Which of the following conditions / properties ensures that the $ntimes n$ matrix $A$ is invertible?












1












$begingroup$



Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?



Possible answers:





  1. $A^Tcdot A$ is invertible

  2. $A^T= A$

  3. $A^2+ A + I = 0$


  4. $AB=BA$, for any invertible matrix $B$




My Answer:
My answer is that number $1$ and $4$ are conditions that ensure that $ntimes n$ matrix $A$ is invertible. Am I correct?






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Suraj You should probably give your response in an answer instead of comments
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:04










  • $begingroup$
    so 3 is the only conditions that ensures that n*n matrix A is invertible
    $endgroup$
    – anders
    Dec 19 '18 at 17:04












  • $begingroup$
    @anders Both $(1),(3)$ ensure the invertibility of $A$
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:05










  • $begingroup$
    oh okey. thank you so much
    $endgroup$
    – anders
    Dec 19 '18 at 17:05










  • $begingroup$
    $(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:06
















1












$begingroup$



Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?



Possible answers:





  1. $A^Tcdot A$ is invertible

  2. $A^T= A$

  3. $A^2+ A + I = 0$


  4. $AB=BA$, for any invertible matrix $B$




My Answer:
My answer is that number $1$ and $4$ are conditions that ensure that $ntimes n$ matrix $A$ is invertible. Am I correct?






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Suraj You should probably give your response in an answer instead of comments
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:04










  • $begingroup$
    so 3 is the only conditions that ensures that n*n matrix A is invertible
    $endgroup$
    – anders
    Dec 19 '18 at 17:04












  • $begingroup$
    @anders Both $(1),(3)$ ensure the invertibility of $A$
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:05










  • $begingroup$
    oh okey. thank you so much
    $endgroup$
    – anders
    Dec 19 '18 at 17:05










  • $begingroup$
    $(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:06














1












1








1





$begingroup$



Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?



Possible answers:





  1. $A^Tcdot A$ is invertible

  2. $A^T= A$

  3. $A^2+ A + I = 0$


  4. $AB=BA$, for any invertible matrix $B$




My Answer:
My answer is that number $1$ and $4$ are conditions that ensure that $ntimes n$ matrix $A$ is invertible. Am I correct?






linear-algebra matrices







share|cite|improve this question











$endgroup$





Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?



Possible answers:





  1. $A^Tcdot A$ is invertible

  2. $A^T= A$

  3. $A^2+ A + I = 0$


  4. $AB=BA$, for any invertible matrix $B$




My Answer:
My answer is that number $1$ and $4$ are conditions that ensure that $ntimes n$ matrix $A$ is invertible. Am I correct?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 17:09









Shubham Johri

5,017717




5,017717










asked Dec 19 '18 at 16:57









andersanders

65




65








  • 1




    $begingroup$
    @Suraj You should probably give your response in an answer instead of comments
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:04










  • $begingroup$
    so 3 is the only conditions that ensures that n*n matrix A is invertible
    $endgroup$
    – anders
    Dec 19 '18 at 17:04












  • $begingroup$
    @anders Both $(1),(3)$ ensure the invertibility of $A$
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:05










  • $begingroup$
    oh okey. thank you so much
    $endgroup$
    – anders
    Dec 19 '18 at 17:05










  • $begingroup$
    $(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:06














  • 1




    $begingroup$
    @Suraj You should probably give your response in an answer instead of comments
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:04










  • $begingroup$
    so 3 is the only conditions that ensures that n*n matrix A is invertible
    $endgroup$
    – anders
    Dec 19 '18 at 17:04












  • $begingroup$
    @anders Both $(1),(3)$ ensure the invertibility of $A$
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:05










  • $begingroup$
    oh okey. thank you so much
    $endgroup$
    – anders
    Dec 19 '18 at 17:05










  • $begingroup$
    $(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 17:06








1




1




$begingroup$
@Suraj You should probably give your response in an answer instead of comments
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:04




$begingroup$
@Suraj You should probably give your response in an answer instead of comments
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:04












