Which of the following conditions / properties ensures that the $ntimes n$ matrix $A$ is invertible?
$begingroup$
Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?
Possible answers:
$A^Tcdot A$ is invertible
- $A^T= A$
- $A^2+ A + I = 0$
$AB=BA$, 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
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
$endgroup$
|
show 1 more comment
$begingroup$
Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?
Possible answers:
$A^Tcdot A$ is invertible
- $A^T= A$
- $A^2+ A + I = 0$
$AB=BA$, 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
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
$endgroup$
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
|
show 1 more comment
$begingroup$
Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?
Possible answers:
$A^Tcdot A$ is invertible
- $A^T= A$
- $A^2+ A + I = 0$
$AB=BA$, 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
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
$endgroup$
Question: Which of the following conditions / properties ensure that the $ntimes n$ matrix $A$ is invertible?
Possible answers:
$A^Tcdot A$ is invertible
- $A^T= A$
- $A^2+ A + I = 0$
$AB=BA$, 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
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
edited Dec 19 '18 at 17:09
Shubham Johri
5,192717
5,192717
asked Dec 19 '18 at 16:57
andersanders
615
asked Dec 19 '18 at 16:57
andersanders
615
asked Dec 19 '18 at 16:57
andersanders
615
615
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$1$- 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
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
$endgroup$
$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14
add a comment |
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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
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
$endgroup$
$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14
add a comment |
$begingroup$
$1$- 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
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
$endgroup$
$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14
add a comment |
$begingroup$
$1$- 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
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
$endgroup$
$1$- 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
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
edited Dec 19 '18 at 17:10
Shubham Johri
5,192717
5,192717
answered Dec 19 '18 at 17:08
user408906
$begingroup$
thank you so much for the help
$endgroup$
– anders
Dec 19 '18 at 17:14
add a comment |
$begingroup$
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
add a comment |
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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