How to check if a matrix is diagonizable?












1














So i have this $3times 3$ matrix



$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$



I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $lambda_1 = −2$ of multiplicity $2,$ and $lambda_2 = 1$ of multiplicity $1.$



In order to ensure that the matrix is diagonalizable we need to find a basis of $mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$



$DeclareMathOperatorNul{Nul}$On the answer sheet is says that we need to find $Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $Nul(-2Id-A)$??



I do not understand this $Nul(A+2Id).$ I always thought that we should compute $Nul(-2Id-A),$ so what do i need to do ?



$Nul(-2Id-A)$ gives me a different result from $Nul(A+2Id).$



Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?



Somebody please help with a clear and simple explanation ! Thanks!






matrices diagonalization







share|cite|improve this question
























  • These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
    – user376343
    Dec 1 at 12:18
















1














So i have this $3times 3$ matrix



$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$



I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $lambda_1 = −2$ of multiplicity $2,$ and $lambda_2 = 1$ of multiplicity $1.$



In order to ensure that the matrix is diagonalizable we need to find a basis of $mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$



$DeclareMathOperatorNul{Nul}$On the answer sheet is says that we need to find $Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $Nul(-2Id-A)$??



I do not understand this $Nul(A+2Id).$ I always thought that we should compute $Nul(-2Id-A),$ so what do i need to do ?



$Nul(-2Id-A)$ gives me a different result from $Nul(A+2Id).$



Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?



Somebody please help with a clear and simple explanation ! Thanks!






matrices diagonalization







share|cite|improve this question
























  • These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
    – user376343
    Dec 1 at 12:18














1












1








1







So i have this $3times 3$ matrix



$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$



I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $lambda_1 = −2$ of multiplicity $2,$ and $lambda_2 = 1$ of multiplicity $1.$



In order to ensure that the matrix is diagonalizable we need to find a basis of $mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$



$DeclareMathOperatorNul{Nul}$On the answer sheet is says that we need to find $Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $Nul(-2Id-A)$??



I do not understand this $Nul(A+2Id).$ I always thought that we should compute $Nul(-2Id-A),$ so what do i need to do ?



$Nul(-2Id-A)$ gives me a different result from $Nul(A+2Id).$



Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?



Somebody please help with a clear and simple explanation ! Thanks!






matrices diagonalization







share|cite|improve this question















So i have this $3times 3$ matrix



$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$



I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $lambda_1 = −2$ of multiplicity $2,$ and $lambda_2 = 1$ of multiplicity $1.$



In order to ensure that the matrix is diagonalizable we need to find a basis of $mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$



$DeclareMathOperatorNul{Nul}$On the answer sheet is says that we need to find $Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $Nul(-2Id-A)$??



I do not understand this $Nul(A+2Id).$ I always thought that we should compute $Nul(-2Id-A),$ so what do i need to do ?



$Nul(-2Id-A)$ gives me a different result from $Nul(A+2Id).$



Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?



Somebody please help with a clear and simple explanation ! Thanks!






matrices diagonalization




matrices diagonalization






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edited Dec 1 at 12:26









I like Serena

3,6721718




3,6721718










asked Dec 1 at 12:10









BM97

688




688












  • These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
    – user376343
    Dec 1 at 12:18


















  • These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
    – user376343
    Dec 1 at 12:18
















These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
– user376343
Dec 1 at 12:18




These spaces are equal - multiplying a vector by -1 doesn't change the eigenspace.
– user376343
Dec 1 at 12:18










2 Answers
2






active

oldest

votes


















0














You can compute $operatorname{Nul}(A+2operatorname{Id})$ or you can compute $operatorname{Nul}(-A-2operatorname{Id})$. It makes no difference, since they are the same space.



By the way, your matrix is diagonalizable.






share|cite|improve this answer





















  • Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
    – BM97
    Dec 1 at 12:31










  • I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
    – José Carlos Santos
    Dec 1 at 12:37





















1














How did you find your roots?



Normally we find them from $det(A-lambda I)=0$. Filling in $lambda=-2$ then gives us $det(A+2I)=0$. The corresponding eigenvector(s) are in $text{Nul}(A+2I)$.



Btw, $text{Nul}(A+2I) = text{Nul}(-2I-A)$.






share|cite|improve this answer





















  • Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
    – BM97
    Dec 1 at 12:19










  • The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
    – I like Serena
    Dec 1 at 12:22













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You can compute $operatorname{Nul}(A+2operatorname{Id})$ or you can compute $operatorname{Nul}(-A-2operatorname{Id})$. It makes no difference, since they are the same space.



