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Question 1: Let $A$ and $B$ be square matrices of the same order. Prove that $I-AB$ is invertible if and only if $I-BA$ is invertible. Proof: Let $C$ be the inverse of $I-AB$. Then $$I-BA=I-BIA=I-BC(I-AB)A=I-BCA(I-BA),$$ which gives us $$(I-BA)(I+BCA)=I.$$ Thus $I-BA$ is invertible with the inverse $I+BCA$. Question 2: Let $A$ be an $m\times n$ and $B$ be ...


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Yes, for $a_1\ne 0$ it does. Hints: Show that for a block-triangular matrix $$ \begin{bmatrix} A & 0\\B & C \end{bmatrix}^{-1}= \begin{bmatrix} A^{-1} & 0\\-C^{-1}BA^{-1} & C^{-1} \end{bmatrix}. $$ Show that for the $n\times n$ matrix $$ C=I-bN, $$ where $N$ is the matrix of all zeros except of ones on the first superdiagonal, the inverse ...


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If we call your matrix $A$, then $$ A^{-1} = \begin{pmatrix} \frac{1}{a_1} & 0 & 0 & 0 & 0 & \dots \\ -\frac{1}{a_1b^2} \sum_{k=2}^Na_kb^k & 1 & b & b^2 & b^3 & \dots \\ -\frac{1}{a_1b^3} \sum_{k=3}^Na_kb^k & 0 & 1 & b & b^2 & \dots \\ -\frac{1}{a_1b^4} \sum_{k=4}^Na_kb^k & 0 & 0 & 1 &...


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First, left-multiply by $D=\operatorname{diag}(1/a_{1},1,1,\ldots,1)$ to get $$ B\equiv DA=\left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0\\ a_{2} & 1 & -b\\ a_{3} & & 1 & -b\\ \vdots & & & 1 & -b\\ a_{N} & & & & 1 \end{array}\right). $$ Expanding along the first row, the determinant of $...


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You have three linear equations in three unknowns. There are a number of well known methods for doing that. Here, the first thing I notice is that there is no "z" in the first equation so it might be simplest to eliminate z from the other two. The second equation has "-Gz" and the third equation has "Fz". Multiplying the second equation by F gives E * F^...


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You've done a lot of the work already; recall that a system of linear equations can be represented as a coefficient matrix multiplied by a vector containing the variables of the problem. ($A\vec{x}=\vec{b}$) Here, $\vec{b}$ represents the right side of all the equations, so $\vec{b}=\begin{bmatrix}A + (R * D) \\ B - (R * E * F) \\ C - (R * E * G)\end{...


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This is actually true if the matrix $A$ is diagonalizable. Such a decomposition always exists for any square complex matrix but $\Lambda$ is in this case only triangular. $\Lambda$ is guaranteed to be diagonal if and only if $A$ is hermitian. This is a consequence of the fundamental theorem of arithmetic. Your professor's claim is correct if $A$ is supposed ...


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You're right; if you copied the problem correctly, this is a mistake. $(Px)^\top y=(x^\top P^\top)y=x^\top(P^\top y)$.


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Think it in the reverse way; Say your field contains $n$ elements. Choose uniformly any number of value from the field, say $p_1,p_2,\cdots , p_k|k\le n$ It is sure that $P(x)=\prod\limits_{i=1}^k(x-p_i)\in\Bbb{F}[x]$


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Let's show that $AB$ and $BA$ have the same eigenvalues. First, let $\lambda$ be a nonzero eigenvalue of $AB$; then $ABv=\lambda v$, for some $v\ne0$. Therefore $BA(Bv)=B(\lambda v)=\lambda(Bv)$ and so $\lambda$ is an eigenvalue of $BA$ (because $Bv\ne0$). If $0$ is an eigenvalue of $AB$, at least one of $A$ and $B$ is not invertible. Thus also $BA$ is not ...


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Hint: prove that $$ \det(\lambda I-AB)=\det(\lambda I-BA). $$ P.S. Matrices has to be square otherwise the statement about eigenvalues is not correct. Edit: There is a nice proof of this fact: If $A$ is invertible then $$ AB=ABAA^{-1}=A(BA)A^{-1}, $$ and, hence, $AB$ and $BA$ are similar. Similar matrices have the same characteristic polynomial. If $A$ ...


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Suppose $I - AB$ is invertible. Suppose $(I - BA)x = 0$. Then : $$BAx = x$$ so $$ABAx = Ax$$ or, what is the same, $$(I - AB)Ax = 0$$ Since $I - AB$ is invertible, this last equality implies $$Ax = 0$$ Hence $x = BAx = 0$. Thus the only solution of $(I - AB)x = 0$ is $x = 0$, so $I - AB$ is invertible.


