Mixing
Rank of block triangular matrix using linearly independent rows/columns
$begingroup$
Prove that $operatorname{rank}begin{pmatrix}A & X \ 0 & B \ end{pmatrix}ge operatorname{rank}(A) + operatorname{rank}(B)$ and the equality is attained if $X=0$ ($A, B, X in M_{m, n} (mathbb C) $) .
I know this is a pretty well-known relation, but I want to prove it using linearly independent rows or columns because all the other proofs I have seen seem way to advanced for me to understand.
linear-algebra matrices
edited Jan 22 at 18:23
Adam Higgins
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
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add a comment |
$begingroup$
Prove that $operatorname{rank}begin{pmatrix}A & X \ 0 & B \ end{pmatrix}ge operatorname{rank}(A) + operatorname{rank}(B)$ and the equality is attained if $X=0$ ($A, B, X in M_{m, n} (mathbb C) $) .
I know this is a pretty well-known relation, but I want to prove it using linearly independent rows or columns because all the other proofs I have seen seem way to advanced for me to understand.
linear-algebra matrices
edited Jan 22 at 18:23
Adam Higgins
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
Prove that $operatorname{rank}begin{pmatrix}A & X \ 0 & B \ end{pmatrix}ge operatorname{rank}(A) + operatorname{rank}(B)$ and the equality is attained if $X=0$ ($A, B, X in M_{m, n} (mathbb C) $) .
I know this is a pretty well-known relation, but I want to prove it using linearly independent rows or columns because all the other proofs I have seen seem way to advanced for me to understand.
linear-algebra matrices
edited Jan 22 at 18:23
Adam Higgins
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
$endgroup$
Prove that $operatorname{rank}begin{pmatrix}A & X \ 0 & B \ end{pmatrix}ge operatorname{rank}(A) + operatorname{rank}(B)$ and the equality is attained if $X=0$ ($A, B, X in M_{m, n} (mathbb C) $) .
I know this is a pretty well-known relation, but I want to prove it using linearly independent rows or columns because all the other proofs I have seen seem way to advanced for me to understand.
linear-algebra matrices
linear-algebra matrices
edited Jan 22 at 18:23
Adam Higgins
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
edited Jan 22 at 18:23
Adam Higgins
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
edited Jan 22 at 18:23
Adam Higgins
613113
edited Jan 22 at 18:23
Adam Higgins
613113
edited Jan 22 at 18:23
Adam Higgins
613113
613113
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
asked Jan 22 at 17:39
JustAnAmateurJustAnAmateur
1096
1096
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
This is a sketch. You should verify each claim using the definition of linear independence.
There exist $text{rank}(A)$ columns of $A$ that are linearly independent
so there exist $text{rank}(A)$ columns of $begin{pmatrix}A\0end{pmatrix}$ that are linearly independent.
There exist $text{rank}(B)$ columns of $B$ that are linearly independent
so there exist $text{rank}(B)$ columns of $begin{pmatrix}X\Bend{pmatrix}$ that are linearly independent.
Together, you obtain $text{rank}(A) + text{rank}(B)$ columns of the full matrix that are linearly independent.
Edit: response to comment
Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.
In any case here is an example.
If $A = begin{pmatrix}1 & 0 \ 0&0 end{pmatrix}$ and $B = begin{pmatrix}0 & 0 \ 1 & 1 end{pmatrix}$
and $X = begin{pmatrix}1&0\0&1end{pmatrix}$
then the inequality is strict.
answered Jan 22 at 18:03
angryavianangryavian
42k23381
$endgroup$
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
add a comment |
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
This is a sketch. You should verify each claim using the definition of linear independence.
There exist $text{rank}(A)$ columns of $A$ that are linearly independent
so there exist $text{rank}(A)$ columns of $begin{pmatrix}A\0end{pmatrix}$ that are linearly independent.
There exist $text{rank}(B)$ columns of $B$ that are linearly independent
so there exist $text{rank}(B)$ columns of $begin{pmatrix}X\Bend{pmatrix}$ that are linearly independent.
Together, you obtain $text{rank}(A) + text{rank}(B)$ columns of the full matrix that are linearly independent.
Edit: response to comment
Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.
In any case here is an example.