By the way, your matrix is diagonalizable.






share|cite|improve this answer





















  • Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
    – BM97
    Dec 1 at 12:31










  • I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
    – José Carlos Santos
    Dec 1 at 12:37


















0














You can compute $operatorname{Nul}(A+2operatorname{Id})$ or you can compute $operatorname{Nul}(-A-2operatorname{Id})$. It makes no difference, since they are the same space.



By the way, your matrix is diagonalizable.






share|cite|improve this answer





















  • Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
    – BM97
    Dec 1 at 12:31










  • I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
    – José Carlos Santos
    Dec 1 at 12:37
















0












0








0






You can compute $operatorname{Nul}(A+2operatorname{Id})$ or you can compute $operatorname{Nul}(-A-2operatorname{Id})$. It makes no difference, since they are the same space.



By the way, your matrix is diagonalizable.






share|cite|improve this answer












You can compute $operatorname{Nul}(A+2operatorname{Id})$ or you can compute $operatorname{Nul}(-A-2operatorname{Id})$. It makes no difference, since they are the same space.



By the way, your matrix is diagonalizable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 12:18









José Carlos Santos

149k22117219




149k22117219












  • Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
    – BM97
    Dec 1 at 12:31










  • I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
    – José Carlos Santos
    Dec 1 at 12:37




















  • Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
    – BM97
    Dec 1 at 12:31










  • I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
    – José Carlos Santos
    Dec 1 at 12:37


















Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
– BM97
Dec 1 at 12:31




Can you tell me what the general equation is ? I am finding so many different answers on the internet that give me different solutions
– BM97
Dec 1 at 12:31












I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
– José Carlos Santos
Dec 1 at 12:37






I don't know what you are talking about. Usually, one computes $operatorname{Nul}(A-lambdaoperatorname{Id})$ for each eigenvalue $lambda$ but, as I wrote, if you wish to compute $operatorname{Nul}(-A+lambdaoperatorname{Id})$, go ahead. It's the same space.
– José Carlos Santos
Dec 1 at 12:37













1














How did you find your roots?



Normally we find them from $det(A-lambda I)=0$. Filling in $lambda=-2$ then gives us $det(A+2I)=0$. The corresponding eigenvector(s) are in $text{Nul}(A+2I)$.



Btw, $text{Nul}(A+2I) = text{Nul}(-2I-A)$.






share|cite|improve this answer





















  • Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
    – BM97
    Dec 1 at 12:19










  • The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
    – I like Serena
    Dec 1 at 12:22


















1














How did you find your roots?



Normally we find them from $det(A-lambda I)=0$. Filling in $lambda=-2$ then gives us $det(A+2I)=0$. The corresponding eigenvector(s) are in $text{Nul}(A+2I)$.



Btw, $text{Nul}(A+2I) = text{Nul}(-2I-A)$.






share|cite|improve this answer





















  • Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
    – BM97
    Dec 1 at 12:19










  • The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
    – I like Serena
    Dec 1 at 12:22
















1












1








1






How did you find your roots?



Normally we find them from $det(A-lambda I)=0$. Filling in $lambda=-2$ then gives us $det(A+2I)=0$. The corresponding eigenvector(s) are in $text{Nul}(A+2I)$.



Btw, $text{Nul}(A+2I) = text{Nul}(-2I-A)$.






share|cite|improve this answer












How did you find your roots?



Normally we find them from $det(A-lambda I)=0$. Filling in $lambda=-2$ then gives us $det(A+2I)=0$. The corresponding eigenvector(s) are in $text{Nul}(A+2I)$.



Btw, $text{Nul}(A+2I) = text{Nul}(-2I-A)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 12:17









I like Serena

3,6721718




3,6721718












  • Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
    – BM97
    Dec 1 at 12:19










  • The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
    – I like Serena
    Dec 1 at 12:22




















  • Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
    – BM97
    Dec 1 at 12:19










  • The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
    – I like Serena
    Dec 1 at 12:22


















Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
– BM97
Dec 1 at 12:19




Nul(A+2I)=Nul(−2I−A) give me too different equations ! But how is that possible?
– BM97
Dec 1 at 12:19












The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
– I like Serena
Dec 1 at 12:22






The equation is $(A+2I)mathbf x=mathbf 0$. This is the same as $-(A+2I)mathbf x=(-2I-A)mathbf x = mathbf 0$ @BM97.
– I like Serena
Dec 1 at 12:22




















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