Please help with the following problem:  Given a $m \times n$ matrix $A$, $m$-vectors $b$ and $y$, and $n$-vectors $c$ and $x$. Write the dual $LP$ problems $P$ and $P^d$ in the standard form.    Whether $x$ (respectively, $y$) is a feasible vector for $P$ (respectively, for $P^d$)? Whether $x$ (respectively, $y$) is an optimal solution for $P$ (respectively, for $P^d$)? Whether the complementary slackness conditions hold for $P$ (respectively, for $P^d$)?    Consider the cases a-g listed below and explain your answers in each case.    a.) $ A = \begin{bmatrix} 1 &amp; 2 \\ 3 &amp; 4 \end{bmatrix}$, $b^T = (8,18)$, $c^T = (2,1)$, $x^T = (6,0)$, $y^T = (0,2/3)$
123389
Given a $m$ x $n$ matrix $A$, $m$-vectors $b$ and $y$, and $n$-vectors $c$ and $x$. Write the dual $LP$ problems $P$ and $P^d$ in the standard form.    Whether $x$ (respectively, $y$) is a feasible vector for $P$ (respectively, for $P^d$)? Whether x (respectively, y) is an optimal solution for $P$ (respectively, for $P^d$)? Whether the complementary slackness conditions hold for $P$ (respectively, for $P^d$)?    Consider the cases a-g listed below and explain your answers in each case.    a.) $ A = \begin{bmatrix} 1 &amp; 2 \\ 3 &amp; 4 \end{bmatrix}$, $b^T = (8,18)$, $c^T = (2,1)$, $x^T = (6,0)$, $y^T = (0,2/3)$    b.) $ A = \begin{bmatrix} 1 &amp; 1 &amp; 2 \\ 2 &amp; 3 &amp; 4 \\ 7 &amp; 6 &amp; 2 \end{bmatrix}$, $b^T = (2,3,8)$, $c^T = (8,9,4)$, $x^T = (6/9,5/9,0)$, $y^T = (0,5/3,2/3)$  c.) $ A = \begin{bmatrix} 1 &amp; 1 &amp; 2 \\ 2 &amp; 3 &amp; 4 \\ 6 &amp; 6 &amp; 2 \end{bmatrix}$, $b^T = (2,3,8)$, $c^T = (8,9,5)$, $x^T = (1,1/3,0)$, $y^T = (1,1,1)$  d.) $ A = \begin{bmatrix} 3 &amp; 2 \\ 1 &amp; -2 \end{bmatrix}$, $b^T = (6,1)$, $c^T = (2,1)$, $x^T = (1,2)$, $y^T = (1,1)$    e.) $ A = \begin{bmatrix} 1 &amp; 1 &amp; 1 &amp; 1 \\ 2 &amp; 1 &amp; -1 &amp; -1 \\ 0 &amp; -1 &amp; 0 &amp; 1 \end{bmatrix}$, $b^T = (40,-5,10)$, $c^T = (1,-3,1,4)$, $x^T = (35/3,0,55/3,10)$, $y^T = (2,2,0)$  f.) $ A = \begin{bmatrix} 2 &amp; -6 &amp; 2 &amp; 7 &amp; 3 &amp; 8 \\ -3 &amp; -1 &amp; 4 &amp; -3 &amp; 1 &amp; 2 \\ 8 &amp; -3 &amp; 5&amp; -2 &amp; 0 &amp; 2 \\ 4 &amp; 0 &amp; 8 &amp; 7 &amp; -1 &amp; 3 &amp; \\ 5 &amp; 2 &amp; -3 &amp; 6 &amp; -2 &amp; -1\end{bmatrix}$, $b^T = (1,-2,4,1,5)$, $c^T = (18,-7,12,5,0,8)$, $x^T = (2,4,0,0,7,0)$, $y^T = (1/3,0,5/3,1,0)$
41568
Let $a$ and $b$ be two unitary vectors in $\mathbb E^3$, and let $Q$ be an orthogonal matrix. Does the following hold?  $$Qa \wedge Qb = \pm Q(a \wedge b)$$
21698
I'm a mathematics undergrad student who finished his first university year succesfully. For this post, it should be interesting to note that I already took a course in group theory.  I also had a course in linear algebra where we treated the following topics:  1) Vector spaces; basis; span; linear dependency; direct sum   2) Linear transformations, matrices, rank  3) Linear varieties, system of linear equations  4) Determinants  5) Eigenvalues and eigenvectors, diagonalisation, triangalisation, cayley-hamilton,  6) Euclidean spaces: inner products, norm, hermitian transformations, orthogonal basis, Gramm-Schmidt, orthogonal transformations  7) Prehilbert spaces  8) Isometries  9) Bilinear forms and quadrics in $2$ and $3$ dimensions  Now, I want to revise these topics, but it would be too boring to reread these books again, so now I look for a book that has all these topics, but also some additional requirements:   Focus on intuition (e.g. connections between linear algebra and geometry). This is a hard requirement for me, since there was no attention given to this in the course I took) Additional topics such as Jordan form of matrices, quotient vector spaces, ... Good exercises (preferred with solutions somewhere) The book must be rigorous   In the course I had, we mainly discussed finite-dimensional vector spaces, but I am also interested in a theory of infinite dimensional vector spaces.  Can someone hint me towards a good book that would suit me? If I have to add any information, please leave a comment and I will edit my post.  Thank you for your time.
