Let $A$ be an $n \times n$ invertible matrix and $v_1,v_2,...,v_m$ an element of $\mathbb{R}^{n}$. Prove if $\{v_1,v_2,...,v_m\}$ spans $\mathbb{R}^{n}$ then $\{Av_1, Av_2,...,Av_m\}$ also spans $\mathbb{R}^{n}$.       I'm not sure how to use the definition of spanning to prove this statement...do I have to get a homogeneous linear system?
123612
Let $A$ be an invertible $n\times n$ matrix. Let $\mathcal C=\{\vec v_1,\ldots,\vec v_k\}$ be a set of vectors with $\def\span{\operatorname{span}}\span(\mathcal C)=\Bbb R^n$. Show that the set $\mathcal D=\{A\vec v_1,\ldots,A\vec v_k\}$ satisfies $\span(\mathcal D)=\Bbb R^n$.  The columns of an $n \times n$ invertible matrix are a basis for $\Bbb R^n$, so multiplying $A$ by $C$ wouldn't change the span. I'm just not sure how to "show" it.
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?
11790
Given a matrix $A$ with dimensions $m \times n$, is $B=AA^T$ invertible if and only if the rows of $A$ are linearly independent?  So far, I've tried writing A as row vectors,   $$A = \begin{bmatrix}v_1\\ v_2\\ \vdots \\ v_m\end{bmatrix}$$  where $B_{i,j} = (v_i v_j) $  The Wikipedia page on the  suggests that the condition I stated above is necessary and sufficient, but does not provide a proof.
55490
In Linear Algebra by Friedberg, Insel and Spence, the definition of span (pg-$30$) is given as:     Let $S$ be a nonempty subset of a vector space $V$. The span of $S$,   denoted by span$(S)$, is the set containing of all linear   combinations of vectors in $S$. For convenience, we define   span$(\emptyset)=\{0\}$.   In Linear Algebra by Hoffman and Kunze, the definition of span (pg-$36$) is given as:     Let $S$ be a set of vectors in a vector space $V$. The subspace   spanned by $S$ is defined to be intersection $W$ of all subspaces of   $V$ which contain $S$. When $S$ is finite set of vectors, $S = \{\alpha_1, \alpha_2, ..., \alpha_n  \}$, we shall simply call $W$ the   subspace spanned by the vectors $\alpha_1, \alpha_2, ..., \alpha_n$.   I am not able to understand the second definition completely. How do I relate "set of all linear combinations" and "intersection $W$ of all subspaces"? Please help.  Thanks.
40437
The problem is from  Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful.      Let $A,B$ be $m \times n$ and $n \times m$ matrices. Prove that $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.    Please provide with only a hint.
101150
I want to see if I am understanding the proof of the replacement theorem correctly.   Let $V$ be a vector space that is spanned by a set $G$ containing $n$ vectors.  Let $L \subseteq V$ be a linearly independent subset containing $m$ vectors.  Then $m\leq n$ and there exists a subset H of G containing exactly $n-m$ vectors s.t. $L \cup H$ generates $V$.   The proof in the book says to start with $m=0$. In that case $L=\varnothing$, the empty set.        Let's assume this theorem is true for some integer $m \geq 0$.   Let $L = \{v_j\}_{j=1}^{m+1}$ and define it as a linearly independent subset of $V$ consisting of $m+1$ vectors.  Since any subset of a linearly independent set is linearly independent as well ($S_1 \subseteq S_2 \subseteq V$), then $\{v_j\}_{j=1}^m$ is linearly independent also.   It then says to use the induction hypothesis to say that $m \geq n$. Well, ok, but that seems sort of pulled out of nowhere to me. The next step is to say that there is therefore another subset, $\{u_k\}_{k=1}^{n-m}$ of G such that $\{v_j\}_{j=1}^m \cup \{u_k\}_{k=1}^{n-m}$ spans $V$.   That being the case there are scalars $(a_j)_{j=1}^m$ and $(b_k)_{k=1}^{n-m}$ which you can multiply by the vectors $v_j$ and $u_k$. You can then add the two sets of vectors, yielding $$\sum_{j=1}^m a_j v_j + \sum_{k=1}^{n-m} b_k u_k = v_{m+1}$$  At this point I am not sure I understand things. Because we are assuming $n-m&gt;0$ -- otherwise $v_{m+1}$ is linearly dependent. But then it says not only is $n&gt;m $ but $n&gt;m+1$, and that is where I am losing the plot to get to the next step.
