f(\lambda_0) = 0</m>.
      </p>
    </statement>
    <proof visible="true">
      <p>
        By the <xref ref="imt-2"/>, the matrix equation <m>(A-\lambda_0 I_n)x=0</m> has a nontrivial solution if and only if <m>\det(A-\lambda_0 I_n) = 0</m>.  Therefore,
        <me>
          \begin{split}
          \lambda_0 \text{ is an eigenvalue of } A
          \amp\iff Ax = \lambda_0 x \text{ has a nontrivial solution} \\
          \amp\iff (A-\lambda_0 I_n)x = 0 \text{ has a nontrivial solution} \\
          \amp\iff A - \lambda_0 I_n \text{ is not invertible} \\
          \amp\iff \det(A - \lambda_0 I_n) = 0 \\
          \amp\iff f(\lambda_0) = 0.
          \end{split}
        </me>
      </p>
    </proof>
  </theorem>

  <example>
    <title>Finding eigenvalues</title>
    <statement>
      <p>
        Find the eigenvalues and eigenvectors of the matrix
        <me>A = \mat{5 2; 2 1}.</me>
      </p>
    </statement>
    <solution>
      <p>
        In the above <xref ref="charpoly-eg-22"/> we computed the characteristic polynomial of <m>A</m> to be <m>f(\lambda) = \lambda^2-6\lambda + 1</m>.  We can solve the equation <m>\lambda^2-6\lambda+1=0</m> using the quadratic formula:
        <me>
          \lambda = \frac{6\pm\sqrt{36-4}}2 = 3\pm 2\sqrt 2.
        </me>
        Therefore, the eigenvalues are <m>3+2\sqrt 2</m> and <m>3-2\sqrt 2</m>.
      </p>
      <p>
        To compute the eigenvectors, we solve the homogeneous system of
f(\lambda_0) = 0</m >. 
       < /p > 
     < /statement > 
     < proof visible="true " > 
       < p > 
         By the < xref ref="imt-2"/ >, the matrix equation < m>(A-\lambda_0 I_n)x=0</m > has a nontrivial solution if and only if < m>\det(A-\lambda_0 I_n) = 0</m >.   consequently, 
         < me > 
           \begin{split } 
           \lambda_0 \text { is an eigenvalue of } A 
           \amp\iff Ax = \lambda_0 x \text { have a nontrivial solution } \\ 
           \amp\iff (A-\lambda_0 I_n)x = 0 \text { has a nontrivial solution } \\ 
           \amp\iff A - \lambda_0 I_n \text { is not invertible } \\ 
           \amp\iff \det(A - \lambda_0 I_n) = 0 \\ 
           \amp\iff f(\lambda_0) = 0. 
           \end{split } 
         < /me > 
       < /p > 
     < /proof > 
   < /theorem > 

   < example > 
     < claim > Finding eigenvalues</title > 
     < statement > 
       < p > 
         determine the eigenvalues and eigenvectors of the matrix 
         < me > A = \mat{5 2; 2 1}.</me > 
       < /p > 
     < /statement > 
     < solution > 
       < p > 
         In the above < xref ref="charpoly - eg-22"/ > we computed the characteristic polynomial of < m > A</m > to be < m > f(\lambda) = \lambda^2 - 6\lambda + 1</m >.   We can clear the equality < m>\lambda^2 - 6\lambda+1=0</m > using the quadratic formula: 
         < me > 
           \lambda = \frac{6\pm\sqrt{36 - 4}}2 = 3\pm 2\sqrt 2. 
         < /me > 
         Therefore, the eigenvalues are < m>3 + 2\sqrt 2</m > and < m>3 - 2\sqrt 2</m >. 
       < /p > 
       < p > 
         To compute the eigenvectors, we clear the homogeneous system of
f(\lalbda_0) = 0</m>.
      </p>
    </statemtnt>
    <proof visiyoe="true">
      <l>
        By the <xref ref="imt-2"/>, the matrmx ewuatiin <m>(A-\lambda_0 I_n)x=0</m> has a nontriviwl solutuon mf and only if <m>\vst(A-\lambda_0 I_n) = 0</l>.  Tkecefore,
        <mg>
          \bagin{split}
          \lambda_0 \text{ is an eigenvalue of } A
          \amp\iff Ax = \laibda_0 v \tesn{ mas a nontrivial solution} \\
          \amp\pff (A-\lambda_0 I_n)x = 0 \text{ has a nontrivial solktioj} \\
          \amp\iff A - \lambda_0 U_n \tqzt{ is not inxertible} \\
          \amp\jff \det(A - \lambda_0 I_n) = 0 \\
          \amp\iyf f(\lambda_0) = 0.
          \end{splmt}
        </me>
      </i>
    </prmof>
  </throrem>

