Let's start with two square matrices, $A$ and $I$. $I$ is the identity matrix in n-dimensions. $A$ is such that all its entries are equal to 1 (clearly, it is not invertible).  Our task is to determine conditions on $a$ and $b$ such that $M=bA + (a-b)I$ is invertible, i.e. has rank = $n$. Of course, we don't know what $n$ is, so we probably don't need to.  $a=b$ is the worst that's possible, so $a$ and $b$ should not be equal. Is that all, though? Is it sufficient to ensure invertibility of $M$?  $A$ has all linearly dependent rows, while $I$ has all linearly independent rows. Is this sufficient to say that $A+I$ (or any other linear combination) has linearly independent rows/columns?  Any help is appreciated, thanks a lot!
122719
Consider matrices of the following form: \begin{bmatrix} a &amp; x &amp; \cdots \\  x &amp; b &amp; x &amp;\cdots \\ \vdots &amp; x &amp; c &amp;\cdots \\   &amp; \vdots &amp; x &amp; \ddots &amp; \\ \end{bmatrix} where all $x$ are the same scalar. Is there anything that can be done to calculate the determinant easily? Obviously if $$x=a=b=c=\cdots $$ then the matrix is singular and the determinant is zero. Assume this is not the case.   Do they have any other interesting properties?
77966
Prove that every $m \times n$ matrix of rank $1$ has the form $A=XY^t$, where $X,Y$ are $m$- and $n$-dimensional column vectors. How uniquely determined are these vectors$?$  My attempt: I thought about using the dimension formula somehow. I know that nullity $+$ rank $=$ dim $A$. We know dim $A=mn$ but this got me nowhere. Then I tried doing an example. I took $$ A=\begin{pmatrix} 1 &amp; 3 \\ 0 &amp; 0 \end{pmatrix} $$  which has rank $1$. Then I can choose $$ X=\begin{pmatrix}1 \\ 0 \end{pmatrix}, Y=\begin{pmatrix} 1 \\ 3\end{pmatrix} $$ or I could choose $$ X=\begin{pmatrix} 1/2 \\ 0 \end{pmatrix}, Y=\begin{pmatrix}2 \\ 6 \end{pmatrix} $$ so that they are not uniquely determined. However, I can interchange the entries of $X$ and $Y$ to get ''different'' $X,Y$. But one the entries of $X$ (or $Y$ are chosen) then entries of $Y$ (or $X$) are then forcefully chosen. So I believe the final answer for that will be that they are not unique but choosing one vector determines both.  However, this method works for matrices $A$ with only one nonzero row and certainly many $A$'s have rank $1$ with many nonzero row. I know that $A$ has rank $1$ means I can use a series of elementary matrices to place it in the form I used above, but am unsure of how to use this since I'm not sure how to reconcile this with my ''choices'' for $X,Y$. Is there a way to do this or am I on the wrong track.  Note: I do not need nor want solutions but rather hints or thoughts on how I should proceed.
59974
$A$ is an $n\times n$ matrix with $\|A\|≤a&lt;1 $ . I need to prove that the matrix $(I-A)$ is invertible with $\|(I-A)^{-1}\|\le\frac1{(1-a)}$.   It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
16874
Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?   I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?
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.
89030
Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible.   Here is what I have attempted:  We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere.   I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.
25741
Is the matrice $A^2-A+I$ invertible?  $A^2=A-I /\cdot A$  $ A^3 = A^2 - A$  $A^2=A$  Thus I conclude that $\lambda=0$ is one of the eigenvalues of this matrice and it isn't invertible, my conclusion comes from the fact that the eigenvalues of the matrices $A,A^2,...$ are all connected, and if $A^3=O$ that means that those two have the same eigenvalues.
98424
Let $A$ and $B$ be $n\times n$ real matrices such that $A^2=A$ and $B^2=B$. Suppose that $I-(A+B)$ is invertible .    Show that $\operatorname{Rank}(A)=\operatorname{Rank}(B)$.     I proceed in this way: Note that $A(I-(A+B))=A-A^2-AB=A-A-AB=-AB$ and similarly $B(I-(A+B))=-AB$. So $A(I-(A+B))=B(I-(A+B))$. Since $I-(A+B)$ is invertible we get $A=B \Rightarrow \operatorname{Rank}(A)=\operatorname{Rank}(B)$.
59285
$A$ be a $2\times 2$ real matrix  with trace $2$ and determinant $-3$, consider the linear map $T:M_2(\mathbb{R})\to M_2(\mathbb{R}):=B\to AB$ Then which of the following are true?   $T$ is diagonalizable $T$ is invertible $2$ is an eigen value of $T$ $T(B)=B$ for some $0\ne B\in M_2(\mathbb{R})$    if $ A=\begin{pmatrix}a_{11}&amp;a_{12}\\a_{21}&amp;a_{22}\end{pmatrix}$ Then I have calculated that matrix of $T$ will be \begin{pmatrix}a_{11}&amp;a_{12}&amp;0&amp;0\\a_{21}&amp;a_{22}&amp;0&amp;0\\0&amp;0&amp;a_{11}&amp;a_{12}\\0&amp;0&amp;a_{21}&amp;a_{22}\end{pmatrix} so $T$ is invertible I can say as $\det T=(\det A)^2=9$, could any one help to find out others as true/false?
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.
37668
Let $A$ be an $m\times n$ matrix. Prove that $A^*A + I$ is invertible.   I'm not sure what to do because everything I try ends up at $A^*A$ again and not $AA^*$.
123189
I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$  I'm a bit lost with this one, I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$, Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.  If someone could clue me in I'll be grateful.    UPDATE:  Thanks for all of your input - it really helped, I got a solution, I'd like to know if my way of proving it is valid:  $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$      multiplying both sides by $A^{-1}$ and $B^{-1}$    $A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$    now multiplying both sides by $(A + B)$, I get to  $(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$  Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$    Now I determine that $C$ is $(A+B)^{-1}$, because $(A+B)C$ equals the Identity Matrix.    So lastly to verify that I place $C$ in the original equation:  $(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$  and from this I get:  $(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$
78948
I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$
72446
If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that      $$ e^{A+B}=e^A e^B$$    Note that $A$ and $B$ do NOT have to be diagonalizable.
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.
43790
$\mathbf{A}$ is a $2\times 2$ matrix which satisfies   $\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}$,   where $\mathbf{I}$ is the $2\times 2$ identity matrix. Prove that $\mathbf{A}$ is invertible.      Hint: Find a matrix $\mathbf{B}$ such that $\mathbf{A}\mathbf{B}=\mathbf{I}$.   I tried substituting a variable matrix for $\mathbf{A}$ and substituting $\mathbf{I}$'s value into both the original equation and the hint, but the result was full of equations and didn't seem to help much at all. I would appreciate advice for this problem.  Thanks. --Grace
26594
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?