even to me. Though it seems obvious, its proof is not. The main problem why the theorem seems strange is because multiplication of matrices, in general, is not commutative
that is why, the theorem is proved by using such lemmas. For otherwise, consider this proof: since $p(\lambda)=|A-\lambdaI|$ is the characteristic polynomial, then, by substituting $A$ in place of $\lambda$, we obtain $p(A)=|A-AI| =0$. But, the proof is wrong, because the lack of commutativity of matrices
@ManeeshNarayanan sorry, not all $B_i$ are necessarily zero by assumption, but $B_r=O$ definitely which in turn implies $B(\lambda)=O$, which implies $C=O$
@PrithiviRaj but it all depends on rank. If all perform poorly, then rank may go up. I all perform well, then rank goes down. So, just wait for final result, when things will become clear.
consider this question: let $T$ be the linear functional from $L^2$ given by $\int_{1/4}^{3/4}3\sqrt{2}f d\mu$ where $\mu$ be lebesgue measure. Then, norm of $T$ is?
it is quite easy. We have $\|Tf\|\leq3\sqrt{2}(int_{1/4}^{3/4})^{1/2}d\mu\|f\|=3\|f\|$ from which $\|T\|=\sup\frac{\|Tf\|}{\|f\|}=3$ quite simple, isnt it?
can you help me? not solving questions. How to improve myself to solve this kind of problems.https://www.isical.ac.in/~admission/IsiAdmission2017/PreviousQuestion/JRF-Math-MTA-2017.pdf @BalarkaSen
Let $X$ be an $ n×n$ complex matrix of rank $1$ and $I $the $n×n$ identity matrix. Show that $det(I + X) = 1 + tr(X)$, where tr(X) denotes the trace of X and det(X) denotes the determinant of X.
My question is what properties does a rank one matrix have? I saw a lot of papers mentioning that the matrix is rank one and so on.
I know rank one of a matrix means that there are no independent columns or rows in that matrix.
I have small doubts regarding the teaching. I know my lectures are utter boring. my students ask short cuts to pass the exam. I am sincerely doing my work. they are complaining me. I told them that I was ready to help to clear their doubts.
My question is what properties does a rank one matrix have? I saw a lot of papers mentioning that the matrix is rank one and so on.
I know rank one of a matrix means that there are no independent columns or rows in that matrix.
