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How to find square root of a matrix if it's eigen valuses are same.
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I came across a question in strang's book stated:
Since it is a matrix with same eigen values that is 0 and eigen vector [1 0]^t
In second part, even with replacing diagonal entries that is 0 to 4 the eigen values remains equal that is 4 and eigen vector [1 0]^t
Since this can't be diagonalized as S^-1 doesn't exist as matrix have dependent eigen vector
How to find it's square root ?
Sorry for the formatting and thanks in advance.
linear-algebra matrices eigenvalues-eigenvectors matrix-equations
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
asked Jul 20 at 19:01
Udolf
355
add a comment |Â
up vote
2
down vote
favorite
I came across a question in strang's book stated:
Since it is a matrix with same eigen values that is 0 and eigen vector [1 0]^t
In second part, even with replacing diagonal entries that is 0 to 4 the eigen values remains equal that is 4 and eigen vector [1 0]^t
Since this can't be diagonalized as S^-1 doesn't exist as matrix have dependent eigen vector
How to find it's square root ?
Sorry for the formatting and thanks in advance.
linear-algebra matrices eigenvalues-eigenvectors matrix-equations
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
asked Jul 20 at 19:01
Udolf
355
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I came across a question in strang's book stated:
Since it is a matrix with same eigen values that is 0 and eigen vector [1 0]^t
In second part, even with replacing diagonal entries that is 0 to 4 the eigen values remains equal that is 4 and eigen vector [1 0]^t
Since this can't be diagonalized as S^-1 doesn't exist as matrix have dependent eigen vector
How to find it's square root ?
Sorry for the formatting and thanks in advance.
linear-algebra matrices eigenvalues-eigenvectors matrix-equations
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
asked Jul 20 at 19:01
Udolf
355
I came across a question in strang's book stated:
Since it is a matrix with same eigen values that is 0 and eigen vector [1 0]^t
In second part, even with replacing diagonal entries that is 0 to 4 the eigen values remains equal that is 4 and eigen vector [1 0]^t
Since this can't be diagonalized as S^-1 doesn't exist as matrix have dependent eigen vector
How to find it's square root ?
Sorry for the formatting and thanks in advance.
linear-algebra matrices eigenvalues-eigenvectors matrix-equations
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
asked Jul 20 at 19:01
Udolf
355
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
edited Jul 20 at 20:01
José Carlos Santos
114k1698177
114k1698177
asked Jul 20 at 19:01
Udolf
355
asked Jul 20 at 19:01
Udolf
355
asked Jul 20 at 19:01
Udolf
355
355
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
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accepted
If $S$ is a $2times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$beginpmatrix0&a\0&0endpmatrix.$$Therefore, $S^2$ is the null matrix.
On the other hand,$$beginpmatrix2&a\0&2endpmatrix^2=beginpmatrix4&4a\0&4endpmatrix.$$So, take $a=frac14$.
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
add a comment |Â
up vote
2
down vote
The minimal polynomial of $A$ is $p(lambda)=lambda^2$. If $A=B^2$, then the minimal polynomial $q(lambda)$ of $B$ must divide $lambda^4$, and must be of order $le 2$, which forces $B^2=0$, which is impossible because $B^2=Ane 0$.
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
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Assume $S^2 = A$. We have $S^2 = A ne 0$ but $S^4 = A^2 = 0$.
Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $le 2$.
answered Jul 20 at 20:49
mechanodroid
22.2k52041
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $S$ is a $2times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$beginpmatrix0&a\0&0endpmatrix.$$Therefore, $S^2$ is the null matrix.
On the other hand,$$beginpmatrix2&a\0&2endpmatrix^2=beginpmatrix4&4a\0&4endpmatrix.$$So, take $a=frac14$.
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
add a comment |Â
up vote
4
down vote
accepted
If $S$ is a $2times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$beginpmatrix0&a\0&0endpmatrix.$$Therefore, $S^2$ is the null matrix.
