Matrix A $\in$ $\mathbb{R}^{n\times n}$ is idempotent if $A^{2} = A$. Describe the Jordan form of A.
How do I do this?
I am able to decompose a matrix to its Jordan form given that the matrix contains numeric entries. But I am having hard time figuring out the correct way to approach the solution in a general case such as this question. Thanks!
$\endgroup$3 Answers
$\begingroup$The minimal polynomial $\mu_A$ of a matrix $A$ divides each polynomial $p$ that satisfies $p(A)=0.$ So $\mu_A(t)$ must divide $t^2-t.$ Therefore, we know that the minimal polynomial of $A$ is either $\mu_A(t)=t$ or $\mu_A(t)=t-1$ or $\mu_A(t)=t^2-t.$ There are no multiple roots in the minimal polynomial, we cannot have any Jordan blocks of size greater than $1,$ and $0$ and $1$ are the only possible eigenvalues. (The multiplicity of an eigenvalue in the minimal polynomial is the size of its largest Jordan block.) This means$$ J=0 \;\; \text{or} \;\; J=I \;\; \text{or} \;\; J = \begin{pmatrix} (0)_{m\times m} & (0)_{m\times (n-m)} \\ (0)_{(n-m)\times m} & I_{n-m} \end{pmatrix} $$
$\endgroup$ $\begingroup$You know that $A^2=A$ or $(A-I)A=0$ and $A(A-I)=0$. So $A=I$ on the range $\mathcal{R}(A)$. And $A=0$ on $\mathcal{R}(A-I)$. And every vector $x$ can be written as a linear sum of $x_0$, $x_1$ with $Ax_0=0$ and $Ax_1=x_1$ because$$ x = (I-A)x+Ax \\ A(I-A)x=0,\;\; (I-A)Ax=0. $$So the Jordan form of $A$ is diagonal, and has $1$'s and $0$'s along the diagonal.
$\endgroup$ $\begingroup$Hint: If $A^2 = A$, you should first show that the Jordan form $J$ of $A$ also satisfies $J^2 = J$. Next, since every Jordan form is composed of Jordan blocks $B$ in block-diagonal form, then you should show that also $B^2 = B$. Finally, investigate Jordan blocks $B$ that satisfy $B^2 = B$.
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