Does a square matrix $X \in\Bbb R^{n\times n}$ always follow that $\dim(\ker(X)) = \dim(\ker(X^T))$? I know it works for a sizable amount, but I am not sure if it works for all of them, is there a way to prove this either way?
122157
How to prove $dim(Ker (A- \lambda I))=dim (Ker(A^t- \lambda I))$ ?  We know both $A$ and $A^t$ have same eigenvalues and $Av= \lambda v$ imply $v^tA^t= \lambda v^t$ but this does not imply that there is a one to one association b/w eigenvectors of $A$ and $A^t$.. So how to proceed?
94698
I'd like a hint rather than a full solution.  The problem I am considering is the following:     $X$ is an $n\times m$ matrix $Y$ is $m\times n$      Show that      $(I - XY)^{-1}\cdot X = X\cdot(I - YX)^{-1}$   The first $I$ is of dimension $n$ and the second $I$ is of dimension $m$.
83978
If $A^2 = I$ where $A$ is an $n\times n$ matrix, then why is $\operatorname{rank} (A+I) + \operatorname{rank} (A-I) =n$.   I am not sure why this seems to be true.
77547
If you have an $m \times n$ matrix and the rows are linearly independent, are the columns linearly independent too?  I know that it's the case for $n \times n$ matrices. I've yet tried to think about it, and I think the answer is yes, for example the matrix: $\begin{bmatrix} 1 &amp; 2 &amp; 3 \\ a &amp; 2a &amp; 3c \end{bmatrix}$ has linearly dependent rows and columns, but I'm not sure that it works for every $m \times n$ matrix.
119996
More specific to my problem, this is a variation on  which was promptly answered with a counterexample.  Let $\mathbb{M}_n$ be the space of $n\times n$ matrices with real entries.  Let $\overline{\mathbb{M}}_n$ be the space of square $n\times n$ matrices with entries $0$ and $1$.  For any $M\in\overline{\mathbb{M}}_n$ we have $$e^M=\exp(M)=\sum_{k=0}^\infty \frac{1}{k!}M^k.$$     Is $\exp:\overline{\mathbb{M}}_n\to\mathbb{M}_n$ injective?   In other words, are there two distinct $M_0,M_1\in \overline{\mathbb{M}}_n$ such that $e^{M_0}=e^{M_1}$?  Since for fixed $n$ the space $\overline{\mathbb{M}}_n$ is finite we can manually check -- it is injective for $n\le4$, but I haven't been able to convince myself for arbitrary $n$.  For $n=4$ there are $2^{n^2}=65536$ matrices to check, and MATLAB's expm and my hastily written matrix building code took half an hour to complete.  For $n=5$ we have $2^{25}&gt;3.3\cdot 10^7$ matrices to test which would take ten days to run.  For larger $n$ a direct test is out of the question.  The smallest nontrivial examples in my problem are $n=6$, and the interesting ones are $n=12$ and larger.
97957
For a vector space $V$ and a linear operator $f : V \to V$, under what conditions does $V = \ker(f) \oplus \operatorname{im}(f)$?  Is it always true, or only in special cases?  Edit: $V$ is finite dimensional.
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?
74126
I would like to know in particular whether a full rank matrix necessarilly has a determinant equal to/different from zero. Also, would the answer change based on the matrix being squared or not?
8479
Given a field $F$ and a subfield $K$ of $F$. Let $A$, $B$ be $n\times n$ matrices such that all the entries of $A$ and $B$ are in $K$. Is it true that if $A$ is similar to $B$ in $F^{n\times n}$ then they are similar in $K^{n\times n}$?   Any help ... thanks!
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.
44785
I'm solving a exercise, and the following question has arisen:     Let $A,B,C$ be non-zero matrices, if $AB=CB$ then is $A=C?$   I already have a partial solution when $A,B,C$ are square matrices and $B$ is invertible: in this case $A=CBB^{-1}=CI_n=C$.  This result is valid in general case? (I mean, when the matrices are arbitrary)  Thanks!
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.
39803
This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216).  I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me.  Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation.  Part (b) asks the reader to use part (a) to show that similar matrices have the same trace.  If $A$ and $B$ are similar, then  $\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$  $= \operatorname{Tr}(BSS^{-1})$  $= \operatorname{Tr}(B)$,  which completes part (b).  Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$.  Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation.  From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space.  Part (d), however, I'm stuck on.  It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices.  I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes.  I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me.  I'm kind of stuck...  Many thanks in advance!
52353
I don't quite understand the proof my notes gave me.  Dimension of $n$ by $n$ matrix is $n^2$. Hence if $k \geq n^2$ then $\mathbf{ \{ I, A, A^2, ..., A^k \} }$ is linearly  dependent. So, there exist scalars $a_i$ not all zero such that $$a_k A^k + \cdots + a_0 I = 0$$  $f(x) = a_k x^k + \cdots + a_0$ is an annihilating polynomial.     I don't understand the bold bit. How do we know that $A, A^2, ...$ are all different? Surely there are matrices which give identity when raised to power of 2,3, etc right?  Thanks.
25677
I have to prove/disprove this:     If $\det(A+X) = \det(B + X)~ \forall X \in M_{n \times n} (\mathbb F) \rightarrow A = B$   I believe it is true but I can not think of a direct way to prove it. Any help is appreciated!
107241
Let $A$ and $B$ are $n\times n$ matrices and $x \in C^{n}$.  If $Ax = Bx$ for all $x$ then $A = B$. To prove this I have selected  $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}. Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$. Hence $A = B.$    Is this proof complete?
30161
If T: $\Bbb R$3→ $\Bbb R$3 is a linear transformation such that:  $$ T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$  T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix} $$  Then the standard matrix for T is...  I'm not exactly sure how to approach this problem. Could anyone explain how to solve this problem?