Mixing
Is $A-BDC$ non-singular in a non-singular block matrix?
$begingroup$
Let $M=begin{bmatrix}A&B\C&Dend{bmatrix}$ be a non-singular block matrix where $Ainmathbb R^{ptimes p}$ and $Dinmathbb R^{qtimes q}$. I suspected that if $p< q$ then $A-BDC$ is a non-singular matrix.
I have tried hard to find a counter-example with no luck. So I tried to prove it. This is where I hit several dead-ends.
First, if you left-multiply $begin{bmatrix}I&-B\0&Iend{bmatrix}$ and right-multiply $begin{bmatrix}I&0\C&Iend{bmatrix}$ to $M$, you'll get
$$begin{bmatrix}A-BDC&B(I-D)\(I+D)C&Dend{bmatrix}$$
but this wouldn't lead to any conclusion.
I also tried to use a pseudo-inverse approach on $[Aquad B]$ but the results are disappointing. So I would appreciate any hints or counter-examples to put an end to this problem.
matrices block-matrices
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
$endgroup$
add a comment |
$begingroup$
Let $M=begin{bmatrix}A&B\C&Dend{bmatrix}$ be a non-singular block matrix where $Ainmathbb R^{ptimes p}$ and $Dinmathbb R^{qtimes q}$. I suspected that if $p< q$ then $A-BDC$ is a non-singular matrix.
I have tried hard to find a counter-example with no luck. So I tried to prove it. This is where I hit several dead-ends.
First, if you left-multiply $begin{bmatrix}I&-B\0&Iend{bmatrix}$ and right-multiply $begin{bmatrix}I&0\C&Iend{bmatrix}$ to $M$, you'll get
$$begin{bmatrix}A-BDC&B(I-D)\(I+D)C&Dend{bmatrix}$$
but this wouldn't lead to any conclusion.
I also tried to use a pseudo-inverse approach on $[Aquad B]$ but the results are disappointing. So I would appreciate any hints or counter-examples to put an end to this problem.
matrices block-matrices
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
$endgroup$
$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16
add a comment |
$begingroup$
Let $M=begin{bmatrix}A&B\C&Dend{bmatrix}$ be a non-singular block matrix where $Ainmathbb R^{ptimes p}$ and $Dinmathbb R^{qtimes q}$. I suspected that if $p< q$ then $A-BDC$ is a non-singular matrix.
I have tried hard to find a counter-example with no luck. So I tried to prove it. This is where I hit several dead-ends.
First, if you left-multiply $begin{bmatrix}I&-B\0&Iend{bmatrix}$ and right-multiply $begin{bmatrix}I&0\C&Iend{bmatrix}$ to $M$, you'll get
$$begin{bmatrix}A-BDC&B(I-D)\(I+D)C&Dend{bmatrix}$$
but this wouldn't lead to any conclusion.
I also tried to use a pseudo-inverse approach on $[Aquad B]$ but the results are disappointing. So I would appreciate any hints or counter-examples to put an end to this problem.
matrices block-matrices
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
$endgroup$
Let $M=begin{bmatrix}A&B\C&Dend{bmatrix}$ be a non-singular block matrix where $Ainmathbb R^{ptimes p}$ and $Dinmathbb R^{qtimes q}$. I suspected that if $p< q$ then $A-BDC$ is a non-singular matrix.
I have tried hard to find a counter-example with no luck. So I tried to prove it. This is where I hit several dead-ends.
First, if you left-multiply $begin{bmatrix}I&-B\0&Iend{bmatrix}$ and right-multiply $begin{bmatrix}I&0\C&Iend{bmatrix}$ to $M$, you'll get
$$begin{bmatrix}A-BDC&B(I-D)\(I+D)C&Dend{bmatrix}$$
but this wouldn't lead to any conclusion.
I also tried to use a pseudo-inverse approach on $[Aquad B]$ but the results are disappointing. So I would appreciate any hints or counter-examples to put an end to this problem.
matrices block-matrices
matrices block-matrices
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
edited Jan 30 at 17:16
polfosol
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
asked Jan 30 at 17:06
polfosolpolfosol
5,92931945
5,92931945
$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16
add a comment |
$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16
$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The block matrix
$$
M = left[!begin{array}{c|cc}0 & 1 & 0 \ hline 1 & 0 & 0 \ 0 & 0 & 1end{array}!right]
$$
is non-singular, yet we get
$$
A - BDC = 0 - begin{bmatrix}1 & 0end{bmatrix}begin{bmatrix}0 & 0 \ 0 & 1end{bmatrix} begin{bmatrix}1 \ 0end{bmatrix} = 0.
$$
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The block matrix
$$
M = left[!begin{array}{c|cc}0 & 1 & 0 \ hline 1 & 0 & 0 \ 0 & 0 & 1end{array}!right]
$$
is non-singular, yet we get
$$
A - BDC = 0 - begin{bmatrix}1 & 0end{bmatrix}begin{bmatrix}0 & 0 \ 0 & 1end{bmatrix} begin{bmatrix}1 \ 0end{bmatrix} = 0.
$$
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
$endgroup$
add a comment |
$begingroup$
The block matrix
$$
M = left[!begin{array}{c|cc}0 & 1 & 0 \ hline 1 & 0 & 0 \ 0 & 0 & 1end{array}!right]
$$
is non-singular, yet we get
$$
A - BDC = 0 - begin{bmatrix}1 & 0end{bmatrix}begin{bmatrix}0 & 0 \ 0 & 1end{bmatrix} begin{bmatrix}1 \ 0end{bmatrix} = 0.
$$
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
$endgroup$
add a comment |
$begingroup$
The block matrix
$$
M = left[!begin{array}{c|cc}0 & 1 & 0 \ hline 1 & 0 & 0 \ 0 & 0 & 1end{array}!right]
$$
is non-singular, yet we get
$$
A - BDC = 0 - begin{bmatrix}1 & 0end{bmatrix}begin{bmatrix}0 & 0 \ 0 & 1end{bmatrix} begin{bmatrix}1 \ 0end{bmatrix} = 0.
$$
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
$endgroup$
The block matrix
$$
M = left[!begin{array}{c|cc}0 & 1 & 0 \ hline 1 & 0 & 0 \ 0 & 0 & 1end{array}!right]
$$
is non-singular, yet we get
$$
A - BDC = 0 - begin{bmatrix}1 & 0end{bmatrix}begin{bmatrix}0 & 0 \ 0 & 1end{bmatrix} begin{bmatrix}1 \ 0end{bmatrix} = 0.
$$
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
answered Jan 30 at 17:25
Misha LavrovMisha Lavrov
48.4k757107
48.4k757107
add a comment |
add a comment |
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$begingroup$
Even in a $2 times 2$ nonsingular matrix you may get $0$ for your expression.
$endgroup$
– GEdgar
Jan 30 at 17:14
$begingroup$
@GEdgar I have made a mistake in wording the question. I should have said $p<q$
$endgroup$
– polfosol
Jan 30 at 17:16