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How to prove that any idempotent matrix is diagonalizable?

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  • $\begingroup$ Do you know how the minimal polynomial relates to diagonalizability? $\endgroup$ – EuYu Dec 10 '13 at 1:41
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    $\begingroup$ Sorry, I don't know. Can you explain in some detail? $\endgroup$ – Lao-tzu Dec 10 '13 at 1:48
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    $\begingroup$ A linear operator is diagonalizable precisely when its minimal polynomial splits into distinct linear factors. This result makes it almost trivial to conclude an idempotent matrix is diagonalizable. If you do not know the result, then it gets a bit trickier. $\endgroup$ – EuYu Dec 10 '13 at 1:53
  • $\begingroup$ Oh, thank you very much! I'll learn your result. $\endgroup$ – Lao-tzu Dec 10 '13 at 1:55
  • $\begingroup$ You should be able to find the theorem in most standard linear algebra books. One reference I am quite fond of is Hoffman and Kunze. $\endgroup$ – EuYu Dec 10 '13 at 1:57
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I will give two proofs of the result, one using minimal polynomials and one without.

Proof 1: Let $A:V\rightarrow V$ be an idempotent operator on the $n$ dimensional space $V$. Suppose that the rank of the operator is $r$. Then there exists $r$ linearly independent vectors $\{\mathbf{u}_1,\ \cdots,\ \mathbf{u}_r\}$ in the image of $A$. Suppose that each $\mathbf{u}_i$ is in the image under $\mathbf{v}_i$: $$\mathbf{u}_i = A\mathbf{v}_i$$ Applying $A$ to the above gives $$A\mathbf{u}_i = \underbrace{A^2\mathbf{v}_i = A\mathbf{v}_i}_\text{idempotent} = \mathbf{u}_i$$ It follows that each $\mathbf{u}_i$ is an eigenvector under the eigenvalue $1$.

On the other hand, we also have $n-r$ vectors forming a basis for the kernel. Together, the basis for the image and the kernel forms a basis of eigenvectors for $V$, that is: $$V = \mathrm{im}(A) \oplus \ker(A)$$ Therefore $A$ is diagonalizable. $\square$

Using minimal polynomials, we can prove the result much quicker.

Proof 2: A linear operator is diagonalizable if and only if its minimal polynomial splits into distinct linear factors. If $A$ is idempotent then $$A^2 = A \implies A(A-I) =0$$ This means that $p(x) = x(x-1)$ is an annihilating polynomial for $A$. It follows that the minimal polynomial must divide $p$ and hence must split. Therefore $A$ must be diagonalizable. $\square$

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  • $\begingroup$ So fascinating proof! Thank you very much! $\endgroup$ – Lao-tzu Dec 10 '13 at 2:21
  • $\begingroup$ could you elaborate "... and hence must split"? $\endgroup$ – Flaudre Nov 28 '16 at 6:30
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    $\begingroup$ @Flaudre The minimal polynomial must divide any annihilating polynomial. Since $p(x) = x(x-1)$ is an annihilating polynomial which factors into distinct linear terms (which I abbreviate by saying it "splits"), it follows that the minimal polynomial must also factor into distinct linear terms, i.e., it splits (into distinct linear terms). $\endgroup$ – EuYu Nov 28 '16 at 7:15
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    $\begingroup$ @Flaudre Not necessarily. All you can say for certain is that it divides $x(x-1)$. It could be $x$, which is just the zero matrix. It could be $x-1$, which is the identity matrix. Or it could be $x(x-1)$, which corresponds to a matrix with both $0$ and $1$ as eigenvalues, for example: $$\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}.$$ All three cases are possible. $\endgroup$ – EuYu Nov 28 '16 at 9:28
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    $\begingroup$ @Idonotknow The decomposition $V = \mathrm{im}(A) \oplus \ker (A)$ is an explicit eigendecomposition for $A$. Every vector in $\mathrm{im}(A)$ is an eigenvector for $A$ with eigenvalue $1$. Every vector in $\ker (A)$ is an eigenvector for $A$ with eigenvalue $0$. Combining basis for $\mathrm{im}(A)$ and $\ker (A)$ gives us a basis for $V$, so we have a basis of eigenvectors for $A$, i.e., $A$ is diagonalizable. $\endgroup$ – EuYu Sep 16 '17 at 10:28
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This is quite easy. For every vector $v$ one has $A^2v=Av$, so $Av$ is in the eigenspace for $\lambda=1$. Also $A(v-Av)=0$, so $v-Av$ is in the eigenspace for $\lambda=0$. (Define the eigenspace for$~\lambda$ as $\ker(A-\lambda I)$, even in case $\lambda$ should be no eigenvalue.) So $v=Av+(v-Av)$ is in the sum of those two eigenspaces; since $v$ was arbitrary, that sum is the whole space. This means $A$ is diagonalisable (eigenvalues $0,1$ only).

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