Editing Cayley–Hamilton theorem (section)

=== A direct algebraic proof ===
This proof uses just the kind of objects needed to formulate the Cayley–Hamilton theorem: matrices with polynomials as entries. The matrix {{math|''t&hairsp;I''<sub>''n''</sub> − ''A''}} whose determinant is the characteristic polynomial of {{mvar|A}} is such a matrix, and since polynomials form a commutative ring, it has an [[Adjugate matrix|adjugate]]
<math display="block">B=\operatorname{adj}(tI_n-A).</math>
Then, according to the right-hand fundamental relation of the adjugate, one has
<math display="block">(t I_n - A)B = \det(t I_n - A) I_n = p(t) I_n.</math>

Since {{math|''B''}} is also a matrix with polynomials in {{math|''t''}} as entries, one can, for each {{math|''i''}}, collect the coefficients of {{math|''t''<sup>&hairsp;''i''</sup>}} in each entry to form a matrix {{math|''B''<sub>''i''</sub>}} of numbers, such that one has
<math display="block">B = \sum_{i = 0}^{n - 1} t^i B_i.</math>
(The way the entries of {{math|''B''}} are defined makes clear that no powers higher than {{math|''t''{{i sup|''n''−1}}}} occur). While this ''looks'' like a polynomial with matrices as coefficients, we shall not consider such a notion; it is just a way to write a matrix with polynomial entries as a linear combination of {{mvar|n}} constant matrices, and the coefficient {{math|''t''<sup>&hairsp;''i''</sup>}} has been written to the left of the matrix to stress this point of view.

Now, one can expand the matrix product in our equation by bilinearity:
<math display="block">\begin{align}
 p(t) I_n &= (t I_n - A)B \\
 &=(t I_n - A)\sum_{i = 0}^{n - 1} t^i B_i \\
 &=\sum_{i = 0}^{n - 1} tI_n\cdot t^i B_i - \sum_{i = 0}^{n - 1} A\cdot t^i B_i \\
 &=\sum_{i = 0}^{n - 1} t^{i + 1} B_i- \sum_{i = 0}^{n - 1} t^i AB_i \\
 &=t^n B_{n - 1} + \sum_{i = 1}^{n - 1} t^i(B_{i - 1} - AB_i) - AB_0.
\end{align}</math>

Writing
<math display="block">p(t)I_n=t^nI_n+t^{n-1}c_{n-1}I_n+\cdots+tc_1I_n+c_0I_n,</math>
one obtains an equality of two matrices with polynomial entries, written as linear combinations of constant matrices with powers of {{math|''t''}} as coefficients.

Such an equality can hold only if in any matrix position the entry that is multiplied by a given power {{math|''t''<sup>&hairsp;''i''</sup>}} is the same on both sides; it follows that the constant matrices with coefficient {{math|''t''<sup>&hairsp;''i''</sup>}} in both expressions must be equal. Writing these equations then for {{math|''i''}} from {{math|''n''}} down to 0, one finds
<math display="block">B_{n - 1} = I_n, \qquad B_{i - 1} - AB_i = c_i I_n\quad \text{for }1 \leq i \leq n-1, \qquad -A B_0 = c_0 I_n.</math>

Finally, multiply the equation of the coefficients of {{math|''t''<sup>&hairsp;''i''</sup>}} from the left by {{math|''A''<sup>''i''</sup>}}, and sum up:

<math display="block">A^n B_{n-1} + \sum\limits_{i=1}^{n-1}\left( A^i B_{i-1} - A^{i+1}B_i\right) -A B_0 = A^n+c_{n-1} A^{n-1} + \cdots + c_1A + c_0I_n.</math>

The left-hand sides form a [[telescoping sum]] and cancel completely; the right-hand sides add up to <math>p(A)</math>:
<math display="block">0 = p(A).</math>
This completes the proof.