New answers tagged

0

As you will encounter in most scenarios when trying to prove that a set is linearly independent, first assume the set is linearly dependent. That is, assume there exists $\alpha,\beta\in\mathbb{R}$ (I assume you're working with matrices with entries in $\mathbb{R}$, but this can be generalized) such that $$ \alpha A +\beta B=0. $$ Then we apply the ...


0

you have $AB=I$, then multiply for the left side of this equation for $C$ you have: $$C(AB)=CI$$ for the property associative of multiplication of matrix we have $C(AB)=(CA)B$,we know that too $CI=C$ and $CA=I$ $$(CA)B=CI<=>IB=C<=>B=C$$


0

Your simple proof just needs a little more justification. To expand upon your proof, you can say Since $CA=I=BA$, we have $CAB=BAB$, which simplifies to $C=B$.


1

$$B \begin{array}[t]{l} = \begin{bmatrix} x_1(w_1 - x^Tw)^2 - x_1^2(w_1 - x^Tw)^2 + x_1x_2(w_1 - x^Tw)^2 \\ x_2(w_2 - x^Tw)^2 - x_2^2(w_2 - x^Tw)^2 + x_2x_1(w_2 - x^Tw)^2 \end{bmatrix} \\[3ex] =\begin{bmatrix} (w_1 - x^Tw)^2 & 0 \\ 0 & (w_2 - x^Tw)^2 \end{bmatrix}\cdot\begin{bmatrix} x_1- x_1^2+x_1x_2 \\ x_2-x_2^2 + x_1x_2\end{bmatrix} \\[3ex] = \...


1

The definition of "rank" for a Euclidean Jordan algebra is as follows, see here: Definition: The maximal number of elements in a Jordan frame is called the rank. (Jordan frame has to do with idempotents).


0

Left muliplication by $A$ is a linear map between spaces of column vectors. There is an obvious one-to-one correspondence between column vectors and row vectors. What is the linear map between spaces of row vectors which corresponds to left multiplication by $A\require{AMScd}$ on column vectors? Stated with commutative diagrams, we want a map linear map $B$ ...


0

Basically we want a coordinate-free proof that if $\mathbf{n}$ is a unit normal vector of an oriented plane $\Pi$ then $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{n}$ is the signed area of the parallelogram spanned by the orthogonal projections of the two vectors $\mathbf{a}$ and $\mathbf{b}$ onto $\Pi$. (My definition of $\mathbf{a}\times\mathbf{b}$ is the ...


1

In general, no. Consider $A=1\oplus\pmatrix{1&1\\ 1&-1}\oplus1$ and $W=1\oplus\pmatrix{1&0\\ 1&1}\oplus1$. We have $\|A\|_2=\sqrt{2}<\frac{1+\sqrt{5}}2=\|A_w\|_2$. As for your proposed conditions, here is a partial answer: When $A$ is entrywise nonnegative, $A^TA\ge A_w^TA_w\ge0$ entrywise. Hence the same order is preserved for their ...


0

I've been looking for your answer and what I found is summarized below: $SL(n,K)$ is commutator group of $GL(n,K)$ in all cases except when $n=2, K= GF(2)$ and $n=2, K=GF(3)$. where $GF(p^n)$ is the finite field of order $p^n$. Now for the proof in short: We know for any group $G/N$ is abelian iff $N$ contains the commutator subgroup. Now the ...


1

Your bound is indeed correct in general. This question (with say the top answer) shows that the sum of the singular values is at least the sum of absolute values of the eigenvalues. On the other hand, a rank-$r$ matrix has exactly $r$ nonzero singular values $\sigma_1,\ldots,\sigma_r$. Since the trace of $A^TA$ is the sum of the squares of the singular ...


1

By elementary row operations reduce as many rows as you possibly can, count the number of non-zero rows. If you are not sure if you have reduced as far as possible, put the matrix into "row echelon" form.


0

Suppose that $A$ is simular ti $B$, they there exist $C$ such that: $B=CAC^{-1}$. So: $tr(B)=tr(CAC^{-1})$, now let me call $X=CA$, and $Y=C^{-1}$, then $tr(B)=tr(XY)=tr(YX)=tr(C^{-1}CA)=tr(A)$.


