If I have a matrix $A$ with two eigenvectors $x$ and $y$. What will be the eigenvectors of

$$A^2 - 3A + 4I ?$$

I know that if we take powers of $A$ then the eigenvectors remain unchanged. But I am not quite sure about the above matrix .

  • $\begingroup$ I think it is worth mentioning that even though the eigenvectors does not change while squaring $A$, the eigenvalue does change from $\lambda$ to $\lambda^2$. $\endgroup$ – Our Jul 29 '17 at 11:49

If we have $Ax=\lambda x$, then this implies $$(A^2-3A+4I)x=A^2x-3Ax+4x=\lambda^2x-3\lambda x+4x=(\lambda^2-3\lambda+4)x$$ hence $x$ is an eigenvector of $A^2-3A+4I$ to the eigenvalue $\lambda^2-3\lambda+4$

  • $\begingroup$ This might sound silly, in 2nd step, you are assuming that eigen vector of A which is x will be eigen vector of the transformed matrix . So how are you proving that x is indeed the eigen vector of transformed matrix. You can also multiply it with y and y will be the eigen vector of new eigen value $\endgroup$ – Zephyr Jul 28 '17 at 16:37
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    $\begingroup$ $A^2x=A\cdot Ax=A\cdot \lambda x=\lambda\cdot Ax=\lambda^2x$ $\endgroup$ – Peter Jul 28 '17 at 16:39
  • $\begingroup$ I know that. I am asking in 2nd step you are multiplying (A^2 - 3A + 4I) with x so you are assuming that x which is the eigen vector of A is the eigen vector of this transformed matrix. You can also multiply it with z right? $\endgroup$ – Zephyr Jul 28 '17 at 16:41
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    $\begingroup$ We want to prove that $x$ is also an eigenvector of $A^2-3A+4I$, so I start with $(A^2-3A+4I)x$. $\endgroup$ – Peter Jul 28 '17 at 16:45
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    $\begingroup$ They are not the same, but with $y$ you can do the same I did with $x$. So $y$ keeps an eigencevtor as well. In fact, I did not notice that there are two eigenvectors. $x$ is just ANY eigenvector and I showed that it keeps an eigenvector. But mixings like $Ax=\lambda y$, if you mean that, are of course wrong in general. $\endgroup$ – Peter Jul 28 '17 at 16:51

If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$,

$Av = \lambda v, \tag{1}$

and $p(A)$ is a polynomial in $A$,

$p(A) = \sum_0^n p_i A^i, \tag{2}$

then $p(\lambda)$ is an eigenvalue of $p(A)$, also with eigenvector $v$, since

$p(A)v = (\sum_0^n p_i A^i)v = \sum_0^n p_i A^iv = \sum_0^n p_i \lambda^i v = (\sum_0^n p_i \lambda^i)v = p(\lambda)v; \tag{3}$

thus if

$p(A) = A^2 -3A + 4I, \tag{4}$


$p(A)v = (\lambda^2 - 3\lambda + 4)v. \tag{5}$

If $x$ and $y$ are two eigenvectors of $A$, the above indicates they will also be eigenvectors of $p(A)$.

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    $\begingroup$ May I suggest adding $=p(\lambda)v$ as a last step of (3) to make the conclusion more explicit? $\endgroup$ – Joonas Ilmavirta Jul 28 '17 at 20:56
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    $\begingroup$ @JoonasIlmavirta: sure, why not, consider it done! $\endgroup$ – Robert Lewis Jul 28 '17 at 21:11

Evaluation of polynomials of a matrix is compatible with change of basis, in other words taking a polynomial of $A$ and then doing a change of basis has the same effect as doing the change of basis directly on$~A$ and then evaluating the same polynomial of the resulting matrix. You can check this by explicit computation using the formula for change of basis (the basic case is where the polynomial is just taking some $k$-th power), but it is more insightful to say that evaluating a polynomial of a linear operator (defined by $A$ relative to a given basis) is well defined, regardless of the coordinates used to describe such an operator, and changing bases is just a switch between two points of view of the same situation.

Then if $A$ is diagonalisable, you can do a change of basis to a situation in which the matrix is actually diagonal, and your question becomes whether a polynomial of a diagonal matrix is still a diagonal matrix, and what can be said of their diagonal entries. Since the set of diagonal matrices is closed under addition, multiplication, and scalar multiplication, the first question has an affirmative answer, and it is clear that evaluating a polynomial $P$ replaces each diagonal entry $a$ by the evaluation $P[a]$.


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