $a^2+b^2=1$ which restrict it to the case $det(A)=1$. We can't write the same mentioned matrix $A$ if $det(A)=-1$, which is the other case of $O(2)$ and determines reflection. EDIT: So, in reference to Will Jagy's answer, I must say that I was wrong about my previous answer (the lines before ...


1

Really it is a very difficult question dated 1980-81 (Am. Math. Monthly). Personally I did not find any solution. The simplest method is as follows. EDIT. Show the result when $A\bar{A}$ has no negative eigenvalues. Show that the multiplicity of a negative eigenvalue of $A\bar{A}$ cannot be $1$. Show that the set of matrices $A$ that satisfy 1. is dense ...


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 ...



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