111738
It is easy to find the minimum of $\|Ax-b\|_2$, when $A$ has full column rank. But how is the case when we add the constraint $\|x\|_2=1$? Or, to be explicit,  $$\min_{\|x\|_2=1}\|Ax-b\|_2=?$$  My idea is to construct the corresponding Lagrange function  $$L(x,\lambda) = (Ax-b)^T(Ax-b) + \lambda(x^T x - 1)$$  Differentiate the function with respect to $x,\lambda$, we have  $$A^T A x - A^T b + \lambda x = 0$$  $$x^T x=1$$  However, I find it hard to move forward to get the answer. Could anyone help me? Thanks a lot!
56592
Let $W = \{ p(B) : p \text{ is a polynomial with real coefficients}\}$, where $$B= \begin{pmatrix} 0 &amp; 1 &amp; 0\\  0 &amp; 0 &amp; 1\\  1 &amp; 0 &amp; 0 \end{pmatrix}$$  Which of the following possibilities presents the tightest bounds on the dimension $d$ of the vector space $W$?     $4 ≤ d ≤ 6$    $6 ≤ d ≤ 9$    $3 ≤ d ≤ 8$    $3 ≤ d ≤ 4$
15141
I need to find $|GL_2(F_p)|$, I am very very uncomfortable in counting, so please help here I proceed,  $\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}$ be such an invertible matrix, so we need and sufficient that $ad\neq bc$, so if i chose all non zero $a,b,c$ arbitrarily and chose $d\neq a^{-1}bc$ and form matrices then the cardinality   will be $(q-1)^3\times (q-2)$, well now if one of $a,b,c,d$ be zero then other three can be anything nonzero so in this case cardinality will be $4(q-1)^3$, so finally if two entries are $0$ then they must be opposite to each other i.e diagonally, but other two must be nonzero, so cardinality in that case $2(q-1)^2$, so total number is just sum of three cases.
57297
Let $A$ be an orthogonal $n × n$ matrix and suppose that $x, y \in \mathbb R^n$   and $\theta \in (0 , \pi)$ satisfy    $$Ax = \cos (θ)x − \sin (θ)y,\\ Ay = \sin (θ)x + \cos (θ) y.$$      Prove that $x$ and $y$ must be perpendicular vectors of equal length, i.e. show that $|x|^2 = |y|^2$   and $x · y = 0$.   We know that because $A$ is orthogonal then$$\langle x,x \rangle =\langle Ax,Ax \rangle, \quad \langle x,y \rangle=\langle Ax,Ay \rangle, \quad \langle y,y \rangle=\langle Ay,Ay \rangle.$$ I was already struggling with showing that $|x|^2 = |y|^2$. So far I just wrote out that   $$|x|^2= \sum_{i=1}^n (Ax_i)^2= \sum_{i=1}^n \cos^2(\theta)x_i^2-2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)y_i^2,\\ |y|^2= \sum_{i=1}^n (Ay_i)^2= \sum_{i=1}^n \cos^2(\theta)y_i^2+2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)x_i^2.$$  I don't really see how I could conclude from this that they are equal.
121997
So it's been a while since I've taken Linear Algebra, but my friend asked me a question, that I couldn't answer.  If a matrix $A$ exists such that $A^3 = I$, does $A$ have to equal the identity matrix $I$?  My first instinct was to say no, but... (edited out my incorrect math)  EDIT: thanks guys for the awesome examples  EDIT2: Followup question: Is there a way to solve for all possibilities of A if given A^3 = I?
