Let $A\in \mathbb R^{n\times n}$ have a right inverse $B\in \mathbb R^{n\times n}$, i.e. $AB=E_{n\times n}$. Is it possible to see only using the definition of the matrix product that $BA=E_{n\times n}$?  Please do not tell me anything about the rank or endomorphisms of finite-dimensional vector spaces or even the Gauss algorithm. Only the definition of $[AB]_{j,\ell} = \sum\limits_{k=1}^n a_{j,k} b_{k,\ell}$ (which, by assumption is $1$ for $j=\ell$ and $0$ else) is allowed.
4880
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
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.
3063
I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.  In the proof of the theorem, they wrote:     $$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$  ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$  [...]   I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?  Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so?
18211
I am having trouble trying to prove that a matrix $A \in M_{n}(\mathbb{R})$ has inverse if and only if it's rank($A$) = n. What I was trying to do is to prove it using the fact that  $$rank(A) = \dim(\text{column-space of} \ A)$$ but I don´t really know how to continue.   Thank you for your help.
99501
$$GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$$  This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as $SL_n(\Bbb{R})$ is $1$ under the determinant, it is the kernel, and by FIT, the above is proved.  I was wondering how to prove this without the isomorphism theorem ?
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.
16310
If $u$ $∈ \mathbb R^{m \times 1}$ and $v ∈ \mathbb R^{n \times 1}$ how do you show that the $(m \times  n)$ matrix $uv^T$ has rank $1$? Would providing an example be sufficient to prove it?
108527
Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
79831
Let $T_1,T_2$ be two linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$. Let $\{x_1,\cdots , x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1x_i \neq 0$ for every $i=1,2,\cdots, n$ and that $x_i \bot \ker T_2$ for every $i=1,2,\cdots, n$. Which of the following is/are necessarily true?   $T_1$ is invertible. $T_2$ is invertible.  both $T_1,T_2$ are invertible.  neither $T_1$ nor $T_2$ is invertible.   My Approch:  Given dimension of $\mathbb{R}=n$ and  $x_i\perp Ker(T_2)$  So, $\text{Ker}T_2=0$  Row space and null space are perpendicular to each other.  From Rank Nullity Theorem:  let T : V → W be a linear map. Then the rank of T is the dimension of the image of T and the nullity of T is the dimension of the kernel of T, so we have  $\text{dim}(\text{Im}(T))+ \text{dim}(\text{Ker}(T))=\text{dim}(V)$  Applying above result we get,  $\text{dim}(\text{Im}(T))+ 0=n$  Therefore, $T_2$ is one-one.  I could not solve this problem completely. what to do next?
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?
108668
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,  $$\mathbf{A}^2=-\mathbf{I}$$  I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?
16070
How do I prove this statement: Let $V$ be an unitary vector space over $\Bbb C$, $(,)$ be an inner product on $V$ and $\Bbb A$ operator $V\rightarrow V$. Then $(\Bbb Av,v)=0$ if and only if $\Bbb A=0$.  I know that on $\Bbb R$ this does not apply, since we are searching for an operator that assigns to my vector a vector that is orthogonal and so every orthogonal projection on a subspace of $\Bbb R$ can be this operator. But how is this possible on $\Bbb C$? Thank you.
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
77656
Let $S$ be the subspace , of  $M_{n \times n}(\mathbb R)$ (the vector space of all $n \times n$ real matrices ) , generated by matrices of the form $AB-BA$ , where $A,B \in M_{n \times n}(\mathbb R)$ , then how do we prove that $\dim S=n^2-1$ ? The only thing that I can determine is that the trace of all matrices of $S$ is $0$ . Please help
42711
Let $R$ be a commutative ring with $1 \neq 0$.  It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.  Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.  If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.  By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?
18794
Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$   I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.  How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?
61681
Let $Z_n$ be the set of all $n \times n$ matrices that commute with all $n \times n $ matrices. Show that $$Z_n = \{\lambda I_n \ | \  \lambda \in \mathbb R\}$$  ($I_n$ is the $n \times n$ identity matrix)  I don't know how to use $E_{ij}$ (matrix with $1$ in $(i,j)$ and $0$ elsewhere) and the elementary matrix $P_{ij}$ to prove this question. Can anyone explain it please?