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$M in M_ntimes n(mathbbC)$ has eigenval's $2, 3, 4, 5$ and determinant $120$. Is this possible with $n > 4$?
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I know that $det(M) = prod_k lambda_k$ if $lambda_k_k=1^n$ are the eigenvalues of $M$ (complex matrix), repeated according to algebraic multiplicity. In this case, if $2, 3, 4, 5$ are the only eigenvalues of $M$, it would seem that $det(M) > 120$ when $n > 4$ ?
The question I am looking at assumes $2, 3, 4, 5$ to be the eigenvalues of $M$ and $120$ to be the value of the determinant. But it doesn't say $M$ is $4times 4$, only $ntimes n$. It's not that this is wrong, but it seems like a typo. It asks if $M$ is diagonalizable (which if I am right that $M$ must be $4times 4$ then it is). Thanks.
linear-algebra eigenvalues-eigenvectors
asked Jul 21 at 20:30
Fred
53349
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up vote
2
down vote
favorite
I know that $det(M) = prod_k lambda_k$ if $lambda_k_k=1^n$ are the eigenvalues of $M$ (complex matrix), repeated according to algebraic multiplicity. In this case, if $2, 3, 4, 5$ are the only eigenvalues of $M$, it would seem that $det(M) > 120$ when $n > 4$ ?
The question I am looking at assumes $2, 3, 4, 5$ to be the eigenvalues of $M$ and $120$ to be the value of the determinant. But it doesn't say $M$ is $4times 4$, only $ntimes n$. It's not that this is wrong, but it seems like a typo. It asks if $M$ is diagonalizable (which if I am right that $M$ must be $4times 4$ then it is). Thanks.
linear-algebra eigenvalues-eigenvectors
asked Jul 21 at 20:30
Fred
53349
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know that $det(M) = prod_k lambda_k$ if $lambda_k_k=1^n$ are the eigenvalues of $M$ (complex matrix), repeated according to algebraic multiplicity. In this case, if $2, 3, 4, 5$ are the only eigenvalues of $M$, it would seem that $det(M) > 120$ when $n > 4$ ?
The question I am looking at assumes $2, 3, 4, 5$ to be the eigenvalues of $M$ and $120$ to be the value of the determinant. But it doesn't say $M$ is $4times 4$, only $ntimes n$. It's not that this is wrong, but it seems like a typo. It asks if $M$ is diagonalizable (which if I am right that $M$ must be $4times 4$ then it is). Thanks.
linear-algebra eigenvalues-eigenvectors
asked Jul 21 at 20:30
Fred
53349
I know that $det(M) = prod_k lambda_k$ if $lambda_k_k=1^n$ are the eigenvalues of $M$ (complex matrix), repeated according to algebraic multiplicity. In this case, if $2, 3, 4, 5$ are the only eigenvalues of $M$, it would seem that $det(M) > 120$ when $n > 4$ ?
The question I am looking at assumes $2, 3, 4, 5$ to be the eigenvalues of $M$ and $120$ to be the value of the determinant. But it doesn't say $M$ is $4times 4$, only $ntimes n$. It's not that this is wrong, but it seems like a typo. It asks if $M$ is diagonalizable (which if I am right that $M$ must be $4times 4$ then it is). Thanks.
linear-algebra eigenvalues-eigenvectors
asked Jul 21 at 20:30
Fred
53349
asked Jul 21 at 20:30
Fred
53349
asked Jul 21 at 20:30
Fred
53349
asked Jul 21 at 20:30
Fred
53349
53349
add a comment |Â
add a comment |Â
1 Answer
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As you observed if $n >4$ then $det(M) neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_lambda$ the algebraic multiplicity of $lambda$. Then $n_lambda geq 1$. You then have
$$120=det(M)= 2^lambda_2cdot 3^lambda_3cdot 4^lambda_4cdot 5^lambda_5 = 120cdot 2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1$$
and
$$2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1 geq 1$$
since $lambda_j-1 geq 0$. You can only get equality if $lambda_j=1$ for all $j=2,3,4,5$.
answered Jul 21 at 20:39
N. S.
97.7k5105197
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As you observed if $n >4$ then $det(M) neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_lambda$ the algebraic multiplicity of $lambda$. Then $n_lambda geq 1$. You then have
$$120=det(M)= 2^lambda_2cdot 3^lambda_3cdot 4^lambda_4cdot 5^lambda_5 = 120cdot 2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1$$
and
$$2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1 geq 1$$
since $lambda_j-1 geq 0$. You can only get equality if $lambda_j=1$ for all $j=2,3,4,5$.
answered Jul 21 at 20:39
N. S.
97.7k5105197
add a comment |Â
up vote
3
down vote
accepted
As you observed if $n >4$ then $det(M) neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_lambda$ the algebraic multiplicity of $lambda$. Then $n_lambda geq 1$. You then have
$$120=det(M)= 2^lambda_2cdot 3^lambda_3cdot 4^lambda_4cdot 5^lambda_5 = 120cdot 2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1$$
and
$$2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1 geq 1$$
since $lambda_j-1 geq 0$. You can only get equality if $lambda_j=1$ for all $j=2,3,4,5$.
answered Jul 21 at 20:39
N. S.
97.7k5105197
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As you observed if $n >4$ then $det(M) neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_lambda$ the algebraic multiplicity of $lambda$. Then $n_lambda geq 1$. You then have
$$120=det(M)= 2^lambda_2cdot 3^lambda_3cdot 4^lambda_4cdot 5^lambda_5 = 120cdot 2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1$$
and
$$2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1 geq 1$$
since $lambda_j-1 geq 0$. You can only get equality if $lambda_j=1$ for all $j=2,3,4,5$.
answered Jul 21 at 20:39
N. S.
97.7k5105197
As you observed if $n >4$ then $det(M) neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_lambda$ the algebraic multiplicity of $lambda$. Then $n_lambda geq 1$. You then have
$$120=det(M)= 2^lambda_2cdot 3^lambda_3cdot 4^lambda_4cdot 5^lambda_5 = 120cdot 2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1$$
and
$$2^lambda_2-1cdot 3^lambda_3-1cdot 4^lambda_4-1cdot 5^lambda_5-1 geq 1$$
since $lambda_j-1 geq 0$. You can only get equality if $lambda_j=1$ for all $j=2,3,4,5$.
answered Jul 21 at 20:39
N. S.
97.7k5105197
answered Jul 21 at 20:39
N. S.
97.7k5105197
answered Jul 21 at 20:39
N. S.
97.7k5105197
answered Jul 21 at 20:39
N. S.
97.7k5105197
97.7k5105197
add a comment |Â
add a comment |Â
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