If I have the following $n \times n$ matrix   $$\begin{bmatrix}      2 &amp;      3 &amp; \cdots &amp;    n+1\\      3 &amp;      4 &amp; \cdots &amp;    n+2\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ n+1    &amp; n+2    &amp; \cdots &amp;     2n\\ \end{bmatrix}$$  how can I determine its rank?  I know that I need to check for the maximum number of linearly independent columns. However, I’m having a hard time determining the linearly independent columns. Thanks for those who’ll help.
64706
Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$.  My work : I noticed that A is symmetric . Hence all of its eigen vectors are real .. That is all i have got .   Your help will be highly appreciated .Thank you .
94384
How can we find the determinant of the following matrix?  $$A = \begin{pmatrix} x_1y_1 &amp; x_1y_2 &amp; x_1y_3 &amp; \cdots &amp; x_1y_{n-1} &amp; x_1y_n \\  x_1y_2 &amp; x_2y_2 &amp; x_2y_3 &amp; \cdots &amp; x_2y_{n-1} &amp; x_2y_n \\  x_1y_3 &amp; x_2y_3 &amp; x_3y_3 &amp; \cdots &amp; x_3y_{n-1} &amp; x_3y_n \\  \vdots &amp;  \vdots &amp;  \vdots &amp;  \ddots &amp;  \vdots &amp;  \vdots \\ x_1y_{n-1} &amp; x_2y_{n-1} &amp; x_3y_{n-1} &amp; \cdots &amp; x_{n-1}y_{n-1} &amp; x_{n-1}y_n \\  x_1y_n &amp; x_2y_n &amp; x_3y_n &amp; \cdots &amp; x_{n-1}y_n &amp; x_ny_n \end{pmatrix}$$  That is, the entry of $A$ is $a_{ij}=x_iy_j$ for $i\leq j$; and $a_{ij}=a_{ji}$ for $i&gt;j$.  I do not have anything new. And I find also it is difficult to find its eigenvalues.
12150
How do I calculate the determinant of the following $n\times n$ matrices  $$\begin {bmatrix} 0 &amp; 1 &amp; \ldots &amp; 1 \\ 1 &amp; 0 &amp; \ldots &amp; 1 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 1 &amp; 1 &amp; ... &amp; 0 \end {bmatrix}$$  and the same matrix but one of columns replaced only with $1$s?  In the  above matrix all off-diagonal elements are $1$ and diagonal elements are $0$.
8035
So I'm complete stuck with something. I know it the following statements are true (or at least the seem to be from the results that I got from messing around with it a bit on MATLAB), but I don't understand why they are true or how to show so. Let $A$ be and $m$X$n$ matrix. Show that:  a) if $x \in N(A^TA)$ then $Ax$ is in both $R(A)$ and $N(A^T)$.   For this one I messed around with it with my own examples and I got $Ax=0$, therefore satisfing the statement, but I don't understand what's actually going on.   b) $N(A^TA)=N(A)$  again, makes sense when I see the results in MATLAB, but don't undestand why it works.  c) $A$ and $A^TA$ have the same rank  d) If $A$ has linearly independent columns, the $A^TA$ is nonsingular.  For the last two I have no idea on how to even start showing the relationship. I feel like I'm missing some crucial relationship between $A$ and $A^TA$, but I'm just not seeing it.   I would greatly appreciate any help of sugestions on how to show that these statements are true.   Thank you very much.
103206
How many &quot;$1$&quot; can be selected at most, provided that only one from each row and column of the attached matrix $A$ is selected?(That is, when one &quot;$1$&quot; is selected, no other element can be selected than the row and column where that &quot;$1$&quot; is located.)  Answer: I think $6$ can be chosen.  How can I give this answer in connection with graph theory? I think it has to do with the Latin Square(or Rectangle). But I can't connect and explain. Please help me. I thought a lot, but I could not establish the relationship completely. Thanks...
11601
I am looking for an intuitive explanation as to why/how row rank of a matrix = column rank. I've read the proof at  and I understand the proof, but I don't "get it". Can someone help me out with this ?   I find it hard to wrap my head around the idea of how the column space and the row space is related at a fundamental level.
108665
$\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$I would like to prove that for every martix, column rank = row rank Let $A\in M_{m\times n}$ be some matrix. fix $R_A$ = vector space of the rows of $A$ , $C_A$ = vector space of the columns of $A$  $$\rank (R_A) = dim \ (\span \{R_1, R_2, \ldots, R_m\})$$ $$\rank (C_A) = dim \ (\span \{C_1, C_2, \ldots, C_n\})$$  let $x$ be some vector. $$x\in \operatorname{Null}(A) \Leftrightarrow \forall i  \ \ (1 \leq i \leq m) : \langle x,R_i\rangle = 0$$ (to be clear - I'm referring to the inner product of $x$ with each row of $A$) Using the rank nullity theorem, $\dim \operatorname{Null}(A) =  n - \rank(R_A)$ as $n$ =  number of columns. what I would like to do is to say: $$\dim \operatorname{Null}(A) =  n - \rank(R_A)$$ $$\dim \operatorname{Null}(A) = n - \rank(C_A)$$ therefore,  $\dim \operatorname{Null}(A) = n - \rank(R_A)= n - \rank(C_A)  \Longrightarrow \rank(R_A) = \rank(C_A) $ it that false? is using the rank- nullity  theorem this way is cheating? or just doesn't prove what needs to be proven formally?
12388
Find the determinant of the following matrix in the terms of $a_1,a_2,\cdots,a_n$ explicitly, $$ \begin{bmatrix} a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_n\\ a_2 &amp; a_3 &amp; a_4 &amp; \cdots &amp; a_1\\ a_3 &amp; a_4 &amp; a_5 &amp; \cdots &amp; a_2\\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ a_n &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_{n-1}\\ \end{bmatrix} $$ When the determinant is zero?
