If $A$ is matrix of $k\times k$ , then is it necessary that $Ax, A^2x , A^3x ...... A^kx$   will be linearly dependent ? $x$ is a vector and $A$ contains all real entries
78290
Problem: Let $A\in M_{n\times n}(\mathbb R)\,$ be a matrix and suppose that a positive number $k\,$ exists such that $A^k = 0\,$ and $A^{k-1} \neq 0$. Suppose that $x=\left[ \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right]$ is a vector in $\mathbb{R^n}$ such that $A^{k-1} x \neq 0$. Prove that the $k\,$ vectors $\,x,Ax,\dots,A^{k-1}x\,$  are linearly independent.     My attempt: Suppose $x + Ax + \dots + A^{k-1}x = 0$. Multiply both sides with $A^{k-1}$. Then we have $A^{k-1}x + A^k (x + Ax + \dots + A^{k-2}x) = 0 \Leftrightarrow A^{k-1}x = 0 \Leftrightarrow x = 0$   which implies $x + Ax + \dots + A^{k-1}x\,$ is linear independent.    This problem looks quite easy but I want my proof to be checked. Is it correct?
89471
Suppose $x\in\mathbb{R}^n$ and $A\in\mathbb{R}^{n\times n}$. What is a sufficient condition for $$x, Ax, A^2x, ..., A^{n-1}x$$ being linearly independent?
5954
The question is:  If A is a square matrix such that $A^2=A$ then $A^n =A$ for all natural numbers $n$  greater than one. What is $A$ if $A \ne 0$ and $A \ne I$.   I figured out an answer but I can't tell if that's the only answer. Let's say that $a_{kk}$ is a value in $A$. Every value in $A$ is $0$ except for $a_{kk}$ and $a_{00}$, they're $1$. I haven't gotten this answer mathematically.  I've tried a few approaches but ended up with the identity matrix.
88413
Let A be $3\times3$ matrix such that $Au$ and $u$ are orthogonal for each column vector $u\in\Bbb R^3$. Prove that $A^T=-A$. I thought  $(A+A^T)u\cdot u=0$ but it doesn't imply $A^T=-A$, so I'm stuck... Edit : Does this property also hold when we replace 3 by an arbitrary $n$?
74313
Let $T:\mathbb{R}^n\to \mathbb{R}^n$ be a linear transformation, where $n\ge2$. For $k\le n$, let $E=\{v_1,v_2,\dots,v_k\}\subset \mathbb{R}^n$ and let $F=\{Tv_1,Tv_2,\dots,Tv_k\}$, then which of the following is true.  $A.$ If $E$ is linearly independent, then $F$ is linearly independent  $B.$ If $F$ is linearly independent, then $E$ is linearly independent.   $C.$ If $E$ is linearly independent, then $F$ is linearly dependent  $D.$ If $F$ is linearly independent, then $E$ is linearly dependent.   Please help me out with this problem. And if some theorem is responsible for it's solution please mention that.
85485
I need to determine if the following set of vectors are are linearly independent or not.  1) $S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3$, it is known that $(1,2,3) \notin \text{span}\{\mathbf u , \mathbf v , \mathbf w\}$  2) $S=\{\mathbf u + \mathbf v,\mathbf u - \mathbf v , \mathbf u - 2\mathbf v +\mathbf w\}\subseteq \mathbb R^n$ where $\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^n$ is a linearly independent set.  3) $S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3$, where it is known that $(1,2,-3),(1,-3,2),(2,-1,5),(0,0,1)$ all belong to $\text{span}\{\mathbf u , \mathbf v , \mathbf w\}$.  I honestly have no idea how to start this, all I know is that the if the vectors $\mathbf u , \mathbf v , \mathbf w$ are linearly independent, then $a\mathbf u + b\mathbf v + c\mathbf w=0$ has only the trivial solution.  Any help here will be appreciated, many thanks!
113065
I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement?   Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}?  Any help would be appreciated! Thank you!
94313
Given that the sequence $(v,Av,A^2v,...,A^nv)$ is linearly dependent (1), $A_{n \times n}$, over $\mathbb{F^n}$ prove that scalars $c_0,c_1,..,c_n$ exist (such that not all of them are zeroes) such that the matrix $M=c_0I+c_1A+..+c_nA^n$ is singular.    First I'm not sure how (*) can be used here exactly. We can say that the sequence $(c_0I,c_1A,..,c_nA^n)$ resembles the sequence (1) where $v=I$. But here we have scalar $c_i$ so it's not exactly the same sequence.   Maybe I could say that the sequence $(c_0I,c_1A,..,c_nA^n)$ is linearly dependent because it has $n+1$ elements while over $dim\mathbb{F^n}=n$ $\quad$(2).  Intuitively I understand that suppose we sum all the matrices $c_iA^i$ and then in the last sum operation we get to some $c_jA^j$ and if it's a linear combination of other matrices then we can have such a $c_j$ that: $$ c_oI+\sum_{i=1}^{n-1} c_iA^i+c_jA^j=0 $$ But I don't think this is enough to prove.
