Values to whom the matrix is diagonalizable and invertible
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I've got this matrix A=$$begin{bmatrix} 2&0&3\0&L&0\4&0&0end{bmatrix}$$
And I must to find the values for what this matrix is diagonalizable and they are 5 and 1 with geometrical and algebrica molteplicity equal 2. And for all of them this matrix is diagonalizable. But I can't understand the second part of this exercise. I must to find the values for what this matrix is invertible and 5|A^-1|>=tr(A)-6. So I find the invertible matrix because the determinant it's different from 0. And this matrix is $$begin{bmatrix} 8&0&4\0&5&0\-3&0&8end{bmatrix}$$ (I used the value L=1 in this case. And |A^-1|=0*0-5*4=-20 so 5|A^-1|=-100 and tr(A)-6=2+1+4=7-6=1 and here I think to do something wrong, and I can't find the correct solution for the second part of this exercise. I try to search on web.
matrix-equations matrix-calculus
asked Dec 16 '18 at 20:20
CiaoCiao
94
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add a comment |
$begingroup$
I've got this matrix A=$$begin{bmatrix} 2&0&3\0&L&0\4&0&0end{bmatrix}$$
And I must to find the values for what this matrix is diagonalizable and they are 5 and 1 with geometrical and algebrica molteplicity equal 2. And for all of them this matrix is diagonalizable. But I can't understand the second part of this exercise. I must to find the values for what this matrix is invertible and 5|A^-1|>=tr(A)-6. So I find the invertible matrix because the determinant it's different from 0. And this matrix is $$begin{bmatrix} 8&0&4\0&5&0\-3&0&8end{bmatrix}$$ (I used the value L=1 in this case. And |A^-1|=0*0-5*4=-20 so 5|A^-1|=-100 and tr(A)-6=2+1+4=7-6=1 and here I think to do something wrong, and I can't find the correct solution for the second part of this exercise. I try to search on web.
matrix-equations matrix-calculus
asked Dec 16 '18 at 20:20
CiaoCiao
94
$endgroup$
add a comment |
$begingroup$
I've got this matrix A=$$begin{bmatrix} 2&0&3\0&L&0\4&0&0end{bmatrix}$$
And I must to find the values for what this matrix is diagonalizable and they are 5 and 1 with geometrical and algebrica molteplicity equal 2. And for all of them this matrix is diagonalizable. But I can't understand the second part of this exercise. I must to find the values for what this matrix is invertible and 5|A^-1|>=tr(A)-6. So I find the invertible matrix because the determinant it's different from 0. And this matrix is $$begin{bmatrix} 8&0&4\0&5&0\-3&0&8end{bmatrix}$$ (I used the value L=1 in this case. And |A^-1|=0*0-5*4=-20 so 5|A^-1|=-100 and tr(A)-6=2+1+4=7-6=1 and here I think to do something wrong, and I can't find the correct solution for the second part of this exercise. I try to search on web.
matrix-equations matrix-calculus
asked Dec 16 '18 at 20:20
CiaoCiao
94
$endgroup$
I've got this matrix A=$$begin{bmatrix} 2&0&3\0&L&0\4&0&0end{bmatrix}$$
And I must to find the values for what this matrix is diagonalizable and they are 5 and 1 with geometrical and algebrica molteplicity equal 2. And for all of them this matrix is diagonalizable. But I can't understand the second part of this exercise. I must to find the values for what this matrix is invertible and 5|A^-1|>=tr(A)-6. So I find the invertible matrix because the determinant it's different from 0. And this matrix is $$begin{bmatrix} 8&0&4\0&5&0\-3&0&8end{bmatrix}$$ (I used the value L=1 in this case. And |A^-1|=0*0-5*4=-20 so 5|A^-1|=-100 and tr(A)-6=2+1+4=7-6=1 and here I think to do something wrong, and I can't find the correct solution for the second part of this exercise. I try to search on web.
matrix-equations matrix-calculus
matrix-equations matrix-calculus
asked Dec 16 '18 at 20:20
CiaoCiao
94
asked Dec 16 '18 at 20:20
CiaoCiao
94
edited Dec 16 '18 at 20:39
Ciao
asked Dec 16 '18 at 20:20
CiaoCiao
94
asked Dec 16 '18 at 20:20
CiaoCiao
94
asked Dec 16 '18 at 20:20
CiaoCiao
94
94
add a comment |
add a comment |
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$begingroup$
Your matrix has characteristic polynomial $P(lambda) =lambda^3 - (2+L) lambda^2 - (12 + 2 L) lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52, left( {L}^{2}-2,L-12 right) ^{2}$, so unless $L$ is one of the roots of that ($1 pm sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
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$begingroup$
Your matrix has characteristic polynomial $P(lambda) =lambda^3 - (2+L) lambda^2 - (12 + 2 L) lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52, left( {L}^{2}-2,L-12 right) ^{2}$, so unless $L$ is one of the roots of that ($1 pm sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
Your matrix has characteristic polynomial $P(lambda) =lambda^3 - (2+L) lambda^2 - (12 + 2 L) lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52, left( {L}^{2}-2,L-12 right) ^{2}$, so unless $L$ is one of the roots of that ($1 pm sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
Your matrix has characteristic polynomial $P(lambda) =lambda^3 - (2+L) lambda^2 - (12 + 2 L) lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52, left( {L}^{2}-2,L-12 right) ^{2}$, so unless $L$ is one of the roots of that ($1 pm sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
$endgroup$
Your matrix has characteristic polynomial $P(lambda) =lambda^3 - (2+L) lambda^2 - (12 + 2 L) lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52, left( {L}^{2}-2,L-12 right) ^{2}$, so unless $L$ is one of the roots of that ($1 pm sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
edited Dec 16 '18 at 20:45
Rebellos
14.8k31248
14.8k31248
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
answered Dec 16 '18 at 20:45
Robert IsraelRobert Israel
323k23213467
323k23213467
add a comment |
add a comment |
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