If $A = begin{pmatrix}1 & 0 \ 0&0 end{pmatrix}$ and $B = begin{pmatrix}0 & 0 \ 1 & 1 end{pmatrix}$
and $X = begin{pmatrix}1&0\0&1end{pmatrix}$
then the inequality is strict.
answered Jan 22 at 18:03
angryavianangryavian
42k23381
$endgroup$
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
add a comment |
$begingroup$
This is a sketch. You should verify each claim using the definition of linear independence.
There exist $text{rank}(A)$ columns of $A$ that are linearly independent
so there exist $text{rank}(A)$ columns of $begin{pmatrix}A\0end{pmatrix}$ that are linearly independent.
There exist $text{rank}(B)$ columns of $B$ that are linearly independent
so there exist $text{rank}(B)$ columns of $begin{pmatrix}X\Bend{pmatrix}$ that are linearly independent.
Together, you obtain $text{rank}(A) + text{rank}(B)$ columns of the full matrix that are linearly independent.
Edit: response to comment
Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.
In any case here is an example.
If $A = begin{pmatrix}1 & 0 \ 0&0 end{pmatrix}$ and $B = begin{pmatrix}0 & 0 \ 1 & 1 end{pmatrix}$
and $X = begin{pmatrix}1&0\0&1end{pmatrix}$
then the inequality is strict.
answered Jan 22 at 18:03
angryavianangryavian
42k23381
$endgroup$
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
add a comment |
$begingroup$
This is a sketch. You should verify each claim using the definition of linear independence.
There exist $text{rank}(A)$ columns of $A$ that are linearly independent
so there exist $text{rank}(A)$ columns of $begin{pmatrix}A\0end{pmatrix}$ that are linearly independent.
There exist $text{rank}(B)$ columns of $B$ that are linearly independent
so there exist $text{rank}(B)$ columns of $begin{pmatrix}X\Bend{pmatrix}$ that are linearly independent.
Together, you obtain $text{rank}(A) + text{rank}(B)$ columns of the full matrix that are linearly independent.
Edit: response to comment
Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.
In any case here is an example.
If $A = begin{pmatrix}1 & 0 \ 0&0 end{pmatrix}$ and $B = begin{pmatrix}0 & 0 \ 1 & 1 end{pmatrix}$
and $X = begin{pmatrix}1&0\0&1end{pmatrix}$
then the inequality is strict.
answered Jan 22 at 18:03
angryavianangryavian
42k23381
$endgroup$
This is a sketch. You should verify each claim using the definition of linear independence.
There exist $text{rank}(A)$ columns of $A$ that are linearly independent
so there exist $text{rank}(A)$ columns of $begin{pmatrix}A\0end{pmatrix}$ that are linearly independent.
There exist $text{rank}(B)$ columns of $B$ that are linearly independent
so there exist $text{rank}(B)$ columns of $begin{pmatrix}X\Bend{pmatrix}$ that are linearly independent.
Together, you obtain $text{rank}(A) + text{rank}(B)$ columns of the full matrix that are linearly independent.
Edit: response to comment
Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.
In any case here is an example.
If $A = begin{pmatrix}1 & 0 \ 0&0 end{pmatrix}$ and $B = begin{pmatrix}0 & 0 \ 1 & 1 end{pmatrix}$
and $X = begin{pmatrix}1&0\0&1end{pmatrix}$
then the inequality is strict.
answered Jan 22 at 18:03
angryavianangryavian
42k23381
edited Jan 22 at 18:42
answered Jan 22 at 18:03
angryavianangryavian
42k23381
answered Jan 22 at 18:03
angryavianangryavian
42k23381
answered Jan 22 at 18:03
angryavianangryavian
42k23381
42k23381
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
add a comment |
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Thank you! What I do not understand is why we get that the rank of the full matrix is $ge$ and not equal to $rank A+rank B$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:11
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$.
$endgroup$
– JustAnAmateur
Jan 22 at 18:28
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
@JustAnAmateur See my edit
$endgroup$
– angryavian
Jan 22 at 18:42
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
$begingroup$
Thank you very much, I had been struggling with proving this result for some time.
$endgroup$
– JustAnAmateur
Jan 22 at 18:47
add a comment |
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