Mixing
which one of the following statement are/is True $?$
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Let $ A$ and $ B$ be $n times n$ matrix.
which one of the following statement are/is True $?$
$a)$ if $A^n = 0$ for some $ n $, then $det A = 0$.
$b)$ if A and B have the same characteristic polynomial, then they are similar.
$c)$ if the eigenvalues of $A$ are $lambda_1,lambda_2,lambda_3,dots,lambda_n$, then A is similar
to the
diagonal matrix diag($lambda_1 ,lambda_2,lambda_3,dots,lambda_n$)
My answer : all options a, b and c..
For option a), take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$
For option b), take $A = beginbmatrix 0 & 1 \ 1& 0 endbmatrix$ and $B = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
For option c), take $A = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
Is it True ??....
linear-algebra
edited Jul 30 at 11:47
Bernard
110k635102
asked Jul 30 at 11:39
Messi fifa
1498
add a comment |Â
up vote
2
down vote
favorite
Let $ A$ and $ B$ be $n times n$ matrix.
which one of the following statement are/is True $?$
$a)$ if $A^n = 0$ for some $ n $, then $det A = 0$.
$b)$ if A and B have the same characteristic polynomial, then they are similar.
$c)$ if the eigenvalues of $A$ are $lambda_1,lambda_2,lambda_3,dots,lambda_n$, then A is similar
to the
diagonal matrix diag($lambda_1 ,lambda_2,lambda_3,dots,lambda_n$)
My answer : all options a, b and c..
For option a), take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$
For option b), take $A = beginbmatrix 0 & 1 \ 1& 0 endbmatrix$ and $B = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
For option c), take $A = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
Is it True ??....
linear-algebra
edited Jul 30 at 11:47
Bernard
110k635102
asked Jul 30 at 11:39
Messi fifa
1498
2
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
1
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $ A$ and $ B$ be $n times n$ matrix.
which one of the following statement are/is True $?$
$a)$ if $A^n = 0$ for some $ n $, then $det A = 0$.
$b)$ if A and B have the same characteristic polynomial, then they are similar.
$c)$ if the eigenvalues of $A$ are $lambda_1,lambda_2,lambda_3,dots,lambda_n$, then A is similar
to the
diagonal matrix diag($lambda_1 ,lambda_2,lambda_3,dots,lambda_n$)
My answer : all options a, b and c..
For option a), take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$
For option b), take $A = beginbmatrix 0 & 1 \ 1& 0 endbmatrix$ and $B = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
For option c), take $A = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
Is it True ??....
linear-algebra
edited Jul 30 at 11:47
Bernard
110k635102
asked Jul 30 at 11:39
Messi fifa
1498
Let $ A$ and $ B$ be $n times n$ matrix.
which one of the following statement are/is True $?$
$a)$ if $A^n = 0$ for some $ n $, then $det A = 0$.
$b)$ if A and B have the same characteristic polynomial, then they are similar.
$c)$ if the eigenvalues of $A$ are $lambda_1,lambda_2,lambda_3,dots,lambda_n$, then A is similar
to the
diagonal matrix diag($lambda_1 ,lambda_2,lambda_3,dots,lambda_n$)
My answer : all options a, b and c..
For option a), take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$
For option b), take $A = beginbmatrix 0 & 1 \ 1& 0 endbmatrix$ and $B = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
For option c), take $A = beginbmatrix 1 & 0 \ 0& 1 endbmatrix$
Is it True ??....
linear-algebra
edited Jul 30 at 11:47
Bernard
110k635102
asked Jul 30 at 11:39
Messi fifa
1498
edited Jul 30 at 11:47
Bernard
110k635102
edited Jul 30 at 11:47
Bernard
110k635102
edited Jul 30 at 11:47
Bernard
110k635102
110k635102
asked Jul 30 at 11:39
Messi fifa
1498
asked Jul 30 at 11:39
Messi fifa
1498
asked Jul 30 at 11:39
Messi fifa
1498
1498
2
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
1
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48
add a comment |Â
2
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
1
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48
2
2
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
1
1
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48
add a comment |Â
1 Answer
1
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(a) is true for a nilpotent matrix has all eigenvalues zero. Since det$A$ is equal to the product of eigenvalue, det$A=0$.
(b) is not true. Take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$ and $B = beginbmatrix 0 & 0 \ 0& 0 endbmatrix$
Both have the same characteristics polynomial but $A$ and $B$ are not similar.
(c) is not true in general. It is true when $A$ is a diagonalizable matrix having eigenvalues $lambda_1, lambda_2,...,lambda_n$.
answered Jul 30 at 13:44
Amit
531411
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
(a) is true for a nilpotent matrix has all eigenvalues zero. Since det$A$ is equal to the product of eigenvalue, det$A=0$.
(b) is not true. Take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$ and $B = beginbmatrix 0 & 0 \ 0& 0 endbmatrix$
Both have the same characteristics polynomial but $A$ and $B$ are not similar.
(c) is not true in general. It is true when $A$ is a diagonalizable matrix having eigenvalues $lambda_1, lambda_2,...,lambda_n$.
answered Jul 30 at 13:44
Amit
531411
add a comment |Â
up vote
2
down vote
accepted
(a) is true for a nilpotent matrix has all eigenvalues zero. Since det$A$ is equal to the product of eigenvalue, det$A=0$.
(b) is not true. Take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$ and $B = beginbmatrix 0 & 0 \ 0& 0 endbmatrix$
Both have the same characteristics polynomial but $A$ and $B$ are not similar.
(c) is not true in general. It is true when $A$ is a diagonalizable matrix having eigenvalues $lambda_1, lambda_2,...,lambda_n$.
answered Jul 30 at 13:44
Amit
531411
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
(a) is true for a nilpotent matrix has all eigenvalues zero. Since det$A$ is equal to the product of eigenvalue, det$A=0$.
(b) is not true. Take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$ and $B = beginbmatrix 0 & 0 \ 0& 0 endbmatrix$
Both have the same characteristics polynomial but $A$ and $B$ are not similar.
(c) is not true in general. It is true when $A$ is a diagonalizable matrix having eigenvalues $lambda_1, lambda_2,...,lambda_n$.
answered Jul 30 at 13:44
Amit
531411
(a) is true for a nilpotent matrix has all eigenvalues zero. Since det$A$ is equal to the product of eigenvalue, det$A=0$.
(b) is not true. Take $A = beginbmatrix 0 & 1 \ 0& 0 endbmatrix$ and $B = beginbmatrix 0 & 0 \ 0& 0 endbmatrix$
Both have the same characteristics polynomial but $A$ and $B$ are not similar.
(c) is not true in general. It is true when $A$ is a diagonalizable matrix having eigenvalues $lambda_1, lambda_2,...,lambda_n$.
answered Jul 30 at 13:44
Amit
531411
answered Jul 30 at 13:44
Amit
531411
answered Jul 30 at 13:44
Amit
531411
answered Jul 30 at 13:44
Amit
531411
531411
add a comment |Â
add a comment |Â
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2
a) is right, b) and c) are wrong. You found a counterexample for b). For c), you might find a counterexample. There is no counterexample for a), you should show that the statement is true for any $ntimes n$ matrix
– Babelfish
Jul 30 at 11:42
1
Hint for a) : $det(AB)=det(A) det(B)$.
– nicomezi
Jul 30 at 11:48