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Example find the eigenvalues of the matrix a 2 2 5 1

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Example : Find the eigenvalues of the matrix A = 2 2 5 - 1 . 1
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The eigenvalues are those λ for which det( A - λ I )= 0. Now det( A - λ I ) = det 2 2 5 - 1 - λ 1 0 0 1 = det 2 2 5 - 1 - λ 0 0 λ = 2 - λ 2 5 - 1 - λ = (2 - λ )( - 1 - λ ) - 10 = λ 2 - λ - 12 . The eigenvalues of A are the solutions of the quadratic equation λ 2 - λ - 12 = 0, namely λ 1 = - 3 and λ 2 = 4. As we have discussed, if det( A - λ I ) = 0 then the equation ( A - λ I ) x = b has either no solutions or infinitely many. When we take b = 0 however, it is clear by the existence of the solution x = 0 that there are infinitely many solutions (i.e., we may rule out the “no solution” case). If we continue using the matrix A from the example above, we can expect nonzero solutions x (infinitely many of them, in fact) of the equation Ax = λ x precisely when λ = - 3 or λ = 4. Let us procede to characterize such solutions. First, we work with λ = - 3. The equation Ax = λ x becomes Ax = - 3 x . Writing x = x 1 x 2 and using the matrix A from above, we have Ax = 2 2 5 - 1 x 1 x 2 = 2 x 1 + 2 x 2 5 x 1 - x 2 , while - 3 x = - 3 x 1 - 3 x 2 . Setting these equal, we get 2 x 1 + 2 x 2 5 x 1 - x 2 = - 3 x 1 - 3 x 2 2 x 1 + 2 x 2 = - 3 x 1 and 5 x 1 - x 2 = - 3 x 2 5 x 1 = - 2 x 2 x 1 = - 2 5 x 2 . This means that, while there are infinitely many nonzero solutions (solution vectors) of the equation Ax = - 3 x , they all satisfy the condition that the first entry x 1 is - 2 / 5 times the second entry x 2 . Thus all solutions of this equation can be characterized by 2 t - 5 t = t 2 - 5 , 2
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where t is any real number. The nonzero vectors x that satisfy Ax = - 3 x are called eigenvectors associated with the eigenvalue λ = - 3. One such eigenvector is u 1 = 2 - 5 and all other eigenvectors corresponding to the eigenvalue ( - 3) are simply scalar multiples of u 1 — that is, u 1 spans this set of eigenvectors. Similarly, we can find eigenvectors associated with the eigenvalue λ = 4 by solving Ax = 4 x : 2 x 1 + 2 x 2 5 x 1 - x 2 = 4 x 1 4 x 2 2 x 1 + 2 x 2 = 4 x 1 and 5 x 1 - x 2 = 4 x 2 x 1 = x 2 . Hence the set of eigenvectors associated with λ = 4 is spanned by u 2 = 1 1 .
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