58821
Let $T_1$ and $T_2$ be two Linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$.  Let $\{x_1,x_2,\cdots,x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1(x_i)\neq 0$ for every $1\leq i\leq n$ and that $x_i\perp Ker (T_2)$ for every $1\leq i\leq n$.   Which of the following is true?   $T_1$ is invertible $T_2$ is invertible Both $T_1$ and $T_2$ are invertible Neither $T_1$ nor $T_2$ is invertible.   As $T_1(x_i)\neq 0$ for each $1\leq i\leq n$ we do not have $T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 $ unless each $a_i=0$ i.e.,$a_1x_1+a_2x_2+\dots+a_nx_n=0$ i.e., $T_1$ is one one thus invertible.  I am not sure if $T_2$ is invertible or not.  we have $x_i\perp Ker (T_2)$ for all $x_i$. Would that be a good idea say something like  $\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)$ and as they span whole space we would have   $\mathbb{R}^n\perp Ker(T_2)$ and thus $Ker(T_2)=0$ so it is injective so it is invertible...  Thus both $T_1$ and $T_2$ are invertible?  I would be so thankful if someone can assure what it has been done here is sufficient/clear.  THank you
38731
If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
17097
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.  The following question was given in a contest I took part. I had an approach but it didn't solve the problem.     Consider $V$ a linear subspace of the real vector space $\mathcal{M}_n(\Bbb{R})$ ($n\times n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?   A quick guess would be $n^2-n$ since if we consider $W$ the set of $n\times n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.  Now the only thing there is to prove is that if $V$ is a subspace of $\mathcal{M}_n(\Bbb{R})$ of dimension $k &gt; n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.  My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
7568
Let $A$ and $B$ be $m\times n$ matrices over $\mathbb{Z}$ such that image($A$) = image($B$), where $A$ and $B$ are considered as maps $\mathbb{Z}^n \rightarrow \mathbb{Z}^m$. Does there exist an invertible $n\times n$ matrix $P$ such that $A=BP$? what happens if we consider the matrices to be over $\mathbb{Q}$ instead of $\mathbb{Z}$?  I would prefer some hints instead of full answer.  Thanks in advance.
113403
Let $V$ be an n-dimensional vector space over $\mathbb{R}$ and $T:V \rightarrow V$ be linear. Call a vector $v \in V$ cyclic if $V$ is spanned by $\{v, \ Tv, \ T^2v,..\}$.  Question:  Show that $v$ is cyclic for $T$ then there is some $k \geq 1$ such that   $$B=\{v, \ Tv,..T^{k-1}v \}$$ is linearly independent in $V$ and that $T^kv$ lies in  the span of $B$.  Thoughts:  If we suppose that no such $k$ exists, then every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent, or it is linearly dependent with $T^lv$ not lying in it span.  There are in effect then two cases to consider.  Suppose every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent. Then $T^{l-1}$ can be expressed as a linear combination of $v,...,T^{l-2}v$. Now, $\{v, \ Tv,..T^{l-2}v \}$ is also linearly dependent by assumption and so we can express $T^{l-2}$ as a linear combination of  $v,...,T^{l-3}v$ etc... Continuing in this way, we show that $T^{l-1}v= \lambda v$ for some $\lambda \in \mathbb{R}$. Thus $$B=\{ \mu v \ | \ \mu \ \text{is an eigenvalue of} \ T^mv \ \text{for some} \ m \}$$ This is 'at best' countably infinite, and therefore could not span $V$ even if it were 1-dimensional, wince our base field is $\mathbb{R}$, which is uncountable.  Suppose then that $B$ is linearly independent for some $k$ but that $T^kv$ does not lie in its span. I can't see how to progress this argument.  Note: This question is for second year Undergraduates, and I think there must be some more simple approach; even the first part surprises me if we need to argue using the uncountability of $\mathbb{R}$, so any alternative approaches are also appreciated.
