The question:

If the eigenvalues of $A$ are 0, 1, and 3, find the eigenvalues of $A-I$. Explain how you obtained them.

My intuition is telling me that I just subtract one from each of the eigenvalues since they are related to the diagonal, and we are subtracting one from each of the members on the diagonal.

  • 1
    $\begingroup$ I'm not so sure about that. If $A$ were diagonal, then it would be the case, but the eigenvalues do not lie on the diagonal in a general $n\times n$ matrix. EDIT: Well, as John has shown below, it actually is the case. $\endgroup$ – JoeDub Nov 27 '13 at 17:15

$x$ is an eigenvalue if $det(A - xI) = 0$. Rewrite slightly to get $$ det((A-I) - (x-1)I) = 0 $$ or $$ det((A-I) - yI) = 0. $$

So: a solution $y$ to the last equation is an eigenvalue of $A-I$. But in that case, $x = y+1$ is an eigenvalue of $A$. So your conjecture is correct.


For any matrix $B$, we have that $\lambda$ is an eigenvalue of $B$ with eigenvector $v$ if and only if, for all scalars $\mu$, $\lambda + \mu$ is an eigenvalue of $B + \mu I$ with the same eigenvector, since

$Bv = \lambda v \tag{1}$

is clearly equivalent to

$(B + \mu I)v = Bv + \mu I v = (\lambda + \mu)v. \tag{2}$

Applying this equivalence to the case at hand shows that the eigenvalues of $A$ are indeed $-1$, $0$, and $2$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!


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