$begingroup$
so 3 is the only conditions that ensures that n*n matrix A is invertible
$endgroup$
– anders
Dec 19 '18 at 17:04






$begingroup$
so 3 is the only conditions that ensures that n*n matrix A is invertible
$endgroup$
– anders
Dec 19 '18 at 17:04














$begingroup$
@anders Both $(1),(3)$ ensure the invertibility of $A$
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:05




$begingroup$
@anders Both $(1),(3)$ ensure the invertibility of $A$
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:05












$begingroup$
oh okey. thank you so much
$endgroup$
– anders
Dec 19 '18 at 17:05




$begingroup$
oh okey. thank you so much
$endgroup$
– anders
Dec 19 '18 at 17:05












$begingroup$
$(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:06




$begingroup$
$(1)$ is true, as $det(A^TA)=det(A^T)cdotdet(A)=det(A)^2ne0impliesdet(A)ne0$.
$endgroup$
– Shubham Johri
Dec 19 '18 at 17:06










1 Answer
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1












$begingroup$

$1$- It implies $A$ is invertible because if we assume $|A|=m$ then because $A^{T}A$ is invertible then $m^2 neq 0$ and hence $|A|=m neq0$ and Hence $A$ is invertible.



for $2$ and $4$ Consider $A=0$



$3$- consider $Acdot(-A-I)=(-A-I)cdot A=-A^2-A=I$. Hence $A$ is invertible in this case.



Thus only $1$ and $3$ implies $A$ is invertible






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you so much for the help
    $endgroup$
    – anders
    Dec 19 '18 at 17:14











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1 Answer
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$begingroup$

$1$- It implies $A$ is invertible because if we assume $|A|=m$ then because $A^{T}A$ is invertible then $m^2 neq 0$ and hence $|A|=m neq0$ and Hence $A$ is invertible.



for $2$ and $4$ Consider $A=0$



$3$- consider $Acdot(-A-I)=(-A-I)cdot A=-A^2-A=I$. Hence $A$ is invertible in this case.



Thus only $1$ and $3$ implies $A$ is invertible






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you so much for the help
    $endgroup$
    – anders
    Dec 19 '18 at 17:14
















1












$begingroup$

$1$- It implies $A$ is invertible because if we assume $|A|=m$ then because $A^{T}A$ is invertible then $m^2 neq 0$ and hence $|A|=m neq0$ and Hence $A$ is invertible.



for $2$ and $4$ Consider $A=0$



$3$- consider $Acdot(-A-I)=(-A-I)cdot A=-A^2-A=I$. Hence $A$ is invertible in this case.



Thus only $1$ and $3$ implies $A$ is invertible






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you so much for the help
    $endgroup$
    – anders
    Dec 19 '18 at 17:14














1












1








1





$begingroup$

$1$- It implies $A$ is invertible because if we assume $|A|=m$ then because $A^{T}A$ is invertible then $m^2 neq 0$ and hence $|A|=m neq0$ and Hence $A$ is invertible.



for $2$ and $4$ Consider $A=0$



$3$- consider $Acdot(-A-I)=(-A-I)cdot A=-A^2-A=I$. Hence $A$ is invertible in this case.



Thus only $1$ and $3$ implies $A$ is invertible






share|cite|improve this answer











$endgroup$



$1$- It implies $A$ is invertible because if we assume $|A|=m$ then because $A^{T}A$ is invertible then $m^2 neq 0$ and hence $|A|=m neq0$ and Hence $A$ is invertible.



for $2$ and $4$ Consider $A=0$



$3$- consider $Acdot(-A-I)=(-A-I)cdot A=-A^2-A=I$. Hence $A$ is invertible in this case.



Thus only $1$ and $3$ implies $A$ is invertible







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 17:10









Shubham Johri

5,017717




5,017717










answered Dec 19 '18 at 17:08







user408906



















  • $begingroup$
    thank you so much for the help
    $endgroup$
    – anders
    Dec 19 '18 at 17:14


















  • $begingroup$
    thank you so much for the help
    $endgroup$
    – anders
    Dec 19 '18 at 17:14
















$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14




$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14


















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