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Are there any general relations between the eigenvalues of a matrix $M$ and those of $M^2$?

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Roughly speaking, the eigenvalues of $f(M)$ are $f(\lambda)$, where the $\lambda$ are eigenvalues of $M$. In this case $f(x) = x^2$. – copper.hat Nov 2 '12 at 18:56
Not just "roughly": it's always the case (with eigenvalues counted by algebraic multiplicity) for any polynomial, or more generally (using the holomorphic functional calculus) for any function $f$ analytic in a neighbourhood of the set of eigenvalues. – Robert Israel Nov 2 '12 at 19:43

Of course if $\lambda$ is an eigenvalue of $M$, then $\lambda^2$ is an eigenvalue of $M^2$: $$M v =\lambda v\quad\Rightarrow\quad M^2v=M\lambda v=\lambda M v =\lambda^2 v.$$

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A useful way to view an eigenspace is that the matrix $M$ just becomes multiplication on the eigenspace.

So, suppose $Mv = \lambda v$. Applying $M$ again just results in another multiplication by $\lambda$, as in $M(M v) = M (\lambda v) = \lambda M v = \lambda^2 v$. Repeating gives $M^kv = \lambda^k v$.

If $p$ is a polynomial, say $p(x) = \sum p_k x^k$, and we let $p(M) = \sum p_k M^k$, then we have $p(M)v = \sum p_k M^k v = \sum p_k \lambda^k v = p(\lambda) v$. So, if $\lambda$ is an eigenvalue of $M$, then $p(\lambda)$ is an eigenvalue of $p(M)$.

The result is true more generally, but that is out of scope here.

If $M$ happened to be invertible, then multiplying an eigenvector by $M^{-1}$ becomes dividing by $\lambda$ (which must be non-zero since I'm assuming invertibility). This follows from $M^{-1} Mv = v = M^{-1} (\lambda v)$, which gives $M^{-1} v = \frac{1}{\lambda} v$. Repeating gives $M^{-k} v = \frac{1}{\lambda^k} v$, so the formula $M^kv = \lambda^k v$ holds for all integers (assuming invertibility).

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Clearly, if $\underline M(u) = \lambda u$, then $\underline M^2(u) = \lambda^2 u$.

However, $\underline M^2$ may have eigenvectors that weren't eigenvectors of the original linear operator. Rotation operators come to mind. A 90 degree rotation has no eigenvectors*, but chaining two of these together makes all vectors in the plane of rotation eigenvectors with eigenvalue $-1$.

*Edit: in the plane of rotation.

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"Clearly" is subjective. Were it clear to the OP, I doubt he would have asked. – Austin Mohr Nov 2 '12 at 18:48
Clearly a rotation has eigenvectors. A rotation is unitary, therefore normal, hence has a full set of eigenvectors. – copper.hat Nov 2 '12 at 18:52
Clarified to say that that particular rotation has no eigenvectors in the plane of rotation. – Muphrid Nov 2 '12 at 18:53
I was childishly picking on the word 'clearly'. – copper.hat Nov 2 '12 at 18:54
What's actually going on in this case (I was going to say "really", but that may not be the appropriate word here) is that the imaginary eigenvalues become real when you square them. – Robert Israel Nov 2 '12 at 19:40

If the eigenvalues are distinct, then the square matrix $A$ is diagonalizable, namely $$A = Q^{-1}DQ.$$

Then, $$A^2 = (Q^{-1}DQ)^2 = Q^{-1}DQQ^{-1}DQ = Q^{-1}D^2Q.$$

The diagonal entries of $D^2$ are the diagonal entries of $D$, squared.

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protected by Zev Chonoles Feb 2 at 3:02

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