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For any $4\times 4$ invertible matrix $A$ the polynomials $$ \begin{bmatrix} p_1(t)\\p_2(t)\\p_3(t)\\p_4(t) \end{bmatrix}=A\begin{bmatrix} 1\\t\\t^2\\t^3 \end{bmatrix} $$ constitute a basis.


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Since this vector space $K[t]_3$ is isomorphic to $K^4$, where $K$ is the field you are taking polynomials over, there is a one-to-one correspondence between bases of $K[t]_3$ and invertible $4\times 4$ matrices over $K$, with the entries in column $k$ giving coefficients of the $k$-th basis polynomial.


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Any $(P_0, P_1, P_2, P_3)$ with $\deg P_i=i$ will be a basis of polynomials of degree three or less. Indeed, it is easy to show that such polynomials are linearly independent.


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Guessing an inverse can sometimes be done if you know similar matrices and their inverses by heart. The matrices I mention in the comment look like this: $${\bf D} = \left[\begin{array}{rrrrr}1&-1&0&0&0\\0&1&-1&0&0\\0&0&1&-1&0\\0&0&0&1&-1\\0&0&0&0&1\end{array}\right]$$ This ...


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Look at the matrix $A$ and notice that adding the second row to the first yields $e_1=(1,0,0,0)$, adding the third to the second yields $e_2$ and adding the last to the third yield $e_3$, after these operations the matrix is reduced to the identity matrix. The inverse simply encodes these operations. It is straightforward to see what is in this case, but the ...


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Disregarding the problem in itself, just answering the question as posed above "The problem": You can vectorize and use the Kronecker product, using some of the formula found here. Also after vectorization the transposition operation can be represented by a permutation matrix as explained here. You will end up with linear equation system $n^2\times n^2$ ...


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By imposing $K^T = -K$, we have $$K_{ij} = -K_{ji} \quad \forall i,j.$$ In particular, when $i = j$, $$K_{ii} = -K_{ii} \implies K_{ii} = 0.$$ By saying "entry can be chosen independently", we mean that an entry can be viewed an independent variable of others. For example, if $K_{12}$ is chosen to be an independent variable, then $K_{21}$ cannot be an ...


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Due to the symmetry condition, choosing a value for some matrix element generally fixes the value of some other matrix element. Therefore that other matrix element cannot be chosen independently. For example, take a symmetric $2\times 2$ matrix, $$M=\begin{pmatrix}*&*\\*&*\end{pmatrix}, M^T=M$$ And say you choose a value for the upper right element (...


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The problem is that looking at the $(3,3)$-entry alone is insufficient to prove that $A$ is singular. It is "natural" to think that a single equation in four unknowns $(x,y,z,t) x_3 = e_3$ is "very likely" to have solution, in contrast to your conclusion of the equation having no solution. ("Usually" when there are more unknowns than equations, there will ...


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You have a linear dependence between rows $r_3$ and $r_4$, we have $r_3=\frac79r_4$. This is sufficient to deduce that the matrix is not invertible. You could also calculate the determinant (this is easy because the matrix is triangle, it's the product of the diagonal coefficients), which is $0$. Hence the matrix is not invertible. There are many ways of ...


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I don't really understand your questions but I can try and clarify what is going on. The polynomials $P = K[x]$ form a commutative algebra and a vector space over $K$ whose basis is given by the monomials $(x^i)_{i=0}^{\infty}$. The elements in the dual space $P^{*}$ are linear functionals $L \colon K[x] \rightarrow K$ and so they "eat" polynomials and ...


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I am not sure what does your teacher means by a proof by row. One possibility is that perhaps compute the RREF and if you see a zero row in the RREF, the matrix is not invertible.


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Not necessarily. Consider $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ with $I=(1,2)$. If $\alpha_1,\alpha_2>0$ and $x=A\alpha$ then $x_1=\alpha_1>0>-\alpha_2=x_2$.


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As $A$ has orthonormal columns, its singular value decomposition must be in the form of $A=U\pmatrix{I\\ 0}V^T$. Hence to maximise $\det(A^T\Lambda A)$ is equivalent to maximise the determinant of $P_k$, the $k\times k$ leading principal submatrix of $P=U^T\Lambda U$. However, by the interlacing property, the $i$-th largest eigenvalue of $P_k$ is bounded ...