0

By your fourth equation, we have $\delta=0$, so by your first equation, we have $\alpha+\beta=\gamma$ and by your second equation, we have $\alpha+3\beta=\gamma$. Subtract the second equation by the first equation to get $2\beta=0 \implies \beta=0$. Now, by the third equation, we have $0=\gamma$ and, going back to the first equation, we get $\alpha=0$. Thus,...


1

For your first question, this situation is similar to the following thread: $A,B,C \in M_{n} (\mathbb C)$ and $g(X)\in \mathbb C[x]$ such that $AC=CB$- prove that $A^jC=CB^j$ and $g(A)C=Cg(B)$ Let $A,B\in M_n(F)$ for some algebraically closed field $F$. Suppose $v\in F^n$ is a nonzero vector such that $A^kv=B^kv$ for $k=1,2,\ldots,n$. Let $m_B(x)$ ...


3

Just look at the column rank of $X$. The matrix has full column rank if and only if it has a nonzero diagonal (not necessarily the main diagonal, but any diagonal of length $n+1$). Therefore $X^\top X$ is positive definite if and only if at least one of $u(0),u(1),\ldots,u(N-n)$ is nonzero, otherwise the matrix product is merely positive semidefinite.


0

There is indeed a geometric interpretation of $\bf{u}\times\bf{v}$ in terms of the areas of the projections of the parallelogram $\bf{P}$ spanned by $\bf{v}$ and $\bf{w}$ onto the coordinate planes. I'll start from scratch. Motivating problem: We wish to create a vector perpendicular to u,v, i.e. construct w s.t. $\bf{w}\cdot \bf{u} = \bf{w}\cdot \bf{...


2

The eigenvalues of both matrices are $1$ and $2$. The difference between $A$ and $B$, if any, lies in the eigenspaces $E_{A,2}$ and $E_{B,2}$ corresponding to the double eigenvalue $2$. Now $$A-2I=\begin{bmatrix}-1&1&5\\0&0&0\\0&0&0 \end{bmatrix}, \qquad B-2I=\begin{bmatrix}-1&5&0\\0&0&7\\0&0&0 \end{bmatrix},$...


5

They are different because A is similar to the Jordan matrix \begin{pmatrix}1&0&0\\ \:0&2&1\\ \:0&0&2\end{pmatrix} and cannot be diagonalized. B is similar to the diagonal matrix \begin{pmatrix}1&0&0\\ \:0&2&0\\ \:0&0&2\end{pmatrix}


2

You can use the characteristic polynomials. If $A$ is similar to $B$, then their characteristic polynomials $f_A(x)$ and $f_B(x)$ are identical. Since $$ f_A(x)=x^n-tr(A)x^{n-1}+\ldots\pm\det(A)\\ $$ and $$ f_B(x)=x^n-tr(B)x^{n-1}+\ldots\pm\det(B) $$ it follows in particular that $tr(A)=tr(B)$.


2

Hint: By very definition, two matrices $A,B$ are similar iff there exists an invertible matrix $S$ such that $A=SBS^{-1}$. Now apply the trace on both sides, and conclude using associativity of the matrix product.


2

If $A$ is invertible, then $A^2$ is invertible too. Thus $$I=(A^2)^{-1}A^2=(A^2)^{-1}0=0,$$ a contradiction.


0

If $\pi_{i=1}^{n}A_i=0$ then atleast one $A_i$ has one eigenvalue equal to $0$, which implies atleast one of the $A_i$ is not invertible.


0

Claim: i) $rank(A^*)=n$ if $rank(A)=n$ ii) $rank(A^*)=1$ if $rank(A)=n-1$ iii) $rank(A^*)=0$ if $rank(A)<n-1$ Proof: i) $det(A^*)\ne0$ since $A^*A=det(A)E$ and $A$ is invertable ii) there exists $i,j$ such that $A_{ij}\ne0$ since $rank(A)=n-1$, so $rank(A^*)\ge1$ since $rank(A)+rank(A^*)\le n$, $rank(A^*)=1$ iii) $A^*=0$ since $A_{ij}=0$ since $...


2

For some reason this elementary question has gotten several answers, so I'll add another. The equation $A^2=0$ implies the kernel of $A$ contains the image of $A$. By the rank-nullity theorem, it follows that the kernel of $A$ cannot be trivial.


2

If $A$ were invertible, then $BA=I$ for some $B$. Then $$ A=IA=(BA)A=BA^2=B0=0 $$ But now $BA=B0=0$, a contradiction.