80609
Let $x = (x_1,\ldots,x_p) \in \mathbb R^p$. I'm solving the constrained optimization problem     $$\begin{align*} \text{min} &amp;\quad \sum_{i=1}^p (y_i-x_i)^2 \\ \text{s.t} &amp;\quad \sum_{i=1}^p y_i - 1 &amp;&amp;=0\\            &amp;\quad\forall  i = \overline{1,p}: -y_i &amp;&amp;\le 0  \end{align*}$$   My attempt:  Let $f(y) = \sum_{i=1}^p (y_i-x_i)^2$, $h(y) = \sum_{i=1}^p y_i - 1$, and $g_i(y) = -y_i$ for all $i = \overline{1,p}$.  We have $f,g_i$ are convex and $h$ is linear. Let $\alpha =(1/p, \cdots, 1/p)$. Then $h(\alpha)=0$ and $g(\alpha) &lt;0$ for all $i = \overline{1,p}$. It follows that Slater's condition is satisfied. By Karush-Kuhn-Tucker conditions, we have $$\begin{aligned} \begin{cases} \forall i = \overline{1,p}:\mu_i &amp;\ge 0 \\ \forall i = \overline{1,p}: g_i(y) &amp;\le 0\\ h(y) &amp;=0 \\ \forall i = \overline{1,p}:\mu_i g_i(y)&amp;=0 \\ \nabla f (y)+ \lambda\nabla h (y)+ \mu_i \nabla g_i (y) &amp;=0 \end{cases} &amp;\iff \begin{cases} \forall i = \overline{1,p}:\mu_i &amp;\ge 0 \\ \forall i = \overline{1,p}:-y_i &amp;\le 0\\ \sum_{i=1}^p y_i - 1&amp;=0 \\ \forall i = \overline{1,p}: -\mu_i y_i &amp;=0 \\ \forall i = \overline{1,p}: 2(y_i - x_i) +\lambda - \mu_i &amp;= 0 \end{cases} \\ \end{aligned}$$  Then I'm stuck at solving the last system of equations.  How can I proceed to solve it? Thank you so much!
34491
I guess that I'm missing here something...  How to prove that, if two matrices are similar, then their minimal polynomials are the same one.   Thanks!
93380
Say I have a matrix A that is m by n and m &lt; n. For context, say I am interested in finding a vector $ \theta_z \in \mathbb{R}^m$ such that:  $$ A^T \theta_z = \theta_x$$  i.e. we have more rows/"equations" than columns/"unknowns".  I was going to do it by multiplying both sides by A and then find the inverse of $AA^T$.  I was told that this only works when we have linearly independent rows. Is this suppose to be obvious? What is an intuitive way to explain this? Also, I am even more interested in a precise and rigorous proof if there is one.   I was also told that when the rows are dependent, then  the system has an infinite number of solutions. However, I was a little skeptical about this, because if we have more rows, then if we did some elimination on the original $ A^T \theta_z = \theta_x$, then we could have at most a rank of size $m$ (one for each column). Which could lead to only 1 solution or zero (none) but never to infinity. So I think that solutions is wrong, i.e. linearly independent rows if anything, shouldn't lead to more than 2 solutions ever, we should only have 1 or zero for the original system $ A^T \theta_z = \theta_x$. Is that correct?
18758
Question:     Let $A$ be a $3 \times 3$ Matrix such that $[-3,4,1]$   is the eigenvector corresponding to eigenvalue $3$, and $[6,-3,2]$ is an eigenvector corresponding to the eigenvalue $2$. If $v$ = $[0,5,4]$, compute $A^{2}v$.      Hint: $[6,-3,2]+2[-3,4,1]$ = $v$   Where I'm stuck:  I know that we can  solve A with the equation:   A = $P^{-1}DP$, where $P$ is the matrix consisting of the eigenvectors and $D$ is the matrix consisting of the eigenvalues.  However, since $A$ is a $3$ x $3$, we should have a $3 \times 3$ matrix $D$ and $ 3 \times 3 $ matrix $P$ as well, consisting of the eigenvalues and eigenvectors respectively - and in this case, we only have two eigenvalues and two eigenvectors. Where can I obtain the other one of each?  Any help will be very appreciated.  Thank you.
85313
Let us look at two linearly independent vectors $v_1$ and $v_2$, which span a plane in $\mathbb{R}^n$ through the origin. Let us form a matrix $V$ out of these vectors so that the first column of $V$ is the vector $v_1$ and the second column is the vector $v_2$. Show that the the matrix $P=V(V^TV)^{-1}V^T$ projects all vectors $x\in\mathbb{R}^n$ orthogonally onto the plane spanned by $v_1$ and $v_2$.
115817
I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
15798
If $A$ is a $n\times m$ matrix, is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true?  I'm thinking that rank-nullity would be the simplest and easiest way to prove this, but would applying the definition of linear independence to the components of $(A^T)$ be more illustrative? I'm a bit hung up on this "necessity" issue, and not sure if it has a particular logical meaning in linear algebra.
115178
Is the set of all $n\times n$ matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all $n\times n$ matrices?   Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with $2\times 2$ matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough.
74827
This statement is used without proof in various sources (incl. Ziegler's &quot;Lectures on polytopes&quot; and Gruber's &quot;Convex and discrete geometry&quot;), yet I don't have any idea how to prove it, or have intuition about it in $n&gt;3$ dimensions. Someone suggested me to use a concept of dual polytope to prove it, I have not explored it thoroughly and maybe someone can help with understanding how duality is convenient here.