There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus,   only $16$ possible $2\times 2$ matrices over $Z_2$, and $6$ of them   have full rank:      $\begin{pmatrix}0&amp;1\\   1&amp;0\end{pmatrix}  \quad \begin{pmatrix}1&amp;1\\   &gt; 1&amp;0\end{pmatrix}  \quad \begin{pmatrix}0&amp;1\\   1&amp;0\end{pmatrix}  \quad &gt; \begin{pmatrix}0&amp;1\\   1&amp;1\end{pmatrix}  \quad \begin{pmatrix}1&amp;1\\   &gt; 0&amp;1\end{pmatrix}  \quad \begin{pmatrix}1&amp;0\\   0&amp;1\end{pmatrix}$      Randomly generate a $n\times n$ matrix over $\mathbb {Z}_2$ (where n   is big, say, $1000$). What's the probability that the matrix has full   rank?   I'd want to talk something regarding to this question.   I found the probability is equal to:  $\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^n} \tag{1}$  On other hand, what's difference between $2^n$ and $2^{n^2}$?  $\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^{n^2}} \tag{2}$  Can you tell whether or not I'm wrong?   Regards!
123306
What is the probability that a random $\{0,1\}$, $n \times n$ matrix is invertible? Assume the 0 and 1 are each present in an entry with probability $\frac{1}{2}$. Is there an explicit formula as a function of $n$?  Does it tend to 1 as $n$ grows large? I'm sure this is all known...  Thanks!
8744
Let $A$ be a $m \times n$ matrix with entries from some field $F$. Define the determinant rank of $A$ to be the largest possible size of a nonzero minor, i.e. the size of the largest invertible square submatrix of $A$. It is true that the determinant rank is equal to the rank of a matrix, which we define to be the dimension of the row/column space.  It's not difficult to see that $\text{rank} \geq \text{determinant rank}$. If some submatrix of $A$ is invertible, then its columns/rows are linearly indepedent, which implies that the corresponding rows/columns of $A$ are also linearly indepedent.  Is there a nice proof for the converse?
95694
If $A \in \mathbb{C}^{n \times n}$ is a non-zero Hermitian matrix, I need to show that  $$\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$$ and reason out when the equality is attained?    Can anyone help me in showing this result?
19620
I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:  Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$
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 $n$.
1386
I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view:      "The rank of a matrix A is the number   of non-zero rows in the reduced   row-echelon form of A".   The lecturer then explained that if the matrix $A$ has size $m  \times n$, then $rank(A) \leq m$ and $rank(A) \leq n$.   The way I had been taught about rank was that it was the smallest of    the number of rows bringing new information the number of columns bringing new information.    I don't see how that would change if we transposed the matrix, so I said in the lecture:  "then the rank of a matrix is the same of its transpose, right?"   And the lecturer said:   "oh, not so fast! Hang on, I have to think about it".   As the class has about 100 students and the lecturer was just substituting for the "normal" lecturer, he was probably a bit nervous, so he just went on with the lecture.  I have tested "my theory" with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't.  So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?  Thanks  :)
32114
Let $k$ and $n$ be positive integers and let $F$ be a field. For matrices $A,B \in M_{k\times n} (F)$, show that the rank of $A+B$ is no more than the sum of the ranks of $A$ and $B$ I believe this question is addressed , but to be honest I don't quite understand the explanations given. Can someone possibly help with a more detailed explanation?
37567
A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&amp;0 \\ 0&amp;-1-\epsilon  \\   \end{pmatrix}$$ (a) exists for all $\epsilon &gt; 0$. (b) does not exist for any $\epsilon &gt; 0$. (c) exists for some $\epsilon &gt; 0$. (d) None of the above is true.  Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix. Could someone please give me a hint on how to go about this problem? Thank you very much for your help in this regard.
87131
"Find rank of matrix $\begin{bmatrix}                     0&amp;0&amp;-3\\                      9&amp;3&amp;5\\                     3&amp;1&amp;1\\                     \end{bmatrix}$ using echelon form?" "I am getting an answer equal to 2."  "When I reduce this I get $\begin{bmatrix}                           1&amp;1/3&amp;0\\                           0&amp;0&amp;1\\                            0&amp;0&amp;0\\ \end{bmatrix}$ so two pivots so rank must me 2 right ? However the book I am using says rank =1."
91358
If $\mathbf {Z}$ is random vector and $A$ is a fixed matrix, could someone explain why $$\mathrm{cov}[A \mathbf {Z}]= A \mathrm{cov}[\mathbf {Z}]A^\top.$$
28059
It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix  $$\begin{pmatrix} B &amp; C\\0 &amp; D \end{pmatrix}$$ where $0$ is the $3 \times 3$ matrix whose entries are all $0$.  From my intuition, I think the eigenvalues of the new $6 \times 6$ matrix are the eigenvalues of $B$ and $D$. But how can I show that?
87011
for (a) it says it can't be rank 1 because there is only one non zero eigenvalue? I understand it can't be 0 or 3 but I don't understand how you can conclude we have 2 just by knowing how many eigenvalues we have.
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.
46102
Assume M is a $2 \times 2$ real matrix such that $MX = XM$ for all real $2\times2$   matrices $X$. Show that $M$ must be some real multiple $q$ of $I$.   I can see that this is logical and have tried a few examples where I have multiplied $qI$ by some random $2\times2$ matrix from both sides and get the same matrix in both cases. How would I go about showing this though? Surely a few examples aren't actually showing anything for the general case, right?  Also, does the property shown in the question hold for all square matrices, or just $2\times2$ ones?
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.
102062
I had a question in a recent assignment that asked if a $3\times 3$ matrix could have a null space equal to its column space... clearly no, by the rank+nullity theorem... but I have a hard time wrapping my head around the concept of such a matrix, no matter what size, even existing. How could this be possible, and does anybody have an example of such a $m\times n$ matrix?
86847
I've just left an exam of algebraic structures, and one of the questions asked to compute the order of $SL_2(\mathbb{Z}/ 5\mathbb{Z})$ (the 2x2 matrices with determinant $1$ over the field $\mathbb{Z}/ 5\mathbb{Z}$). The answer should be $120$, and a friend of mine could achieve that, but by methods of raw counting. My question is: Is there a way to compute the order of $SL_2(\mathbb{Z}/ n\mathbb{Z})$ for all $n$?  OBS: Although $120=5!$, I believe this is purely a coincidence.