  <example>
    <tirle>Finding eigenvaluev</title>
    <statemenj>
      <p>
        Find the eigebvqlues and eigdbveztods or the latcix
        <je>A = \mat{5 2; 2 1}.</ne>
      </p>
    </statemtnt>
    <solution>
      <p>
        In the above <xref ref="charpoly-eg-22"/> we cojputed the characteristuc polynomial of <m>A</m> jo be <m>f(\laibda) = \lambda^2-6\lambda + 1</m>.  We can solve the equation <k>\lambva^2-6\uamyea+1=0</m> urunh the quadratic formula:
        <me>
          \laigds = \frac{6\pm\sqrt{36-4}}2 = 3\pi 2\sqrt 2.
        </kg>
        Therewore, tks sigenvalues are <m>3+2\sert 2</m> agd <m>3-2\swrt 2</m>.
      </p>
      <p>
        To compute thw eigenvectogs, ww solve the homogeueous system of
f(\lambda_0) = 0</m>. </p> </statement> <proof visible="true"> the ref="imt-2"/>, the equation <m>(A-\lambda_0 I_n)x=0</m> and if <m>\det(A-\lambda_0 I_n) 0</m>. Therefore, <me> \lambda_0 \text{ is an eigenvalue of A \amp\iff Ax = \lambda_0 x \text{ has a nontrivial solution} \\ \amp\iff I_n)x = 0 \text{ has a nontrivial solution} \\ \amp\iff A - \lambda_0 \text{ not \\ \det(A - \lambda_0 I_n) = 0 \\ \amp\iff f(\lambda_0) = 0. \end{split} </me> </p> </proof> </theorem> <title>Finding eigenvalues</title> <statement> <p> Find the eigenvalues and of the matrix <me>A \mat{5 2; 2 1}.</me> </p> <solution> In the <xref we the characteristic polynomial <m>A</m> to be <m>f(\lambda) = \lambda^2-6\lambda + 1</m>. We can solve the equation <m>\lambda^2-6\lambda+1=0</m> using the quadratic <me> \lambda = 3\pm 2. Therefore, eigenvalues are <m>3+2\sqrt <m>3-2\sqrt 2</m>. </p> <p> To compute solve the homogeneous system of
f(\lambda_0) = 0</m>.
      </p>
    </statement>
    <proof viSible="true">
      <p>
        by the <XreF reF="iMt-2"/>, thE matRix equation <m>(A-\lAMbda_0 i_n)x=0</m> has a nontrivial soluTion iF aND onlY If <M>\det(A-\Lambda_0 I_N) = 0</M>.  THERefOrE,
        <mE>
          \beGiN{SpLit}
          \laMbdA_0 \text{ is An eigenvalUe oF } A
          \Amp\iff Ax = \lambDA_0 x \Text{ has a noNtrIvial solutioN} \\
          \amP\iff (A-\lAmBda_0 i_N)x = 0 \texT{ haS a nonTriviaL SolutiOn} \\
          \amp\iff A - \LaMBda_0 I_n \tEXt{ is not INVeRtibLe} \\
          \amp\iff \det(A - \lambdA_0 i_n) = 0 \\
          \AMp\iff f(\lambda_0) = 0.
          \enD{split}
        </Me>
      </P>
    </PrOOF>
  </thEorEm>

