Let $A \in M_5(\mathbb R)$ ; if what is rank $adj(A)$ when rank of $A$ is $3$ or $5$ respectively ? In general , if $A \in M(n,\mathbb R)$ , and rank of $A$ is $m(&lt;n)$ , then what is the rank of the adjugate of $A$ ?  I think that if  rank $A=n$ , then det $A \ne 0$  , then det $(adjA)=(det A)^{n-1} \ne 0$ , then $adj A$  should also have rank $n$ ; but I can't figure out what if rank$A$ is less than $n$ . Please help. Thanks in advance
61323
Let $A$ be a square and singular matrix of order $n$.  Is $\operatorname{adj}(A)$ necessarily singular? What would be the rank of $\operatorname{adj}(A)$?
114634
How can I prove that $\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$, if $A$ and $B$ are any two $n\times n$-matrices. Here, $\operatorname{adj} A$ means the  of the matrix $A$.   I know how to prove it for non singular matrices, but I have no idea what to do in this case.
18794
Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$   I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.  How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?
67141
I have the following proof that I would like to be walked through because I'm not intuitively seeing what to do:  If $A$ is $n\times n$, prove $\det\left(\operatorname{adj}(A)\right) = \det(A)^{n-1}$.  I know the property of $A\operatorname{adj}(A) = \det(A)I$ is important but I don't know how to apply it to get an answer. Any help is much appreciated.  Thanks,
13172
Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$, where $A,B \in M_{m\times n}(\mathbb{F})$.   I'm trying to think in terms of linear transformations. We can define $T_a, T_b:\mathbb{F}^n\rightarrow \mathbb{F}^m$. I know that $\dim_{\mathbb F}\operatorname{Im} T_a, \dim_{\mathbb F}\operatorname{Im} T_b \le m$.  What should I do next?
74337
Let $A,\ B,\ C,\ D \in \mathcal{M}_n(\mathbb{C})$. If $\operatorname{rank}\left( \begin{bmatrix} A &amp;B \\ C &amp;D \end{bmatrix}\right)=n$, prove that $\det(AD)=\det(BC)$.
52087
I'm trying to prove the following: Let $A$ be a  $k\times k$ matrix, let $D$ have size $n\times n$, and $C$ have size $n\times k$. Then,  $$\det\left(\begin{array}{cc} A&amp;0\\ C&amp;D \end{array}\right) = \det(A)\det(D).$$  Can I just say that $AD - 0C = AD$, and I'm done?
73332
Let $A\in M_{n\times n}(\Bbb R)$ so that $I\notin span(A,A^2,...,A^n)$. Prove that $\det(A)=0$.   I was thinking of showing $A$ is not invertible, meaning it has an eigenvalue of $0$. Since no matter the power you give to $A$, it is still not the identity, you can deduce that at least one eigenvalue is indeed $0$. However this doesn't work if you have different eigenvalues in the Matrix, so I got stuck.  Help will be most appreciated.
53789
Let $A,B$ be matrices $m\times n$ $(A, B \in M_{m\times n}(R))$. How can we prove that $$\operatorname{rank}(A+ B) \le \operatorname{rank}(A) + \operatorname{rank}(B)\ ?$$
82606
Can anyone prove/disprove the following statement?  Given a square matrix of size $n\ge2$, if all of its $(n-1)$-rowed minors are zero, then $\operatorname{rank}(A)\leq n-2$.  I'm having trouble connecting the information on the minors to what I have to prove.  Thank you in advance.
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?