On the other hand,$$beginpmatrix2&a\0&2endpmatrix^2=beginpmatrix4&4a\0&4endpmatrix.$$So, take $a=frac14$.
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $S$ is a $2times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$beginpmatrix0&a\0&0endpmatrix.$$Therefore, $S^2$ is the null matrix.
On the other hand,$$beginpmatrix2&a\0&2endpmatrix^2=beginpmatrix4&4a\0&4endpmatrix.$$So, take $a=frac14$.
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
If $S$ is a $2times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$beginpmatrix0&a\0&0endpmatrix.$$Therefore, $S^2$ is the null matrix.
On the other hand,$$beginpmatrix2&a\0&2endpmatrix^2=beginpmatrix4&4a\0&4endpmatrix.$$So, take $a=frac14$.
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
answered Jul 20 at 19:07
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
2
down vote
The minimal polynomial of $A$ is $p(lambda)=lambda^2$. If $A=B^2$, then the minimal polynomial $q(lambda)$ of $B$ must divide $lambda^4$, and must be of order $le 2$, which forces $B^2=0$, which is impossible because $B^2=Ane 0$.
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
add a comment |Â
up vote
2
down vote
The minimal polynomial of $A$ is $p(lambda)=lambda^2$. If $A=B^2$, then the minimal polynomial $q(lambda)$ of $B$ must divide $lambda^4$, and must be of order $le 2$, which forces $B^2=0$, which is impossible because $B^2=Ane 0$.
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The minimal polynomial of $A$ is $p(lambda)=lambda^2$. If $A=B^2$, then the minimal polynomial $q(lambda)$ of $B$ must divide $lambda^4$, and must be of order $le 2$, which forces $B^2=0$, which is impossible because $B^2=Ane 0$.
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
The minimal polynomial of $A$ is $p(lambda)=lambda^2$. If $A=B^2$, then the minimal polynomial $q(lambda)$ of $B$ must divide $lambda^4$, and must be of order $le 2$, which forces $B^2=0$, which is impossible because $B^2=Ane 0$.
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
answered Jul 20 at 20:59
DisintegratingByParts
55.6k42273
55.6k42273
add a comment |Â
add a comment |Â
up vote
1
down vote
Assume $S^2 = A$. We have $S^2 = A ne 0$ but $S^4 = A^2 = 0$.
Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $le 2$.
answered Jul 20 at 20:49
mechanodroid
22.2k52041
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
add a comment |Â
up vote
1
down vote
Assume $S^2 = A$. We have $S^2 = A ne 0$ but $S^4 = A^2 = 0$.
Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $le 2$.
answered Jul 20 at 20:49
mechanodroid
22.2k52041
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume $S^2 = A$. We have $S^2 = A ne 0$ but $S^4 = A^2 = 0$.
Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $le 2$.
answered Jul 20 at 20:49
mechanodroid
22.2k52041
Assume $S^2 = A$. We have $S^2 = A ne 0$ but $S^4 = A^2 = 0$.
Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $le 2$.
answered Jul 20 at 20:49
mechanodroid
22.2k52041
answered Jul 20 at 20:49
mechanodroid
22.2k52041
answered Jul 20 at 20:49
mechanodroid
22.2k52041
answered Jul 20 at 20:49
mechanodroid
22.2k52041
22.2k52041
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
add a comment |Â
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
Can you please elaborate the part "but x^2 does not annihilate S , which is impossible because the degree of the minimal polynomial of S is <=2"
– Udolf
Jul 21 at 4:16
1
1
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
@Udolf We know that $S^4 = 0$ so the polynomial $x^4$ annihilates $S$. The minimal poylnomial of $S$ divides $x^4$, so it can be $x$ or $x^2$ (the minimal polynomial has degree at most $2$, because by Hamilton-Cayley it divides the characteristic polynomial which has degree $2$ in this case). But $S ne 0$ and $S^2 ne 0$ so neither of those annihilates $S$. This is a contradiction.
– mechanodroid
Jul 21 at 9:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
Thank you for the explanation.
– Udolf
Jul 21 at 13:39
add a comment |Â
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