I have a problem in which I am trying to prove over GF(2) that a binary symmetric matrix (A) with a diagonal of ones has a rank always equal to the rank of its augmented matrix with a ones vector (C) $$   C=\left[\begin{array} \\     1 \\     \vdots \\     1   \end{array}\right] $$  To clarify, such matrix is constructed like so: $$   A=\left[\begin{array}{rrrr} 1 &amp; a_{1,1} &amp; a_{1,2} &amp; \dots &amp; a_{1,n} \\ a_{1,1} &amp; 1 &amp; a_{2,1} &amp; \ddots &amp; \vdots \\ a_{1,2} &amp; a_{2,1} &amp; \ddots &amp; a_{n-1,n-1} &amp; a_{n-1,n} \\ \vdots &amp; \ddots &amp; a_{n-1,n-1} &amp; 1 &amp; a_{n,n} \\ a_{1,n} &amp; \dots &amp; a_{n-1,n} &amp; a_{n,n} &amp; 1   \end{array}\right] $$  For example, a 3 by 3 matrix like this has a rank of 2: $$   A=\left[\begin{array}{rrr}     1 &amp; 1 &amp; 0 \\     1 &amp; 1 &amp; 0 \\     0 &amp; 0 &amp; 1   \end{array}\right] $$ When we augment it with a ones vector, we get this matrix which also has a rank of 2: $$   A|C=\left[\begin{array}{rrr|r}     1 &amp; 1 &amp; 0 &amp; 1 \\     1 &amp; 1 &amp; 0 &amp; 1 \\     0 &amp; 0 &amp; 1 &amp; 1   \end{array}\right] $$ Cleary rank(A) = rank(A|C) over GF(2).  Why is this always true for such type of matrices?  If you have a proof, an idea, or a suggestion on how to proceed, please let me know. Any help is appreciated.
104463
Let $A$ be a (skew-) symmetric matrix over $\mathbb{Z}/2$. (In fact, I would take $A$ as the linking matrix of an oriented framed link in $S^3$ or the matrix representing the intersection form on a closed smooth 4-manifold. The following statement however does seem to hold in general.) I am interested in the following linear system over $\mathbb{Z}/2$, $$a_{i1}x_1+a_{i2}x_2\cdots+a_{in}x_n=a_{ii},\quad i=1,\cdots,n.$$ This system is known to always have a solution. (c.f. Saveliev's Lectures on the Topology of 3-Manifolds.) But I cannot see why is this true unless $A$ is nonsingular over $\mathbb{Z}/2$. Is there a general method to deal with these kinds of linear systems?
40247
So far, I've deduced that if the rank of A is n, then all the columns of A are linearly independent since A has n columns. As a result, m must be greater than or equal to n.  In the case that m = n, then A is a square matrix and we can just use matrix operations to distribute the inverse operation over the parenthesis; then we can apply the associative property and compute A*A-1 and AT-1 * AT first, eventually giving that the expression inside the norm is just the identity matrix, where the 2-norm of the identity matrix is obviously 1.  I'm having trouble for the case where m > n. I've identified that the expression inside the norm is actually the matrix of orthogonal projection, but I'm not sure if that will help me with the proof. Help is greatly appreciated, thanks.  This problem is from the text of "Matrix Computations" 3rd edition by Golub after covering a section on orthogonality and SVDs for some background.
95169
I'm studying for my exam of linear algebra.. I want to prove the following corollary:     Given $A \in{R^{n\times n}}$, there is a solution $x$ to $Ax = y$ for all $y$, if and only if $A$   has rank $m$ (full row rank).   I know that the rank of a matrix is the maximum number of columns (rows respectively) that are linearly independent and is defined by:     $\operatorname{Img} (A) = \operatorname{Rg} (A):= y \in{C^m}:y = Ax, x \in{C^n}$   My problem is that I can not find a way to relate the two concepts in order to reach a formal proof. Any help?
