Characteristic Polynomial of an idempotent matrix.

characteristic polynomiallinear algebramatrices

I have to make an argument that the characteristic polynomial of an idempotent matrix A is precisely (maybe a constant
multiple of) $\lambda^{n-r}(1-\lambda)^{r}$, where r is the rank of A. I know that for an idempotent matrix A, the eigenvalue is either 0 or 1 and that the 1-eigenspace of A is the image of A. Furthermore, I was given the hint to use the geometric multiplicity of the eigenvalue 1 and to entertain the questions: What is the geometric multiplicity of 0 as an eigenvalue? What is the sum of these two multiplicities? Is it possible that either eigenspace is deficient?

Update: Following Steven's suggestions, I tackled the problem as follows:

  1. I found that geometric multiplicities as follows:
    $$
    \begin{align*}
    gmu(0)&=dim(ker(A-(0)I)) \\
    &=dim(ker(A)) \\
    &=n-rank(A-I) \\
    &=n-r
    \end{align*}
    $$
    and
    $$
    \begin{align*}
    gmu(1)&=dim(ker(A-(1)I)) \\
    &=dim(E_{1}) \\
    &=dim(im(A)) \\
    &=n
    \end{align*}
    $$

  2. I then use these multiplicities to argue that the minimal polynomial must be of the form:
    $$
    f_{A}(\lambda)=(\lambda)^{n-r}(1-\lambda)^{r}
    $$
    where the roots are either 0 or 1, as suggested by the answer below. Then, matrix A is diagonalizable and similar to a diagonal matrix D with r $1$'s and (n-r) $0$'s as its diagonal entries.

Best Answer

Question: "How can one approach this argument? Any help would be appreciated."

Answer: Let $A$ be an $n \times n$ matrix with $A^2=A$. Let $v \neq 0$ be an eigenvector with eigenvalue $\lambda$. It follows $Av=\lambda v$ and

$$\lambda v=Av=A^2v=A(Av)=A(\lambda v)=\lambda^2v$$

hence $\lambda^2=\lambda$ and hence $\lambda =0,1$. Hence the eigenvalues of $A$ are $0,1$. It follows the characteristic polynomial (which has degree $n$) must look as follows:

$$\chi_A(t)=t^{n-r}(t-1)^r$$

for some integer $r$, since the eigenvalues of $A$ equal the roots ot $\chi_A(t)$.

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