1817
There is an invertible $n×n$ matrix with each entry $0$ or $1$. Now we negate ($x\to 1-x$) the diagonal of the matrix, and it is still invertible. How many such matrices exist for a given $n$?  Can we derive a general formula? If $n=2$ there are $2$ such matrices, if $n=3$ there are $48$. But I have no idea whether a formula exists for higher $n$.
3284
If X and Y are two sets of vectors in a vector space V, and if X $\subset$ Y, then is span X $\subset$ span Y? If so, why is or isn't the span of X a subset of the span of Y?  EDIT:  Thank you for the hints! Here is the proof I came up with; please let me know if it is correct.    The spanning set of X can be written as: Span(X) = {$a_1 x_1 + a_2 x_2 + ... + a_n x_n$} where all $x_i$ are vectors and all $a_i$ are scalars. Since X$\subset$Y then all $x_i$ are also in Y. Thus, Span(x) is a linear combination of vectors from Y So Span(x) $\subset$ Span(Y).
101311
Let $V$ be the vector space made by the $n \times n$ square matrices.  a) Prove that $S=\{A\in V|A^t=A\}$ is a subspace of $V$  b) Prove that $T=\{A\in V|A^t=-A\}$ is a subspace of $V$  c) Prove that $V=S\oplus T$    SOLUTION  a) Let $O$ be the zero matrix, then $O \in S$ because $O^t=O$.  Then, let $A, B \in S$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(A^t)+k(B^t)=rA^t+kB^t$$ then $rA+kB \in S \implies$ $S$ is a subspace of $V$  b) Let $O$ be the zero matrix, then $O \in T$ because $O^t=O$.  Then, let $A, B \in T$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(-A^t)+k(-B^t)=-rA^t-kB^t=-(rA^t+kB^t)$$ then $rA+kB \in T \implies$ $T$ is a subspace of $V$  c) I have no idea how to do it.  I HOPE YOU CAN HELP ME VERYFING THE a), b) parts and with the part c).
8349
Let $M(n;\mathbb{R})$ denote the set of all $n \times n$ matrices with real entries (identified with $\mathbb{R}^{n^{2}}$ and endowed with its usual topology) and let $GL(n;\mathbb{R})$ denote the group of invertible matrices. Let $G$ be a subgroup of $GL(n;\mathbb{R})$. Define $$H = \biggl\{ A \in G  \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}$$  Then is $H$ normal in $G$?  For proving $H$ normal i should verify two things:   First $H$ is a subgroup. For all $A \in G, B \in H$, we must have $A \cdot B \cdot A^{-1} \in H$.   That's it i  am not able to proceed any further.
16846
Let $S$ and $W$ be subsets of a vector space $V$. Show that if $S$ is a subset of $W$, then $\mathrm{span}(S)$ is a subspace of $\mathrm{span}(W)$.  Ok I'm finally understanding what each of these things mean.. but I'm running out of ideas on how to actually show it without using arbitrary vector space examples like Rn. Since sets and subspaces are kind of new its hard to figure out how to write correct proofs with them.   I can see how if $V = \mathbb{R}^3$ and $S = \{(1,1,1)\}$ and $W = \{(1,1,1), (1,1,2)\}$ then $\mathrm{span}(S) = R1(line), span(W) = R2(plane)$ and that R1 is a subspace of R2... im just having trouble showing this officially when all vector spaces and sets are completely in general....  I could get as far as writing out $S=(u_1,u_2,\dots,u_n)$, $u_i \in V$ and same with $W$, but I can't really figure out what else to do in this sort of proof.
26512
Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'.