Matrix Eigenvalue formula $|lambda I-A |$ or $|A-lambda I|$?












0














my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.



Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?



Thanks



enter image description here






matrices eigenvalues-eigenvectors







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  • 2




    That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
    – Crostul
    Nov 29 '18 at 22:44












  • $|lambda I - A| = |A- lambda I|$.
    – amWhy
    Nov 29 '18 at 22:46










  • okay thanks, just realized that
    – JIM BOY
    Nov 29 '18 at 22:46










  • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
    – JIM BOY
    Nov 29 '18 at 22:49






  • 4




    @amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
    – Arthur
    Nov 29 '18 at 22:50


















0














my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.



Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?



Thanks



enter image description here






matrices eigenvalues-eigenvectors







share|cite|improve this question


















  • 2




    That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
    – Crostul
    Nov 29 '18 at 22:44












  • $|lambda I - A| = |A- lambda I|$.
    – amWhy
    Nov 29 '18 at 22:46










  • okay thanks, just realized that
    – JIM BOY
    Nov 29 '18 at 22:46










  • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
    – JIM BOY
    Nov 29 '18 at 22:49






  • 4




    @amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
    – Arthur
    Nov 29 '18 at 22:50
















0












0








0







my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.



Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?



Thanks



enter image description here






matrices eigenvalues-eigenvectors







share|cite|improve this question













my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.



Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?



Thanks



enter image description here






matrices eigenvalues-eigenvectors




matrices eigenvalues-eigenvectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 22:38









JIM BOYJIM BOY

356




356








  • 2




    That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
    – Crostul
    Nov 29 '18 at 22:44












  • $|lambda I - A| = |A- lambda I|$.
    – amWhy
    Nov 29 '18 at 22:46










  • okay thanks, just realized that
    – JIM BOY
    Nov 29 '18 at 22:46










  • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
    – JIM BOY
    Nov 29 '18 at 22:49






  • 4




    @amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
    – Arthur
    Nov 29 '18 at 22:50
















  • 2




    That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
    – Crostul
    Nov 29 '18 at 22:44












  • $|lambda I - A| = |A- lambda I|$.
    – amWhy
    Nov 29 '18 at 22:46










  • okay thanks, just realized that
    – JIM BOY
    Nov 29 '18 at 22:46










  • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
    – JIM BOY
    Nov 29 '18 at 22:49






  • 4




    @amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
    – Arthur
    Nov 29 '18 at 22:50










2




2




That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44






That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44














$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46




$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46












okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46




okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46












But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49




But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49




4




4




@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50






@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50












3 Answers
3






active

oldest

votes


















1














Both are actually equivalent!



Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:



$$Avec{x} = lambda vec{x}$$



Then,



$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$



Then we try to find $lambda$ such that $det(A - lambda I) = 0$.



But wait! We can do this a different way, as:



$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$



and thus we seek $lambda$ such that $det(lambda I - A) = 0$



Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.






share|cite|improve this answer





























    1














    If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.






    share|cite|improve this answer





















    • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
      – JIM BOY
      Nov 29 '18 at 22:50










    • They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
      – José Carlos Santos
      Nov 29 '18 at 22:51










    • Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
      – Eevee Trainer
      Nov 29 '18 at 22:52





















    0














    Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      Both are actually equivalent!



      Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:



      $$Avec{x} = lambda vec{x}$$



      Then,



      $$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$



      Then we try to find $lambda$ such that $det(A - lambda I) = 0$.



      But wait! We can do this a different way, as:



      $$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$



      and thus we seek $lambda$ such that $det(lambda I - A) = 0$



      Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.






      share|cite|improve this answer


























        1














        Both are actually equivalent!



        Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:



        $$Avec{x} = lambda vec{x}$$



        Then,



        $$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$



        Then we try to find $lambda$ such that $det(A - lambda I) = 0$.



        But wait! We can do this a different way, as:



        $$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$



        and thus we seek $lambda$ such that $det(lambda I - A) = 0$



        Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.






        share|cite|improve this answer
























          1












          1








          1






          Both are actually equivalent!



          Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:



          $$Avec{x} = lambda vec{x}$$



          Then,



          $$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$



          Then we try to find $lambda$ such that $det(A - lambda I) = 0$.



          But wait! We can do this a different way, as:



          $$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$



          and thus we seek $lambda$ such that $det(lambda I - A) = 0$



          Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.






          share|cite|improve this answer












          Both are actually equivalent!



          Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:



          $$Avec{x} = lambda vec{x}$$



          Then,



          $$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$



          Then we try to find $lambda$ such that $det(A - lambda I) = 0$.



          But wait! We can do this a different way, as:



          $$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$



          and thus we seek $lambda$ such that $det(lambda I - A) = 0$



          Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 22:45









          Eevee TrainerEevee Trainer

          5,3811836




          5,3811836























              1














              If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.






              share|cite|improve this answer





















              • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
                – JIM BOY
                Nov 29 '18 at 22:50










              • They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
                – José Carlos Santos
                Nov 29 '18 at 22:51










              • Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
                – Eevee Trainer
                Nov 29 '18 at 22:52


















              1














              If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.






              share|cite|improve this answer





















              • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
                – JIM BOY
                Nov 29 '18 at 22:50










              • They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
                – José Carlos Santos
                Nov 29 '18 at 22:51










              • Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
                – Eevee Trainer
                Nov 29 '18 at 22:52
















              1












              1








              1






              If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.






              share|cite|improve this answer












              If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 29 '18 at 22:48









              José Carlos SantosJosé Carlos Santos

              152k22123225




              152k22123225












              • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
                – JIM BOY
                Nov 29 '18 at 22:50










              • They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
                – José Carlos Santos
                Nov 29 '18 at 22:51










              • Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
                – Eevee Trainer
                Nov 29 '18 at 22:52




















              • But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
                – JIM BOY
                Nov 29 '18 at 22:50










              • They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
                – José Carlos Santos
                Nov 29 '18 at 22:51










              • Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
                – Eevee Trainer
                Nov 29 '18 at 22:52


















              But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
              – JIM BOY
              Nov 29 '18 at 22:50




              But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
              – JIM BOY
              Nov 29 '18 at 22:50












              They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
              – José Carlos Santos
              Nov 29 '18 at 22:51




              They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
              – José Carlos Santos
              Nov 29 '18 at 22:51












              Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
              – Eevee Trainer
              Nov 29 '18 at 22:52






              Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
              – Eevee Trainer
              Nov 29 '18 at 22:52













              0














              Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.






              share|cite|improve this answer


























                0














                Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.






                  share|cite|improve this answer












                  Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 22:54









                  FJ.marsanFJ.marsan

                  414




                  414






























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                      Ellipse (mathématiques)

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                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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                      Quarter-circle Tiles

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                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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