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I am not aware of any relevant research. Yet, for any $n\ge3$, there always exists a matrix that is non-triangular but whose eigenvalues are any $n$ given scalars $\lambda_1,\lambda_2,\ldots,\lambda_n$. The construction is recursive. First, we begin with a triangular matrix $$A_2=\pmatrix{\lambda_1&1\\ 0&\lambda_2}.$$ Now, if $n\ge3$ is odd, we ...


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The standard notation of the base matricx $B$ to $A$ ($B\dashrightarrow A$) is $S=M(id,A,B)$.


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matrix example of this type : Let $A$ and $B$ are two Jordan matrices form not diagonalizable, and $M$ the matrix obtained by concatenating $A$ and $^tB$, then $M$ is a matrix with two block in its diagonal and this block one are upper triangular matrix and the other are lower triangular matrix. Example $M =\left( \begin{array}{cccc} \alpha & 1 ...


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Let $u_1,\dots,u_n$ denote the rows of $A^{-1}$, and let $v_1,\dots,v_n$ denote the columns of $A$. Note that $A^{-1}A = I$, so that $u_i \cdot v_i = 1$ for all $i$ and $u_i \cdot v_j = 0$ for all $i \neq j$. Suppose that both have non-negative entries. Claim: Each column of $A$ contains at most one non-zero entry. Proof: Suppose that $v_j$ contains two ...


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Since $\{x_1,\dots,x_n\}$ is linearly dependent there exists $a_1,\dots,a_n\in F$ not all $0$, such that $a_1x_1+\dots+a_nx_n=0$. Therefore if $a_i$ is the scalar different than $0$ we have $x_i=-\frac{a_1}{a_i}x_1-\dots-\frac{a_n}{a_i}x_n$. Evaluating $w$ and using the statement you found in wikipedia one has $$w(x_1,\dots,x_n)=w(x_1,\dots,x_i,\dots,x_n)=w(...


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Consider the following example, where $AB$ has not even real eigenvalues. Consider the positive definite matrix: $$A = \begin{bmatrix} 2&1&2\\ 1&1&1\\ 2&1&3 \end{bmatrix} $$ with eigenvalues $\lambda_1 = 0.308$, $\lambda_2=0.6431,$ $\lambda_3=5.0489.$ Also, consider the upper triangular matrix: $$B=\begin{bmatrix} 1&1&1\\ 0&...


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We first see that $X{\bf e}_k = {\bf e}_{k+1}$ and $D{\bf e}_k = k {\bf e}_{k-1}$ so $X$ acts as multiplication by $x$ and $D \sim \frac{d}{dx}$. The commutator of $[X,D] = \text{diag}(-1,-1,-1,\ldots,-1,n)$ so $[X,D] = -1$ (and higher order correlators vanish) when acting on ${\bf e}_k$ for $k<n$. By Baker–Campbell–Hausdorff we therefore have $$e^{t(X-D)...


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I'm getting $P^{-1}AP=J$ where \begin{align*} P &= \left[\begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right] & J &= \left[\begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & ...


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I don't know what you mean by derive here, but by definition the two products give different types of geometric objects. The exterior product of two vectors (in $\mathbf{R}^n$, for any $n$) is a bivector, whereas the cross product of two vectors (in $\mathbf{R}^3$ only) is another vector. For $n=3$, there happens to be an isomorphism between bivectors and ...


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I don't think that the slope-like expressions such as $\frac{x_a-x_c}{y_a-y_c}$ will be particularly helpful in this problem. Instead, this is most easily expressed in vector notation: $$B = A + \frac{r}{|C-A|} (C-A) $$ To make it more explicit, it can be expanded out: $$(x_b,y_b) = (x_a,y_a) + \frac{r}{\sqrt{(x_c-x_a)^2 + (y_c-y_a)^2}} (x_c-x_a,y_c-y_a) $$


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The answer to your question an obvious "no". It should be easy to generate a random counterexample. E.g. suppose $$ A=B=\pmatrix{1&-1\\ 1&-1\\ &&1},\ D_1=\pmatrix{1\\ &1\\ &&0},\ D_2=\pmatrix{0\\ &1\\ &&1}. $$ Then $S=\operatorname{null}(A)$ consists of scalar multiples of $(1,1,0)^T$ and $p=\dim S=1$. However, $$ \Phi=...


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I believe the exterior products (wedge products) of two vectors $u$ and $v$ are denoted by $u\wedge v$. Note: the $LaTeX$ command for wedge products is also \wedge. I have borrowed this picture from Wikipedia's page on exterior algebra as it explains the difference very clearly. $a\times b$ is the cross product of the two vectors $a$ and $b$. The cross ...


