Suppose matrix A has rank r. Pick any r independent rows and r independent columns from A. Must the submatrix formed from these rows and columns be invertible?
74369
Let $A$ be an $m \times n$ matrix of rank $r$, let $I$ be a set of row indices such that the corresponding rows of $A$ are independent and let $J$ be a set of $r$ column indices such that the corresponding columns of $A$ are independent.  Let $M$ denote the $r \times r$ submatrix obtained by taking rows from $I$ and columns from $J$.  Then $M$ is invertible.  So far I've got: Let $B$ be the $m \times r$ matrix obtain from taking the rows from $J$.  $B$ is row-reducible to having a pivot in each column or else $Ax = 0$ has more than one solution, a contradiction.  Also, the dimension of the row space of $B$ equals $r$ as well.
106222
Let $A$ be a $7\times 7$ matrix such that it has rank $3$ and $a$ be $7\times 1$ column vector.  Then least possible rank of $A+(a a^T)$ is? ($a^T$ is transpose of the vector)  Intuitively I think it's $2$ as $a a^T$ has rank $1$, and the most it could affect is one row of $A$.
57335
A nonsingular matrix is one which is invertible, and hence the determinant is not equal to $0$. So at first I thought about having $\det(A_{1})+\det(A_{2})=\det(A_{1}+A_{2})$ and then the resulting sum being a invertible matrix, but this is not generally the case.  Then I thought about eigenvalues, as the determinant is the product of the eigenvalues and that using that I could showing that the spectrum of the sum is equal to the sum of the spectrums.   Is that a better way?
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.
98354
I'm given a matrix A, its dimensions are n x n.  I am required to prove that an invertible matrix B exists, such that the product of the matrices BA is triangular.  Any help?
101079
Let $A$ be an $n\times n$ matrix.  If $A$ is invertible then one of it's eigenvalues is 0.  If $A$ is diagonalizable then it has $n$ linearly independent eigenvectors.   Are these two statements true? Is there any connection between them? May I infer one from the other?
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.
43040
Let ${A}$ be a $2 \times 2$ matrix. For every two-dimensional vector ${v}$, there exists a two-dimensional vector ${w}$ such that ${A} {w} = {v}.$ Show that ${A}$ is invertible.  I have no idea on how to start this, but I have a hint:  Which two vectors ${v}$ are we most interested in if we are trying to find the matrix ${B}$ such that ${A}{B}= {I}$?  Any help is appreciated!
35038
I ran into this problem recently.  If $A$ is a $n\times n$ matrix with $1$s in the diagonal and all off diagonal entries have absolute value less than $1$, is $A$ invertible?  It it definitely true for a $2\times 2$ matrix and seems to be true for larger matrices, just by trying to compute the determinant, however, my expression seems to get messy, and I was wondering if there was a easier method to solve this. If not, direction on how to argue from definition of determinant would be appreciated. Although I can see that an argument might come from there, I am having problems with that too.  Thank you for any help, and any help is much appreciated.
2545
(a) Given the column vector $a$ as an $n\times 1$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  (b) Given the row vector $a^T$ as an $1\times n$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  I currently get stucked on this question, since I couldn't find the eigenvalues of the $n\times n$ matrix $aa^T$ in part (a), and $a^Ta$ in part (b) to find the columns of $U$ and $V$, respectively. Can someone please help me on this problem?
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How can I understand that $A^TA$ is invertible if $A$ has independent columns? I found a similar , phrased the other way around, so I tried to use the theorem  $$ rank(A^TA) \le min(rank(A^T),rank(A)) $$  Given $rank(A) = rank(A^T) = n$ and $A^TA$ produces an $n\times n$ matrix, I can't seem to prove that $rank(A^TA)$ is actually $n$.  I also tried to look at the question another way with the matrices  $$ A^TA = \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix}   \begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix} = \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix} $$  But I still can't seem to show that $A^TA$ is invertible. So, how should I get a better understanding of why $A^TA$ is invertible if $A$ has independent columns?
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?
8559
Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.  I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction  n=2 we have the $A=\begin{bmatrix} 0 &amp; 1\\  1 &amp; 0 \end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1  Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$  Now to show for  a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
69198
How to prove a general matrix invertible given by as below?  How to prove that $A^TA+I$ is always invertible for $\forall A \in \mathbb{R}^{n\times n}$?
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?
24346
If matrix $A$ is an $n\times n$ matrix such that $A^2 -A -2I=0$.  How can I show that $A$ is invertible and that $A^{-1} = \frac12(A-I)$? Also, how do I show that one of the eigenvalues of $A$ is $2$ or $ -1$? I appreciate any help on this question.
87552
Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$).  The hint they provided is that for any invertible matrix B, rank(AB) = rank (A) and rank(BA) = rank(A).