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Find the values of $a$ and $b$ such that the following matrices are similar
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Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$
I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.
linear-algebra matrices
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 12:22
Messi fifa
1718
add a comment |Â
up vote
1
down vote
favorite
Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$
I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.
linear-algebra matrices
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 12:22
Messi fifa
1718
2
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
1
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$
I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.
linear-algebra matrices
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 12:22
Messi fifa
1718
Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$
I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.
linear-algebra matrices
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
asked Jul 22 at 12:22
Messi fifa
1718
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
edited Jul 22 at 12:47
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 22 at 12:22
Messi fifa
1718
asked Jul 22 at 12:22
Messi fifa
1718
asked Jul 22 at 12:22
Messi fifa
1718
1718
2
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
1
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28
add a comment |Â
2
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
1
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28
2
2
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
1
1
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
add a comment |Â
up vote
2
down vote
Calculate the characteristic polynomials:
$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$
$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and
$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.
answered Jul 22 at 12:42
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
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 two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
add a comment |Â
up vote
4
down vote
accepted
If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
edited Jul 22 at 13:11
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
answered Jul 22 at 12:34
José Carlos Santos
113k1698177
113k1698177
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
add a comment |Â
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45
1
1
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12
add a comment |Â
up vote
2
down vote
Calculate the characteristic polynomials:
$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$
$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and
$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.
answered Jul 22 at 12:42
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
Calculate the characteristic polynomials:
$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$
$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and
$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.
answered Jul 22 at 12:42
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Calculate the characteristic polynomials:
$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$
$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and
$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.
answered Jul 22 at 12:42
mechanodroid
22.2k52041
Calculate the characteristic polynomials:
$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$
$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and
$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.
answered Jul 22 at 12:42
mechanodroid
22.2k52041
answered Jul 22 at 12:42
mechanodroid
22.2k52041
answered Jul 22 at 12:42
mechanodroid
22.2k52041
answered Jul 22 at 12:42
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
2
down vote
Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
add a comment |Â
up vote
2
down vote
Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
answered Jul 22 at 12:54
Marc van Leeuwen
84.8k499212
84.8k499212
add a comment |Â
add a comment |Â
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2
$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25
ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27
1
$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28