Lemma: If A is an nxn matrix and p(x) is any polynomial, then the eigenvalues of p(A) are p(lambda), where lambda is an eigenvalue of A. The proof is not hard, but we will try to justify the lemma with an example instead.

In[22]:=

  Clear[A];
  A={{5,-6,-6},{-1,4,2},{3,-6,-4}};
  MatrixForm[A]
  Eigenvalues[A]

Out[22]=

  5    -6   -6
  
  -1   4    2
  
  3    -6   -4

Out[23]=

  {1, 2, 2}

Let p(x)=x^3-4x^2-7

In[24]:=

  p[x_]:=x^3-4 x^2-7

p(A)=A^3-4A^2-7I

In[25]:=

  pA=A.A.A-4 A.A -7 IdentityMatrix[3];
  MatrixForm[pA]
  Eigenvalues[pA]

Out[25]=

  -30   30    30
  
  5     -25   -10
  
  -15   30    15

Out[26]=

  {-15, -15, -10}

In[27]:=

  Eigenvalues[A]
  p[Eigenvalues[A]]
  Eigenvalues[pA]

Out[27]=

  {1, 2, 2}

Out[28]=

  {-10, -15, -15}

Out[29]=

  {-15, -15, -10}

Did it work?

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