109117
If $S = \{ A_{m \times n} : m &gt; n, A_{ij} \in \{0,1\} \}, $ is the set of all $m \times n$ binary matrices, and I choose a random matrix $r \in S$, what would be the probability that $r$ would have full column rank (each of the n columns would be linearly independent)?  Several papers simply claim that this probability is very high, but I'm curious as to what the exact probability would be, and how to compute it.
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?
77302
How to evaluate  $$\det\begin{bmatrix}1+x_1y_1&amp;\cdots &amp; 1+x_1y_n\\ \vdots &amp; \ddots &amp; \vdots \\ 1+x_ny_1 &amp; \cdots &amp; 1+x_ny_n\end{bmatrix}$$?  I tried factorisation but I have no idea. I could not factor anything out... I need a hint to kick off.   Edit: A conjecture due to multilinearity: if we let $v_j=(x_1y_j,\cdots,x_ny_j)^T$, then my guess would be $$\det\begin{bmatrix}1+x_1y_1&amp;\cdots &amp; 1+x_1y_n\\ \vdots &amp; \ddots &amp; \vdots \\ 1+x_ny_1 &amp; \cdots &amp; 1+x_ny_n\end{bmatrix}=\det(1,v_2,\cdots,v_n)+\det(v_1,1,\cdots,v_n)+\cdots+\det(v_1,v_2,\cdots,1)=\color{red}{0}$$ where $1$ denote $(1,\cdots,1)^T\in \Bbb K^n$,$\Bbb K$ the underlying number field.
33605
\begin{array}{rrrrr|r}     b &amp; a &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; b &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; a &amp; b &amp; \cdot \cdot \cdot &amp; a \\     \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ a &amp; a &amp; a &amp; \cdot \cdot  \cdot &amp; b   \end{array}  I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.  I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it.
84962
While I was solving problem to determine if given set of vectors are linearly independent or not, the solution given was that the coefficient matrix A of the homogeneous linear system of those vectors   k1a1 + k2b1 + k3c1 = 0  k1a2 + k2b2 + k3c2 = 0  k1a3 + k2b3 + k3c3 = 0  was a 3x3 square matrix, and determinant(A) = 0  Thus it has a non trivial solution and hence the vectors are linearly dependent.  I am unaware about how if A is nxn square matrix and if determinant(A) = 0, then the system is having a non trivial solution.
490
How to show that the determinant of the following $(n\times n)$ matrix  $$\begin{pmatrix} 5 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 2 &amp; 5 &amp; 2 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; 2 &amp; 5 &amp; 2 &amp; 0 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots&amp; \vdots&amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 0 &amp; \cdots &amp; \cdots &amp; 0 &amp; 2 &amp; 5 &amp; 2 \\ 0 &amp; \cdots &amp; \cdots &amp; \cdots &amp; \cdots &amp; 2 &amp; 5 \end{pmatrix}$$  is equal to $\frac13(4^{n+1}-1)$?    More generally:     How does one compute the determinant of the following  (where the three diagonals are constant)?      $$M_n(a,b,c) = \begin{pmatrix} a &amp; b &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ c &amp; a &amp; b &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; c &amp; a &amp; b &amp; 0 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots&amp; \vdots&amp; \vdots &amp; \vdots&amp; \vdots &amp; \vdots \\ 0 &amp; \cdots &amp; \cdots &amp; 0 &amp; c &amp; a &amp; b \\ 0 &amp; \cdots &amp; \cdots &amp; \cdots &amp; \cdots &amp; c &amp; a \end{pmatrix}$$      Here $a,b,c$ can be taken to be real numbers, or complex numbers.   In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that $$m_{ij} = \begin{cases} a &amp; i = j, \\ b &amp; i = j - 1, \\ c &amp; i = j + 1, \\ 0 &amp; \text{otherwise.} \end{cases}$$  There does not seem to be an easy pattern to use induction: the matrix is not a diagonal  of the type $M = \bigl(\begin{smallmatrix} A &amp; C \\ 0 &amp; B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so  gets complicated quickly.  Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:  $$\begin{align} \det(M_1(a,b,c)) &amp; = \begin{vmatrix} a \end{vmatrix} = a  \\ \det(M_2(a,b,c)) &amp; = \begin{vmatrix} a &amp; b \\ c &amp; a \end{vmatrix} = a^2 - bc \\ \det(M_3(a,b,c)) &amp; = \begin{vmatrix} a &amp; b &amp; 0 \\ c &amp; a &amp; b \\ 0 &amp; c &amp; a \end{vmatrix} = a^3 - 2abc \end{align}$$  But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
102335
$$     \begin{bmatrix}     -2 &amp; -2 &amp; 0 &amp; 0 \\     -b &amp; -b &amp; 0 &amp; 0 \\      0 &amp; 0 &amp; -4 &amp; 4 \\      0 &amp; 0 &amp; b &amp; -b \\     \end{bmatrix} $$  I have this matrix and I need to find the values for $b$ that this matrix won't  be diagonalizable.  For what I got this matrix is diagonalizable for every $b$ value.  Please write me also the final answer, thank you very much.
115178
Is the set of all $n\times n$ matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all $n\times n$ matrices?   Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with $2\times 2$ matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough.
30552
I know that I can find a basis for the column space of a matrix $A$ by reducing the matrix to reduced row echelon form $J$. The columns of $A$ corresponding to the linearly independent columns of $J$ then form a basis for $Col(A)$, because linear dependence is preserved under elementary row operations. I can't figure out why this is true though, and a google search returns nothing, so I'm sure it's simple. Can someone give me a proof?