Mixing
$A$ square matrix, $mathrm{rank}(A)=3,$ characteristic polynomial of A is $x^2(x-1)(x-2) Rightarrow A$ is...
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I've been trying to prove or disprove the following statement:
Let $A$ be a square matrix such that $mathrm{rank}(A)=3$. Prove or disprove that if the characteristic polynomial of $A$ is $x^2(x-1)(x-2)$, then A is diagonalizable.
So I can see that the size of $A$ must be $4times 4$. However, the fact that the rank of $A$ is $3$ made it almost impossible for me to find a (really) specific matrix, such that its characteristic polynomial is the one requested above. I couldn't find a counterexample, either.
Thanks!
linear-algebra diagonalization
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
$endgroup$
add a comment |
$begingroup$
I've been trying to prove or disprove the following statement:
Let $A$ be a square matrix such that $mathrm{rank}(A)=3$. Prove or disprove that if the characteristic polynomial of $A$ is $x^2(x-1)(x-2)$, then A is diagonalizable.
So I can see that the size of $A$ must be $4times 4$. However, the fact that the rank of $A$ is $3$ made it almost impossible for me to find a (really) specific matrix, such that its characteristic polynomial is the one requested above. I couldn't find a counterexample, either.
Thanks!
linear-algebra diagonalization
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
$endgroup$
add a comment |
$begingroup$
I've been trying to prove or disprove the following statement:
Let $A$ be a square matrix such that $mathrm{rank}(A)=3$. Prove or disprove that if the characteristic polynomial of $A$ is $x^2(x-1)(x-2)$, then A is diagonalizable.
So I can see that the size of $A$ must be $4times 4$. However, the fact that the rank of $A$ is $3$ made it almost impossible for me to find a (really) specific matrix, such that its characteristic polynomial is the one requested above. I couldn't find a counterexample, either.
Thanks!
linear-algebra diagonalization
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
$endgroup$
I've been trying to prove or disprove the following statement:
Let $A$ be a square matrix such that $mathrm{rank}(A)=3$. Prove or disprove that if the characteristic polynomial of $A$ is $x^2(x-1)(x-2)$, then A is diagonalizable.
So I can see that the size of $A$ must be $4times 4$. However, the fact that the rank of $A$ is $3$ made it almost impossible for me to find a (really) specific matrix, such that its characteristic polynomial is the one requested above. I couldn't find a counterexample, either.
Thanks!
linear-algebra diagonalization
linear-algebra diagonalization
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
asked Jan 13 at 21:08
Amit ZachAmit Zach
595
595
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The statement is false. For instance, the matrix
$$
A = pmatrix{0&1&0&0\0&0&0&0\0&0&1&0\0&0&0&2}
$$
Has the appropriate rank and eigenvalues, but fails to be diagonalizable.
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
What about
$$A=
begin{pmatrix}
0&1&0&0\
0&0&0&0\
0&0&1&0\
0&0&0&2
end{pmatrix}$$
Which is not diagonalizable as its minimal polynomial do not have only simple roots.
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
If $A$ is $ntimes n$ and diagonalizable, then $x^mmid det(A-xI)$ if and only if $operatorname{rk}Ale n-m$.
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is false. For instance, the matrix
$$
A = pmatrix{0&1&0&0\0&0&0&0\0&0&1&0\0&0&0&2}
$$
Has the appropriate rank and eigenvalues, but fails to be diagonalizable.
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
The statement is false. For instance, the matrix
$$
A = pmatrix{0&1&0&0\0&0&0&0\0&0&1&0\0&0&0&2}
$$
Has the appropriate rank and eigenvalues, but fails to be diagonalizable.
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
The statement is false. For instance, the matrix
$$
A = pmatrix{0&1&0&0\0&0&0&0\0&0&1&0\0&0&0&2}
$$
Has the appropriate rank and eigenvalues, but fails to be diagonalizable.
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
The statement is false. For instance, the matrix
$$
A = pmatrix{0&1&0&0\0&0&0&0\0&0&1&0\0&0&0&2}
$$
Has the appropriate rank and eigenvalues, but fails to be diagonalizable.
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
answered Jan 13 at 21:13
OmnomnomnomOmnomnomnom
128k791184
128k791184
add a comment |
add a comment |
$begingroup$
What about
$$A=
begin{pmatrix}
0&1&0&0\
0&0&0&0\
0&0&1&0\
0&0&0&2
end{pmatrix}$$
Which is not diagonalizable as its minimal polynomial do not have only simple roots.
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
What about
$$A=
begin{pmatrix}
0&1&0&0\
0&0&0&0\
0&0&1&0\
0&0&0&2
end{pmatrix}$$
Which is not diagonalizable as its minimal polynomial do not have only simple roots.
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
What about
$$A=
begin{pmatrix}
0&1&0&0\
0&0&0&0\
0&0&1&0\
0&0&0&2
end{pmatrix}$$
Which is not diagonalizable as its minimal polynomial do not have only simple roots.
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
What about
$$A=
begin{pmatrix}
0&1&0&0\
0&0&0&0\
0&0&1&0\
0&0&0&2
end{pmatrix}$$
Which is not diagonalizable as its minimal polynomial do not have only simple roots.
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Jan 13 at 21:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
26.8k22157
add a comment |
add a comment |
$begingroup$
If $A$ is $ntimes n$ and diagonalizable, then $x^mmid det(A-xI)$ if and only if $operatorname{rk}Ale n-m$.
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
$begingroup$
If $A$ is $ntimes n$ and diagonalizable, then $x^mmid det(A-xI)$ if and only if $operatorname{rk}Ale n-m$.
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
$begingroup$
If $A$ is $ntimes n$ and diagonalizable, then $x^mmid det(A-xI)$ if and only if $operatorname{rk}Ale n-m$.
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
If $A$ is $ntimes n$ and diagonalizable, then $x^mmid det(A-xI)$ if and only if $operatorname{rk}Ale n-m$.
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
edited Jan 13 at 21:33
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
answered Jan 13 at 21:13
Saucy O'PathSaucy O'Path
5,9691627
5,9691627
add a comment |
add a comment |
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