12

$A^2=0$ $\Rightarrow$ $\det(A^2)=\det(A)^2=0$ $\Rightarrow$ $\det(A)=0$.


16

Unfortunately, $A^2=0$ does not imply that $A=0$. Consider for instance $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Instead, if $A$ is invertible then it has an inverse $B$. What happens if you multiply both sides of $A^2=0$ by $B$ (on the left, say)?


0

One way to see, you can compare entries of two matrices $(AB)^t$ and $B^tA^t$. Firstly, we have $$AB_{ij} = \sum_{k} a_{ik}b_{kj}.$$ When to transpose a matrix, we switch the rows and the columns. $$(AB)^t_{ij} = AB_{ji} = \sum_{k} a_{jk}b_{ki}.$$ Now, let see the entry ij of $B^tA^t$: $$B^tA^t_{ij} = \sum_{k} B^t_{ik}A^t_{kj} = \sum_{k} b_{ki}a_{jk}.$$ ...


2

you have the next property too $$(AB)^{-1}=B^{-1}A^{-1}$$ then the inverse of $(AB)^{T}$ is $\left[(AB)^{T}\right]^{-1}$ $$\left[(AB)^{T}\right]^{-1}=\left[B^{T}A^{T}\right]^{-1}=\left[A^{T}\right]^{-1}\left[B^{T}\right]^{-1}=\left[A^{-1}\right]^{T}\left[B^{-1}\right]^{T}$$


1

Taking determinants of both sides and using the multiplicativity of $\det()$, you get $64\det(A) = \det(A)^7$. So $\det(A) = 0$ or $\det(A)^6 = 64$. (Note that $\det(A)$ can be $0$, so you CAN'T multiply by $A^{-1}$; $A$ may not be invertible.) Now simply solve $x^6 = 64$. You didn't say if $A$ has elements in $\mathbb C$ or $\mathbb R$. Solve in the proper ...


2

If $4A=A^7$, then $\det(4A)=\det(A^7)=[\det(A)]^7=4^3\det(A)$ Since $\det(cA)=c^n\det(A)$ Let $\det(A)=x$, then we must solve $64x=x^7$


1

you use $\det(xA)=x^3 \det(A)$ and $\det(A^7)=(\det(A))^7$


0

Square the matrix once to get $$M^2=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$ Then, $$M^4=\begin{bmatrix}0&1\\-1&0\end{bmatrix}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}.$$ $$M^6=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-M^2.$$ $$M^8=-M^...


0

I am not exactly sure what your function is supposed to do or why you want this particular function, but here is approximately what pseudocode could look like: subroutine subMatrix(M, i, j, size) : N = matrix(size, size) // a new size x size matrix. for x = i; x < i+size; i = i+1 : for y = j; y < j+size; j=j+1 : N[x,y] ...


2

You have correctly computed $$ A= [p]_\alpha^\alpha= \left[\begin{array}{rrr} 0 & 1 & 2 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array}\right] $$ where $\alpha=\{1,x,x^2\}$. You wish to find all $a$, $b$, $c$ for which $$ B= [p]_\beta^\beta= \left[\begin{array}{rrr} a & 1 & 0 \\ b & 0 & 1 \\ c & 0 & 0 \end{array}\right]...


2

The answer to your question is positive. For a self-contained proof, see Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem.


1

we begin by 3) we have $\sigma(P)=\{1,-1\}$ and for all polynomials $f$ $$ f(\sigma(P))=\sigma(f(P)) $$ so if we take $f=1+2 x$ $Q=F(P)$ and then $$ \sigma(Q)=\{f(1),f(-1) \}=\{3,-1\} $$ so in particular $0\not\in\sigma(Q)$ and then $Q$ is invertible so this give as 2) but 2) implies that $det(Q)\neq0$, and we know that $\det(P)=\{ \textrm{Product of ...


0

This is only a partial answer. But maybe it will be helpful to someone. $(1)$ I was only able to show that $M_n\leq p$, where $p$ is the first prime larger than $n^2!n$. $(2)$ For the existence of $n^2$ numbers $a_1, \ldots, a_{n^2}\in [a,b]$ such that every matrix whose entries form a permutation of the set $\{a_1, \ldots, a_{n^2}\}$ has a ...