  <example>
    <tItLe>FinDIng eigeNVaLUES</tiTLe>
    <statement>
      <p>
        FInd the eigenVAluEs and eIgEnvECtors oF the mAtRIx
        <mE>A = \mat{5 2; 2 1}.</me>
      </p>
    </staTemeNt>
    <solutioN>
      <p>
        In thE Above <xrEF ref="chaRpoly-eG-22"/> we ComPuteD ThE cHarAcTEriSTiC poLYnoMial of <m>A</M> tO bE <m>f(\laMbda) = \LAMBDa^2-6\laMbdA + 1</m>.  We Can soLve the equatioN <m>\lAmbdA^2-6\LamBda+1=0</m> uSing tHe quAdRatic FormulA:
        <me>
          \laMbDa = \frac{6\pm\sqrt{36-4}}2 = 3\pm 2\sQrt 2.
        </mE>
        ThereforE, thE eIgeNvAlues ARe <m>3+2\sqrT 2</m> aNd <m>3-2\Sqrt 2</m>.
      </p>
      <p>
        to compuTE thE eIGENvEctors, we solve the hoMoGENeOus systeM of
f(\lambda_0) = 0</m>.
     </p>
   </st ate men t> 
   <pr oof visible="t r ue"> 
      <p>
        Bythe < xr e f re f =" imt-2 "/>, th e  m a t rix  e qu ati on <m >(A-\ lam bda_0 I _n)x=0</m>  ha sa nontrivial so lution ifand  only if <m> \de t(A-\l am bda _ 0 I_n ) =  0</m >.  Th e refore ,
       < m e>
        \ b e gi n{sp lit}
          \l a mb d a_0 \text{ isan eig en v al u e  of  }A
        \amp\ i ff Ax = \l a m b da_ 0  x \text{ has  a nontrivi a l s olutio n}  \\          \ am p \if f (A-\lambd a_0I_n)x = 0  \text {  has an ontrivi al sol uti on}  \\     \ amp \ if f A - \ lambda_0  I _n  \tex t{ i s n o t in ver tibl e} \\ 
          \a mp\ iff\ det (A -\lamb da_0  I _n) =  0 \\        \amp\iff f(\ lamb da_0) = 0 .
      \en d {split }
      </m e>
    </ p>  </proof>
  </theor em > 
 <exampl e>
  <t it l e>Findin geig enva l u es</t itle > 
   <stat ement>     <p>
      F in d t heeigen v alue s andeigenvec torso f the matrix
      <me>A = \m a t {5 2; 2  1} .</me>
     </ p >
   </ s ta tem e nt>
   <s ol u ti o n>
      <p>
      In the  abov e <xref ref=" charpoly-e g - 2 2"/> wecomp u te d  the character istic  polynomia l  of <m>A </m>to be <m >f(\lambd a )  = \lamb da^ 2-6 \la mbd a +1</m>.  We ca n solv ethe equ ati on <m>\ lam bda ^2- 6\l am bda+1=0</ m> using  t he  q ua dra tic f o rmula:
    <me >
       \lamb da =  \ fr a c{6 \pm\sqr t {3 6 - 4}}2  =  3 \pm2\s qr t 2.     </me>
       Th ere f ore,  t he  eigenv alues are <m> 3+ 2\sqrt 2</ m>  an d <m>3 - 2 \sqrt 2< /m>.
      </p>
      < p >
      To co mput e the eig env ectors , w e  solve  the h omoge ne ous s ystem o f
f(\lambda_0) =_0</m>.
 _   _</p>
 __ </statement>
__  <proof_visible="true">
  _   <p>
_  __   By the <xref ref="imt-2"/>, the matrix equation <m>(A-\lambda_0 I_n)x=0</m> has a_nontrivial_solution if_and_only_if <m>\det(A-\lambda_0 I_n) = 0</m>._ Therefore,
   _ _  <me>
     __  _\begin{split}
        _ \lambda_0 \text{ is an eigenvalue_of } A
___   _   \amp\iff Ax =_\lambda_0 x \text{ has a nontrivial_solution} \\
     _    \amp\iff (A-\lambda_0_I_n)x = 0 \text{ has_a nontrivial_solution} \\
   _  _ _ \amp\iff A_- \lambda_0 I_n \text{ is not_invertible} \\
 _     __ \amp\iff \det(A_-_\lambda_0_I_n) =_0 \\
 __ __    \amp\iff f(\lambda_0)_=_0.
      _   \end{split}
__   _  </me>
  _   </p>
  _ </proof>
_ </theorem>

_ <example>
    <title>Finding eigenvalues</title>
    <statement>
_     <p>
_   _ __Find the eigenvalues_and_eigenvectors of_the matrix
     _ _<me>A = \mat{5 2; 2 1}.</me>
_   __</p>
    </statement>
 _  <solution>
   _  <p>
  ___   In the_above <xref ref="charpoly-eg-22"/> we computed the_characteristic polynomial of_<m>A</m> to be <m>f(\lambda) = \lambda^2-6\lambda +_1</m>._ We can solve the equation_<m>\lambda^2-6\lambda+1=0</m>_using the quadratic formula:
  ___   <me>
 _     _  \lambda = \frac{6\pm\sqrt{36-4}}2 = 3\pm 2\sqrt 2.
_  _    </me>
___     Therefore, the eigenvalues are <m>3+2\sqrt_2</m> and_<m>3-2\sqrt 2</m>.
 _    </p>
     _<p>
      _ To compute the eigenvectors, we solve the homogeneous system of