98004
how do you prove if a subset of a set of vectors is linearly dependent, then the entire set is linearly dependent.  i know that linear dependence means that it can be written as a linear combination. so i can take 1 vectors from the set and write it as a linear combination of some of the vectors find it to be linearly dependent and do so for the whole and find it to be linearly dependent.  i do not know how i can prove  without actually resorting to an example and you cant prove by example.
70275
So, $A$ is a $n \times n$ matrix with integer entries.  The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .  I know that $A^{-1}= {\rm adj}(A)/{\rm det}(A)$ but I have no idea where to go from there for the forward direction.  Any help here would be greatly appreciated.  For the backwards direction, I think I am okay.  I plugged the 1 and the -1 options into $ {\rm adj}(A)/{\rm det}(A)$ So, I know that $A^{-1}= {\rm adj}(A)$ or $-{\rm adj}(A)$.  Then I said that because the ${\rm adj}(A)$ is simply the matrix of co-factors of $A$, and because $A$ has all integer entries then ${\rm adj}(A)$ will have all integer entries, which will mean that $A^{-1}$ will have all integer entries.  Is that okay? Or am I missing something else?   Thank you!!``
97335
Prove that if $A$ is nonsingular, then the sequence $X_{k+1}=X_k+X_k(I-AX_k)$ where $A$ and $X_k$ are $n\times n$ matrices with $k=0,1,2,...$ converges to $A^{-1}$ if and only if $ρ(I-X_0A)&lt;1$.  I'm really stuck on this problem. Any hints are greatly appreciated.
65673
Basically does a vector set that is linearly independent in $R^n$ automatically span $R^n$?  My initial thought is yes, but is there some counterexample that can disprove this?
111356
Given:$ A_{n\times n} , B_{n\times n} .$  Is it necessary true that there is similarity between $AB$ and $BA$?  I'm not quite sure how to check it is true for every non-singular M matrix. $M^{-1}ABM = BA$  Anyone have any ideas what am I missing to understand in similarity? If it's not necessary true then give a counter example.
73890
Let $A \in Mat_{n \times n}(F)$ a matrix which satisfies $A^n=0$ and $A^{n-1} \ne 0$ for some positive integer $n$. Let $v \in F^n$ be such that $A^{n-1}v \ne 0$.  Prove that $\{v, Av, A^2v,...,A^{n-1}v\}$ is a basis for $F^n$.     So, from definition of $A$, I know that $A^iv \ne 0$ for all $0\le i \le n-1$. I know I need to prove the set is linearly independent and span $F^n$ to prove it's a basis, but I just don't know how.
88616
Given a matrix $A$ in $\mathbb R^{m\times n}$, if I want to know whether or not there is an $x \in \mathbb R^{n}$ such that  $$ Ax &gt; 0, $$ (meaning all elements in $Ax$ are positive). Is there some easy way to verify whether or not such $x$ exists?
62613
Let $A$ and $B$ be $n\times n$ matrices such that $ABAB=0_{n\times n}$.  Can we conclude that  $BABA$ must be  $0_{n\times n}$?   I fail to give a counterexample when $n=2$， so I guess the answer is "yes"
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The question was given in the early chapters of Linear Algebra by Hoffman &amp; Kunze, so I am trying to give a proof with only the tools given to me so far - which are mainly row reduction and knowledge of matrix multiplication, row reduced echelon forms, row equivalence and linear independence.  I attempted a proof as per the following:     Consider $A$ as a collection (not sure if this would be the ideal expression) of $1 \times n$ row vectors, and $B$ as a collection of $n \times 1$ column vectors. Then we have that:      $$ A=\begin{bmatrix} r_1 \\ \vdots \\ r_m \end{bmatrix},\ B=\begin{bmatrix} c_1 &amp; \cdots &amp; c_m \end{bmatrix}. $$   Thus it follows that:   $$ AB =\begin{bmatrix} r_1\cdot c_1 &amp; \cdots &amp; r_1\cdot c_m \\ \vdots &amp; ~ &amp; \vdots \\ r_m\cdot c_1 &amp; \cdots &amp; r_m \cdot c_m  \end{bmatrix} $$   Clearly, by inspection, the rows are linearly dependent.      Since the rows of $AB$ are linearly dependent, it naturally follows that the reduced row echelon form of $AB$ contains zero rows. Hence, $AB$ is not invertible.   Would this be a mathematically sufficient proof?