Find all real valued $ntimes n$ matrices $A$ such that for any $vinmathbbR^n$ there exists $yinmathbbR^n$ with $Av=v+y$
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I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?
Thanks in advance.
linear-algebra matrices
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up vote
0
down vote
favorite
I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?
Thanks in advance.
linear-algebra matrices
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?
Thanks in advance.
linear-algebra matrices
I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?
Thanks in advance.
linear-algebra matrices
linear-algebra matrices
asked Sep 10 at 18:55
Jeff
1449
1449
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1 Answer
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Isn't there something missing from the question? Some condition, maybe?
Every real valued matrix is good: let $y= Av-v$.
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Isn't there something missing from the question? Some condition, maybe?
Every real valued matrix is good: let $y= Av-v$.
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
add a comment |Â
up vote
2
down vote
accepted
Isn't there something missing from the question? Some condition, maybe?
Every real valued matrix is good: let $y= Av-v$.
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Isn't there something missing from the question? Some condition, maybe?
Every real valued matrix is good: let $y= Av-v$.
Isn't there something missing from the question? Some condition, maybe?
Every real valued matrix is good: let $y= Av-v$.
answered Sep 10 at 19:04
A. Pongrácz
4,465725
4,465725
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
add a comment |Â
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
No, I took it straight off of an old qualifying exam. I guess it is just that simple?
â Jeff
Sep 10 at 19:06
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
â A. Pongrácz
Sep 10 at 19:11
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
â Jeff
Sep 10 at 19:12
add a comment |Â
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