Eigenvalue of a Polynomial [duplicate]

Let $$A: \mathbb C^4 \to \mathbb C^4$$ be a linear operator and let $$f(x)$$ be a polynomial with complex coefficents. If $$c$$ is an eigenvalue for $$f(A)$$, does there exists a eigenvalue $$a$$ of $$A$$ such that $$f(a) = c$$?

Please, explain why this is true or false.

marked as duplicate by Arnaud D., Mostafa Ayaz, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 14 '18 at 20:21

• What have you done so far? What is the condition for $c$ to be an eigenvalue of $f(A)$? – SZN Nov 26 '18 at 19:47
• Hint: Express $A$ in Jordan normal form and then compute the diagonal elements of $f(A)$. – Connor Harris Nov 26 '18 at 20:03
The statement is true. One proof is as follows: let $$f(z) - c = (z-z_1)\cdots(z-z_d)$$ be a factorization into linear factors. Each $$z_i$$ satisfies $$f(z_i) = c$$. If $$f(A)$$ has $$c$$ as an eigenvector, then $$f(A) - cI$$ is not invertible. Applying our above factorization, this means that the matrix product $$(A - z_1 I) \cdots (A - z_d I)$$ fails to be invertible. Thus, $$(A - z_i I)$$ is non-invertible for some $$i$$. That is, $$z_i$$ must be an eigenvector of $$A$$.