33147
I'm sure I've done this before in abstract algebra. Regardless it's escaped me now.   I have proved that for $T:U\rightarrow V$, with $dim(U)=m$ and $dim(V)=n$ that $rank(T)\le m$ which is obvious, but I have proved it none the less.  I want to.... this is where I get stuck. I'm not quite sure how to say it.  Suppose we have a $T$ for example that takes $U\subset\mathbb{R}^3\rightarrow V\subset\mathbb{R}^3$, we could still have a $T$ that maps a plane to something (a plane, line or point) in this case rather than $rank(T)\le 3$ I can state $rank(T)\le 2$.  (Correct my notation here, I don't like writing subset, I mean number of dimensions of the space!)  To show this I can say $F:P\subset\mathbb{R}^2\rightarrow U\subset\mathbb{R}^3$.  Then let $G:P\rightarrow V$ and that $G=TF$, now $rank(G)\le dim(P) = 2$  Now, if $U,P,Q,R$ are spaces of dimension $\le 2$ and P,Q,R are subspaces of U  $A:U\rightarrow P$  $B=A:P\rightarrow Q$  $C=A:Q\rightarrow R$  We can now say $A^2=BA$ then $A^3=CBA$  We know that for $A$ rank(A)$\le 2$, this thus provides an upper-bound for rank($A^3$)  I want to show now that if I keep applying a linear transformation that the rank is a monotonically decreasing ($\le$) sequence. I am unsure on a proof that the rank of a transformation cannot be greater than the domain's dimensions. Although this is trivial.  (NOTE:  I have proven that if a set of n vectors, R, span a vector space V, and you take a set W of m linearly independent vectors in V that m$\le$n)  If rank(A)=2 then rank($A^2$)=2 and so rank($A^3$)=2  if rank(A)=1 then rank($A^2$)=0 and rank($A^3$)=0  if rank(A)=0 then rank($A^2$)=0 and rank($A^3$)=0  But again, I can't prove this, or at least I am unsure of how to write it.  something to clarify:  For a map $T:U\rightarrow V$ how do we distinguish (using notation) whether or not we actually use all dim(U)?  For example we might have a T that maps a plane in R^3 to a line in R^3, this can be expressed as a composition of maps, one that takes a 2 dimensional vector space (coordinates of the plane in 3 space) to a 3 dimensional point in T's domain, which T then maps to a line.  We have rank - the dimensions of the image - to make this distinction on the map's target (the rank of T is 1 if it maps something to a line, but the dimensions of the target is 3)  How do we write this? What do I say to describe this?   How do I distinguish between the number of vectors in a basis and the number of dimensions the space that basis is in has?  Sorry for the length of this! I tried to talk though my problem in the hope of seeing it myself, no such luck, thanks.
41998
Problem:  Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow  rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS} $ is the Hilbert Schmidt Norm.  Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.  Proof of the implication $\Leftarrow $:   If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$ If $rank(A)=1$, than: $ A=\begin{pmatrix} A_{1}\\  \alpha _{2}A_{1}\\  ...\\  \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$   and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$  Proof of the implication $\Rightarrow $   We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$  On the other hand: $ A=\begin{pmatrix} A_{1}\\  A_{2}\\  ... \\ A_{n}  \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\  \left \langle A_{2},x \right \rangle\\  ...\\   \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} $$  Where $ \left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$  Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} $ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$  Please let me know if my solution makes sense.
83671
Let $A$ be $m \times n$ matrix, and $B$ be $n \times m$ matrix. Then  Show that $$\det\begin{bmatrix}I_{n} &amp; B\\ A &amp; I_{m} \end{bmatrix}=\det\begin{bmatrix}I_{m} &amp; A\\ B &amp; I_{n}\end{bmatrix}$$  Show that $$\det(I_{m}-AB)=\det(I_{n}-BA)$$  Is it true that $I_{m}-AB$ and $I_{n}-BA$ have the same rank?    Someone please show me a way to start proving. I tried to use formulas for calculating block matrix with square matrices, but it didn't help me on this one. I really don't know where to start.
78925
Let $A$ be the matrix described when $A \in M_n(\mathbb{R})$.  Let $a,b \in \mathbb{R}$.  Prove that:   Iv'e tried to factor out $a+b$, but got nowhere.  A hint or a solution would be very appreciated.  Thanks a lot!
21427
Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?     If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.   If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.  I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.
95482
Suppose A is a 3*3  symmetric matrix s.t.  $$\begin{pmatrix} x &amp; y &amp; 1 \\   \end{pmatrix} A  \begin{pmatrix} x  \\  y\\ 1\end{pmatrix}  = xy -1 $$  let p be the no. of positive eigen value of A and q = rank(A)-p, then p=? q=?  what matrix A should be I can't find??