Since A is normal, it is diagonalizable by a unitary matrix U, which means that

A = UDU^*,

where D is the diagonal matrix of eigenvalues of A. Since the only eigenvalues of A are 5 and 47, we have

D = diag(5, 5, …, 5, 47, 47, …, 47)

where the number of 5’s and 47’s are determined by their respective multiplicities.

Using this representation of A, we can compute A^2 as

A^2 = UD^2U^*,

and hence

A^2 – 52A + 235I = UD^2U^* – 52UDU^* + 235I.

Since U and U^* are inverses of each other, we can rewrite this as

A^2 – 52A + 235I = U(D^2 – 52D + 235I)U^*.

Now we just need to compute D^2 – 52D + 235I. Since D is diagonal, we have

D^2 = diag(5^2, 5^2, …, 5^2, 47^2, 47^2, …, 47^2),

and

52D = diag(52×5, 52×5, …, 52×5, 52×47, 52×47, …, 52×47)

= diag(260, 260, …, 260, 2444, 2444, …, 2444).

Therefore,

D^2 – 52D + 235I

= diag(5^2 – 260 + 235, 5^2 – 260 + 235, …, 5^2 – 260 + 235, 47^2 – 2444 + 235, 47^2 – 2444 + 235, …, 47^2 – 2444 + 235)

= diag(0, 0, …, 0, 0, 0, …, 0)

= 0.

Therefore, we have shown that A^2 – 52A + 235I = 0, as desired.