1

Q1: Let $\lambda=0$ and have rank(A)=r. We can also reach the conclusion that $rank(B)\leq r$. If rank(B)=r'>r, then we know that B has an submatrix $B_0$ of order r', $det(B_0)\neq 0$. We consider the corresponding submatrix of $A+\lambda B$ (we suppose it is $A_0+\lambda B_0$), and we know that $det(A_0+\lambda B_0)$ is a polynomial of degree r', which ...


2

Your mistake is when you made the substitution $(R_i)_j=a_{ij}$. You should have $$\sum_{i=1}^nv_i\sum_{j=1}^n(R_i)_jv_j=\sum_{i=1}^nv_i\sum_{j=1}^na_{ij}v_j.$$ Now neither $a_{ij}$ nor $v_j$ can be pulled out of the sum over $j$ as both depend on $j$. The formula isn't going to "simplify" beyond this.


2

I think what you're saying is something like this: $$\sum_{j=1}^n (R_k)_jv_j=\left(\sum_{j=1}^n (R_k)_j\right) \left( \sum_{j=1}^n v_j \right)$$ This is your mistake: summations are not distributive over multiplication. This is like saying the following: $$\sum_{j=1}^n (1)(1)=\left(\sum_{j=1}^n 1\right) \left( \sum_{j=1}^n 1\right)$$ Clearly, the left side ...


0

A bilinear form $f:V\times V \to \mathbb{K}$ takes values in the underlying field of scalars $\mathbb{K}$ of $V$, and is entirely determined by its action on pairs of basis elements of $V$. If $V$ is finite-dimensional, $\dim V = n$, then each choice of basis $(e_i)_{i=1}^n$ for $V$ induces an isomorphism between the space of bilinear forms and the space of $...


1

Tang and Saad have a method that uses random vectors (not necessarily projections): A Probing Method for Computing the Diagonal of the Matrix Inverse


1

It seems that $A$ dictates the evolution of the system over time. My best guess is that the relevant piece of information here is $A + A^T = 0 \implies$ $A$ has imaginary eigenvalues $\implies$ energy is conserved On the other hand, $A + A^T \preceq 0 \implies$ $A$ has eigenvalues with a non-positive real part $\implies $ energy decays with a steady state ...


1

recall if $f/V\times V \rightarrow W$ is a bilinear map and $W=K$ the field of scalar, then $f$ is sayd a bilinear fom. if $dimV=1$ there are no degenerate bilinear map othere the zero map, so all non zero bilinear map are non degenerat. if $dimV>1$ the example above in comment is on, anothere tack $f(e_1,e_j)=0$ and somme one $f(e_i,e_j)\not=0$ for ...


0

f := proc(N::posint, k::posint:=9) local A, j, gen; gen := 2*(rand(0..1)-1/2)*rand(1..k); A := Matrix(LinearAlgebra:-RandomMatrix(N,generator=gen,shape=triangular[lower])); for j from 1 to floor(N/2) do A[j+1..N-j,j], A[j,j+1..N-j] := Vector(N-2*j), A[j+1..N-j,j]; A[j,N-j+1] := gen(); end do; return A; end proc: f(1); ...


1

A (adjA)= |A| I here adj A = [a -a a -a ] is not in G. so A^-1 = adjA/|A| doesnot mean in G


3

The two not obvious points needed to prove are : $a,b \in E \implies ab \in E$ Every $a \in E$ such that $a \neq 0$ is invertible. Let's start with a useful lemma: Lemma: Let $P \in K[X]$. Then $P(A) \in E$. Proof: By Euclidean division, there exists $Q,R \in K[X]$ with $\deg R \leq n-1$ such that $P=\mu_A Q+R$. Then $P(A)=\mu_A(A) Q(A)+R(A)=R(A) \in E$...


0

your system can not be overdetermined system, but it can be underdetermined or not; and it would have not solution iff $Y$ is not in image of $X$ and it have infenitely of solution iff $Y$ is not in the image of $X$, so the solvabilite of $Y=X\beta$ depending on the $Y$.


0

The usual case is you have $n$ unknowns and $m$ equations, or short $$ A x = y $$ with $A \in \mathbb{F}^{m \times n}$, $x \in \mathbb{F}^n$ and $y \in \mathbb{F}^m$. You do your Gauss elimination and count the non-zero rows of the result and end up with $r = \DeclareMathOperator{rank}{rank}\rank A$ non-zero rows, where $r \in \{ 0, \dotsc, n \}$. An ...



Top 50 recent answers are included