True or False ( Invertibility of $A$)
$begingroup$
Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.
$A^2 + A = –I$
$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.
Is this explanation enough to prove the statement true?
linear-algebra matrices
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
$endgroup$
add a comment |
$begingroup$
Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.
$A^2 + A = –I$
$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.
Is this explanation enough to prove the statement true?
linear-algebra matrices
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
$endgroup$
1
$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27
add a comment |
$begingroup$
Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.
$A^2 + A = –I$
$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.
Is this explanation enough to prove the statement true?
linear-algebra matrices
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
$endgroup$
Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.
$A^2 + A = –I$
$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.
Is this explanation enough to prove the statement true?
linear-algebra matrices
linear-algebra matrices
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
asked Dec 9 '18 at 16:55
MathsaddictMathsaddict
3459
3459
1
$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27
add a comment |
1
$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27
1
1
$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27
$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27
add a comment |
2 Answers
2
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$begingroup$
I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
$endgroup$
add a comment |
$begingroup$
Short Answer: Yes.
Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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$begingroup$
I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
$endgroup$
add a comment |
$begingroup$
I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
$endgroup$
add a comment |
$begingroup$
I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
$endgroup$
I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 16:58
GenericMathematicianGenericMathematician
863
863
add a comment |
add a comment |
$begingroup$
Short Answer: Yes.
Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
$begingroup$
Short Answer: Yes.
Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
$begingroup$
Short Answer: Yes.
Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
$endgroup$
Short Answer: Yes.
Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
answered Dec 9 '18 at 17:05
bruderjakob17bruderjakob17
1997
1997
add a comment |
add a comment |
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$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27