Rank matrix sequence
$begingroup$
Let $A in M_{n}(C)$. I have to show that the sequence $(a_{k})_{k geq 0}$, where $a_{k}=$rank $( A^{k+1})- $ rank$( A^k )$ is increasing.
I have tried using this :
$detAneq 0 Rightarrow $ rank$ A=$rank$A^{k}=nRightarrow a_{k}=0$.
Now if $detA= 0$ i don't know to prove that the sequence is increasing.
matrix-rank
asked Jan 23 at 19:51
mathlearningmathlearning
1717
$endgroup$
add a comment |
$begingroup$
Let $A in M_{n}(C)$. I have to show that the sequence $(a_{k})_{k geq 0}$, where $a_{k}=$rank $( A^{k+1})- $ rank$( A^k )$ is increasing.
I have tried using this :
$detAneq 0 Rightarrow $ rank$ A=$rank$A^{k}=nRightarrow a_{k}=0$.
Now if $detA= 0$ i don't know to prove that the sequence is increasing.
matrix-rank
asked Jan 23 at 19:51
mathlearningmathlearning
1717
$endgroup$
$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16
add a comment |
$begingroup$
Let $A in M_{n}(C)$. I have to show that the sequence $(a_{k})_{k geq 0}$, where $a_{k}=$rank $( A^{k+1})- $ rank$( A^k )$ is increasing.
I have tried using this :
$detAneq 0 Rightarrow $ rank$ A=$rank$A^{k}=nRightarrow a_{k}=0$.
Now if $detA= 0$ i don't know to prove that the sequence is increasing.
matrix-rank
asked Jan 23 at 19:51
mathlearningmathlearning
1717
$endgroup$
Let $A in M_{n}(C)$. I have to show that the sequence $(a_{k})_{k geq 0}$, where $a_{k}=$rank $( A^{k+1})- $ rank$( A^k )$ is increasing.
I have tried using this :
$detAneq 0 Rightarrow $ rank$ A=$rank$A^{k}=nRightarrow a_{k}=0$.
Now if $detA= 0$ i don't know to prove that the sequence is increasing.
matrix-rank
matrix-rank
asked Jan 23 at 19:51
mathlearningmathlearning
1717
asked Jan 23 at 19:51
mathlearningmathlearning
1717
asked Jan 23 at 19:51
mathlearningmathlearning
1717
asked Jan 23 at 19:51
mathlearningmathlearning
1717
asked Jan 23 at 19:51
mathlearningmathlearning
1717
1717
$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16
add a comment |
$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16
$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16
$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16
add a comment |
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$begingroup$
To show that this sequence is increasing, it is equivalent to showing that $A^{k+2} + A^{k} geq 2A^{k+1}$. (Why?) This question may help math.stackexchange.com/questions/1146133/…
$endgroup$
– math.h
Jan 23 at 20:16