Suppose square matrix $A$ with order-n, and $A^*$ is it's adjugate matrix, when $n&gt;2$, prove: $\left(A^*\right)^*=\left|A\right|^{n-2}A$  Proof:     when $A$ is invertible, $A^{-1}=\frac{1}{|A|}A^*$, so $A^*=|A|A^{-1}$, and then   \begin{align}  \left(A^*\right)^*   = \left|\left(|A|A^{-1}\right)\right|          \left(|A|A^{-1}\right)^{-1}   = \left|A\right|^n\left|A\right|^{-1}A|A|^{-1}   = \left|A\right|^{n-2}A \end{align}   how about the case $A$ is not invertible?   $|A|=0$, then seems we should show $\left(A^*\right)^*=0$
122817
I had my exam of linear algebra today and one of the questions was this one.  Given $ A \in \mathbb{R}^{n \times n}$, prove that:  $$\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A.$$  Of course I was not able to prove this identity, otherwise I wouldn't post it here. But I'm still curious how one can prove this identity.  Could someone point me in the right direction?
112623
I know this proof is short but a bit tricky. So I suppose that $AB$ is invertible then $(AB)^{-1}$ exists. We also know $(AB)^{-1}=B^{-1}A^{-1}$. If we let $C=(B^{-1}A^{-1}A)$ then by the invertible matrix theorem we see that since $CA=I$(left inverse) then $B$ is invertible. Would this be correct?  Edit Suppose $AB$ is invertible. There exists a matrix call it $X$ such that $XAB=I$. Let $C=XA$ Then $CB=I$ and it follows that $B$ is invertible by the invertible matrix theorem.
22526
Let $A=[a_{ij}]_{n×n}$ be the matrix defined by letting $a_{ij}$ be the rational number such that $$a_{ij}=\gcd(i,j ).$$ How prove that $A$ is invertible, and compute $\det(A)$? thanks in advance
87100
Let A = $(a_{ij})_{n\times n}$ such that $AB = BA$ for every square matrix $B$ of order $n$.  $(I)$ Prove that $A$ is a diagonal matrix.  [Hint: Let $B_i$ denote the matrix whose $(i,i)$ entry is $1$, and $0$ elsewhere. Show that $a_{ij} = 0$ whenever $i ≠ j$.]  $(II)$ Prove that $A$ is a scalar matrix.  [Hint: For $i ≠ j$, let $B_{ij}$ denote the matrix whose $(i,j)$-entry is $1$ and $0$ elsewhere.Show that, $a_{ii} = a_{jj}$ for all $i ≠ j$.]  Can someone help me with this question using basic linear algebra concepts? Thanks.
100974
look here my friend. How to prove the following equation? Or give a counter-example. Thank you so much  $$\text{rank}(A^+)\leq \text{rank}(A)$$  where $\text{rank}(A^+)$ represents the positive components of matrix $A$, e.g. $$\left[\begin{matrix}2 &amp; -1.5 &amp; 0\\-2.3 &amp; 2 &amp; 4.1\end{matrix}\right]^+=\left[\begin{matrix}2 &amp; 0 &amp; 0\\0 &amp; 2 &amp; 4.1\end{matrix}\right].$$
122746
If $A$ and $B$ be two invertible $n \times n$ real matrices and $A + B$ is invertible, how to prove that $A^{-1} + B^{-1}$ is also invertible?
40757
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$, thus $\det(A) =0$.    I proved that way, but looks like too simple so I think maybe there is a trick that I missed. Or I proved it right?   Modification: ($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$ ($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)    Now I think I complete the proof.
17900
I'm trying to prove that for a certain matrix $A$, and its conjugate transpose $A^*$, we have $⟨Ax,y⟩=⟨x,A^*y⟩$, where $⟨⟩$ represent the inner product. So here it's simply the dot product in $R^n$.  Does anyone have an idea of how to prove this?  Thank you!
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
114672
$$A=\left[\frac{1}{i+j}\right]=\left(\begin{matrix}\frac{1}{1+1}&amp;\frac{1}{1+2}&amp;\cdots&amp;\frac{1}{1+n}&amp;\\\frac{1}{2+1}&amp;\frac{1}{2+2}&amp;\cdots&amp;\frac{1}{2+n}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\\frac{1}{n+1}&amp;\frac{1}{n+2}&amp;\cdots&amp;\frac{1}{n+n}\end{matrix}\right)$$  My textbook exercise says yes, but I can't prove it.
85447
The following is the problem at hand:  $A^4 = 2A^2.$ Prove that $(I-A^2) = (I-A^2)^{-1}$  My attempts at a solution:  $I = A^{-1} * A,$ therefore we can start with $(A^{-1}A - A^2) = (A^{-1}A - (1/2)A^4) = (A^{-1}A - A^2) = (A^{-1}A - (1/2)(A^{-4})^{-1}) = ???$  I'm having problems with the fundamentals of this question. Sure, I can play around with the individual matrices on the left side, but that doesn't help to handle the inverse of a sum of matrices, like how the transpose of a sum is equal to the sum of the transpose of its individual matrix terms. How do you approach this without delving into more complex matrix operations?
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.
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 $AA^*$.
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.
68302
Is there square matrix $A$ of size $3$ with real entries such that  $$ \operatorname{tr}(A)=0\text{ and }A^2+A^T=I. $$  I have proved that there is not with size $2$ using definition of  "trace", but for size $3$ it becomes complicated.  Here is the sketch of the proof for $2$. $$ a_{11}+a_{22}=0,\\ a_{11}^2+a_{12}a_{21}+a_{11}=1,\\ a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\ a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\ a_{12}a_{21}+a_{22}^2+a_{22}=1. $$ Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together.   Thanks!
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 $\det(AD)=\det(BC)$.
28670
I have problems to show that if $A$ is a $2 \times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.