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Whether or not the points lie on a graph we have for a set of $n$ points $$ \bar{x} = \Sigma x_i/n, \bar{y} = \Sigma y_i/n, $$ For a triangle $n=3.$


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Let $V, W$ are vectors spaces over the same field $K$ and $B$ a base of $V$, a bilinear map $f:V\times V\rightarrow W$ is nondegenerate iff there is no $v\not=0$ in $V$ s.t $f(v,e)=0 $ for all $e\in B$. So $f$ is well a $\Bbb{R}$-bilinear map, normaly if we want to check that $f$ is nondegenrate we must resolve $f(A,e)=0$ for all $e $ in the canonical real ...


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The value of the product in part $b$ is a vector, specifically $$\begin{bmatrix} 4 \\ -1 \\ 0 \\ 2 \end{bmatrix}$$ Now we want to find $a_1, a_2, a_3$ such that $$a_1 v_1 + a_2v_2 +a_3v_3 = a_1\begin{bmatrix}\frac{\sqrt 3}{3}\\{-}\frac{\sqrt 3}{3}\\{-}\frac{\sqrt 3}{3}\\0 \end{bmatrix} + a_2\begin{bmatrix} \frac{2\sqrt{15}}{9}\\\frac{\sqrt{15}}{9}\\\frac{\...


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Every field is a vector space over itself, not just the binary field. Look at @Felicity's answer -- if $\vec{u}$ and $\vec{v}$ are elements of the field itself, then you can see that all of the 10 vector space axioms are satisfied trivially.


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This is all you need. Follow each axiom and see if your binary field fits into each axiom.


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The answer in the linked question still applies. Consider the matrix \begin{align*} y &= \begin{pmatrix} 0 & \cdots & 0 & 1 \\ 0 & \cdots & 0 & 0\\ \vdots & \ddots & 0 & 0\\ 0 & \cdots & 0 & 0 \end{pmatrix} \in M_n(\mathbb{R}). \end{align*} If $y = x^n$ for some $x\in M_n(\mathbb{R})$, then all the ...


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There is no need to consider eigenvalues or eigenvectors. You have: $$ A = B^{-1} D B $$ with $D$ being a diagonal matrix. It follows that for any $n\in\mathbb{N}$ $$ A^n = (B^{-1}DB)\cdot(B^{-1}DB)\cdot\ldots\cdot(B^{-1}DB)= B^{-1} D^n B$$ holds, so for every polynomial $p\in\mathbb{R}[x]$ $$ p(A) = B^{-1} p(D) B$$ holds too. Since $p(D)$ is a diagonal ...


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It is false. Consider the matrix.$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ It is not the square of a complex matrix, so $p(X)=X^2$ is not surjective.


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If we can go directly to matrix $U$ starting from $A$ (through row operations without altering rows), then it is possible for $A$ to be written as $A=LU$. In other case, no. $$\begin{bmatrix} 1&1&0\\ 1&1&2\\ 1&2&1\end{bmatrix}\overset{R'_2:=R_2-R_1}{\longrightarrow} \begin{bmatrix} 1&1&0 \\ 0&0&2 \\ 1&2&1 \end{...


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We can write $p$ into this form:$$p(x) = a_mx^m + a_{m−1}x^{m−1} + · · · + a_1x + a_0$$ Now let $\Bbb v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Since $A^k\Bbb v = \lambda ^k\Bbb v$ for every $k$, we see that$$p(A)v = a_mA^m\Bbb v + a_{m−1}A^{m−1}\Bbb v + · · · + a_1A\Bbb v + a_0I\Bbb v \\= a_mλ^m\Bbb v + a_{m−1}λ^{m−1}\Bbb v + · · · + a_1λ\Bbb v ...


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Hint: 1) $$\left|\frac{x-2}{x+3} \right|=\frac{x-2}{x+3} \Leftrightarrow x\in (-\infty;-3) \cup [2;+\infty)$$ 2)$$\left|\frac{x-2}{x+3} \right|=-\frac{x-2}{x+3}\Leftrightarrow x\in (-3;2)$$


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$$\frac{x-2}{x+3}\ge 0\stackrel{\text{Mult. by}\;(x+3)^2}\iff (x-2)(x+3)\ge0\;,\;\;x\neq-3\iff$$ $$x<-3\;\;\text{or}\;\;x\ge2$$ and then $$\left|\frac{x-2}{x+3}\right|e^{|x-2|}=\begin{cases}\frac{x-2}{x+3}e^{x-2},&x\ge2\\{}\\\frac{x-2}{x+3}e^{-(x-2)},&x<-3\end{cases}$$ And thus $$-3<x<2\implies \left|\frac{x-2}{x+3}\right|e^{|x-2|}=-...



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