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.
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.
107921
Let , $S$ be the set of $3\times 3$ real matrices $A$ with $$AA^T=\left(\begin{matrix}1&amp;0&amp;0\\0&amp;0&amp;0\\0&amp;0&amp;0\end{matrix}\right).$$Then the set $S$ contains :   (A) a nilpotent matrix.  (B) a matrix of rank $1$.  (C) a matrix of rank $2$.  (D) a non-zero skew-symmetric matrix.  Attempt :  Clearly , $\left(\begin{matrix}1&amp;0&amp;0\\0&amp;0&amp;0\\0&amp;0&amp;0\end{matrix}\right)\in S$. So, (B) is TRUE.   Again $rank(AA^T)=rank(A)$. As , $rank(AA^T)=1$ , so $S$ does not contain a matrix of rank $2$. So (C) is FALSE.  Again we know the rank of a non-zero skew-symmetric matrix can never be $1$. So , $S$ does not contain a non-zero skew-symmetric matrix. So (D) is FALSE.  But I am unable to understand the option (A) whether it is correct or NOT ? Any hint ?
106152
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?
82544
How do we prove that  $\operatorname{rank}(A) = \operatorname{rank}(AA^T) = \operatorname{rank}(A^TA)$ ?  Is it always true?
102089
I have looked this up on several sites but they confused me because some of the given information was wrong / unclear / contradicting whatever.  I hope you can tell me all / most important ways to calculate the rank of a matrix.  As example, I take the matrix  $$A = \begin{pmatrix} 1 &amp; 2 &amp; 3\\  0 &amp; 5 &amp; 4\\  0 &amp; 10&amp; 2 \end{pmatrix}$$  Now several sites included that info so it must be true: If, we are looking at this example, there is no line with zeroes only, the rank of this matrix will be $3$. (?)  Here is the problem. It will cost time to form this matrix to see if there will be lines with zeroes only. For this I can use Gaussian Elimination.  I have tested it with that Gauss and I couldn't get a line with zeroes only, so I conclude that this matrix $rank(A)=3$.  This however seems very inefficient way, I hope you can tell me better ways?
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 $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$
49640
Let $P,Q$ real $n\times n$ matrices  such that $P^2=P$ , $Q^2=Q$ and $I-P-Q$ is an invertible matrix.  Prove that $P$ and $Q$ have the same rank.  Some help with this please , happy year and thanks.
36832
If $A $ is  an  $n \times n$  matrix  over $\mathbb C$ such that $A^2=A$ then is it true  that  $\operatorname{trace} A = \operatorname{rank} A$?
76941
Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)?
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  :)
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?
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?
66956
What is a smart way to see that $$A\stackrel{\cdot}{=} \begin{bmatrix} 1/2 &amp; 1/3 &amp; \cdots &amp; 1/(n+1) \\ 1/3 &amp; 1/4 &amp; \cdots &amp; 1/(n+2) \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 1/(n+1) &amp; 1/(n+2) &amp; \cdots &amp; 1/(2n) \end{bmatrix} $$is non-singular?  I computed $\det A$ for $n=1,2$ and $3$ but I failed to see the pattern.  The context is as follows: in $\Bbb R[x] $ with inner product given by $\langle p(x),q(x)\rangle \stackrel{\cdot}{=}\int_0^1p(x)q(x)\,{\rm d}x$, I want to see that if $U\stackrel{\cdot}{=} \{p(x) \in \Bbb R[x]\mid p(0)=0\}$, then $U^{\perp} =\{0\}$ (and hence $U^{\perp\perp}=\Bbb R[x]\neq U$). I took $f(x) \in U^\perp$ and computed  $\langle f(x),x^k\rangle$ for $k\geq 1$. I obtained a homogeneous system for the coefficients of $f(x)$ which has $A $ as the matrix associated. So if $A$ is non-singular I'm done.