For $ntimes n$ matrix $A$, there exists $B$ such that $AB=BA=0$ and $rk(A) + rk(B) = n$
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For every $ntimes n$ matrix $A$, there exists $B$ such that $AB=BA=0$ and
$rk(A) + rk(B) = n$
The question is to prove or disprove this.
I think it is true. Take a $ntimes n$ matrix $tilde A$ in Echelon form where $a_1, cdots a_k$ are linearly independent row vectors
$$tilde A = beginbmatrix
-- &a_1 & -- \
;&cdots\
-- &a_k & -- \
0& cdots & 0 \
0& cdots & 0 \
endbmatrix$$
We can define
$$tilde B = beginbmatrix
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 & |&cdots &|\
0 &cdots & 0 & b_1& cdots& b_n-k\
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 &|&cdots &|\
endbmatrix$$
where column vectors $b_1,cdots b_n-k$ are linearly independent vectors from
$$textspana_1, cdots a_k^perp.$$
Now clearly $tilde A tilde B = tilde B tilde A = 0$.
Now for general $A$, we know row operation and swapping two rows are given by invertible matrices multiplied on the left side, so there exists a invertible matrix $P$ such that
$$A = P tilde A$$
where $tilde A$ is in Echelon form. Now just define
$$B = tilde B P^-1$$
we have $AB = BA = 0$ and $$rk(A)+rk(B) = rk(tilde A)+rk(tilde B) = n.$$
Is this correct?
linear-algebra proof-verification
asked Aug 22 at 9:00
Xiao
4,50011334
add a comment |Â
up vote
2
down vote
favorite
For every $ntimes n$ matrix $A$, there exists $B$ such that $AB=BA=0$ and
$rk(A) + rk(B) = n$
The question is to prove or disprove this.
I think it is true. Take a $ntimes n$ matrix $tilde A$ in Echelon form where $a_1, cdots a_k$ are linearly independent row vectors
$$tilde A = beginbmatrix
-- &a_1 & -- \
;&cdots\
-- &a_k & -- \
0& cdots & 0 \
0& cdots & 0 \
endbmatrix$$
We can define
$$tilde B = beginbmatrix
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 & |&cdots &|\
0 &cdots & 0 & b_1& cdots& b_n-k\
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 &|&cdots &|\
endbmatrix$$
where column vectors $b_1,cdots b_n-k$ are linearly independent vectors from
$$textspana_1, cdots a_k^perp.$$
Now clearly $tilde A tilde B = tilde B tilde A = 0$.
Now for general $A$, we know row operation and swapping two rows are given by invertible matrices multiplied on the left side, so there exists a invertible matrix $P$ such that
$$A = P tilde A$$
where $tilde A$ is in Echelon form. Now just define
$$B = tilde B P^-1$$
we have $AB = BA = 0$ and $$rk(A)+rk(B) = rk(tilde A)+rk(tilde B) = n.$$
Is this correct?
linear-algebra proof-verification
asked Aug 22 at 9:00
Xiao
4,50011334
What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
1
This looks correct!
– Peter Melech
Aug 22 at 9:15
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For every $ntimes n$ matrix $A$, there exists $B$ such that $AB=BA=0$ and
$rk(A) + rk(B) = n$
The question is to prove or disprove this.
I think it is true. Take a $ntimes n$ matrix $tilde A$ in Echelon form where $a_1, cdots a_k$ are linearly independent row vectors
$$tilde A = beginbmatrix
-- &a_1 & -- \
;&cdots\
-- &a_k & -- \
0& cdots & 0 \
0& cdots & 0 \
endbmatrix$$
We can define
$$tilde B = beginbmatrix
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 & |&cdots &|\
0 &cdots & 0 & b_1& cdots& b_n-k\
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 &|&cdots &|\
endbmatrix$$
where column vectors $b_1,cdots b_n-k$ are linearly independent vectors from
$$textspana_1, cdots a_k^perp.$$
Now clearly $tilde A tilde B = tilde B tilde A = 0$.
Now for general $A$, we know row operation and swapping two rows are given by invertible matrices multiplied on the left side, so there exists a invertible matrix $P$ such that
$$A = P tilde A$$
where $tilde A$ is in Echelon form. Now just define
$$B = tilde B P^-1$$
we have $AB = BA = 0$ and $$rk(A)+rk(B) = rk(tilde A)+rk(tilde B) = n.$$
Is this correct?
linear-algebra proof-verification
asked Aug 22 at 9:00
Xiao
4,50011334
For every $ntimes n$ matrix $A$, there exists $B$ such that $AB=BA=0$ and
$rk(A) + rk(B) = n$
The question is to prove or disprove this.
I think it is true. Take a $ntimes n$ matrix $tilde A$ in Echelon form where $a_1, cdots a_k$ are linearly independent row vectors
$$tilde A = beginbmatrix
-- &a_1 & -- \
;&cdots\
-- &a_k & -- \
0& cdots & 0 \
0& cdots & 0 \
endbmatrix$$
We can define
$$tilde B = beginbmatrix
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 & |&cdots &|\
0 &cdots & 0 & b_1& cdots& b_n-k\
0 &cdots & 0 &|&cdots &|\
0 &cdots & 0 &|&cdots &|\
endbmatrix$$
where column vectors $b_1,cdots b_n-k$ are linearly independent vectors from
$$textspana_1, cdots a_k^perp.$$
Now clearly $tilde A tilde B = tilde B tilde A = 0$.
Now for general $A$, we know row operation and swapping two rows are given by invertible matrices multiplied on the left side, so there exists a invertible matrix $P$ such that
$$A = P tilde A$$
where $tilde A$ is in Echelon form. Now just define
$$B = tilde B P^-1$$
we have $AB = BA = 0$ and $$rk(A)+rk(B) = rk(tilde A)+rk(tilde B) = n.$$
Is this correct?
linear-algebra proof-verification
asked Aug 22 at 9:00
Xiao
4,50011334
edited Aug 22 at 9:10
asked Aug 22 at 9:00
Xiao
4,50011334
asked Aug 22 at 9:00
Xiao
4,50011334
asked Aug 22 at 9:00
Xiao
4,50011334
4,50011334
What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
1
This looks correct!
– Peter Melech
Aug 22 at 9:15
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58
add a comment |Â
What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
1
This looks correct!
– Peter Melech
Aug 22 at 9:15
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58
What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
1
1
This looks correct!
– Peter Melech
Aug 22 at 9:15
This looks correct!
– Peter Melech
Aug 22 at 9:15
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58
add a comment |Â
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What is $n$ in the statement you want to prove?
– Taroccoesbrocco
Aug 22 at 9:09
@Taroccoesbrocco sorry $A,B$ are $ntimes n$ matrices
– Xiao
Aug 22 at 9:10
1
This looks correct!
– Peter Melech
Aug 22 at 9:15
math.stackexchange.com/questions/298836/…
– Jneven
Aug 22 at 11:58