A problem in my textbook proceeds as follows:
Find $\det(A)$ given that $A$ has $p(\lambda)$ as its characteristic polynomial
a) $p(\lambda)=\lambda^3-2\lambda^2+\lambda+5$
b) $p(\lambda)=\lambda^4-\lambda^3+7$
What I did was:
a) Since $\det(\lambda I-A)=\lambda^3-2\lambda^2+\lambda+5$, then $\det(-A)=5$. Hence, $\det(A)=-5$. b) Since $\det(\lambda I-A)=\lambda^4-2\lambda^3+7$, then $\det(-A)=7$. Hence, $\det(A)=7$.
Am I correct here?
$\endgroup$ 43 Answers
$\begingroup$You are almost correct but not quite.
Hint. If $$A=\pmatrix{1&2\cr 1&1\cr}$$ write down $-A$ and find the determinants of $A$ and $-A$, doing all the working and not attempting to use any short cuts. Then repeat for $$A=\pmatrix{1&2&3\cr 1&1&2\cr -1&0&1\cr}\ .$$ What do you notice?
$\endgroup$ 7 $\begingroup$In general when the characteristic polynomial is written in simplest form the constant term of the polynomial is equal to ${-1}^n \cdot det(A)$. Where $n$ is the dimension of the matrix.
$\endgroup$ $\begingroup$Let $A = I$.
Then $\det(\lambda I - A) = \det [(\lambda - 1) I] = (\lambda - 1)^n = \ldots + (-1)^n$, for $A$ being $n\times n$.
By your logic, we'd conclude that $\det (-A) = (-1)^n$. But it's nonsense to say that. $A$ is the identity; the determinant of $I$ does not depend on its dimensionality.
Moreover, does it make sense to say that $\det(-I) = \det I$? I don't think so.
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