Problem is:

Let $\mathbf{A}$ be $n\times n$ matrix with real entries such that $\mathbf{A}^{2} = \mathbf{A}$. If $\mathbf{I}$ denotes the identity matrix, then how do I prove the result: $$\operatorname{Rank}(\mathbf{A}-\mathbf{I})=\operatorname{Nullity}(\mathbf{A})$$

I have trivially shown by taking $\mathbf{A}=\mathbf{I}$, but in general I do not know how to prove this, thanks for help.

  • $\begingroup$ Was it in JAM last year? $\endgroup$ – Swapnil Tripathi Dec 22 '14 at 20:34

Hint: $$ \mathbf{A}^{2} = \mathbf{A} \quad \Leftrightarrow \quad \mathbf{A}^{2} - \mathbf{A} = \mathbf{0}_{n\times n} \quad \Leftrightarrow \quad \mathbf{A} \left(\mathbf{A}-\mathbf{I}\right) = \mathbf{0}_{n\times n}. $$ Let $\mathbf{B} = (\mathbf{A} - \mathbf{I})$. The above implies that every vector in the range of $\mathbf{B}$ lies in the nullspace of $\mathbf{A}$. Can you take it from there?

Edit: I am expanding the answer because it seems we need to clarify a few things. Why is it that the range of $\mathbf{B}$ lies in the nullspace of $\mathbf{A}$? Let $\mathbf{w}$ be any vector in the range of $\mathbf{B}$, $\mathcal{R}(\mathbf{B})$. This means that we can write $\mathbf{w}$ as a linear combination of the columns of $\mathbf{B}$, i.e., there exists a $\mathbf{c}$ such that $\mathbf{w}=\mathbf{B}\mathbf{c}$. But, $$ \mathbf{A}\mathbf{w} =\mathbf{A}\mathbf{B}\mathbf{c} = \mathbf{0}_{n\times n} \mathbf{c} = \mathbf{0}, $$ which implies that $\mathbf{w}$ belongs to the nullspace of $\mathbf{A}$, $\mathcal{N}(\mathbf{A})$. Hence, we have shown that $$ \mathbf{w} \in \mathcal{R}(\mathbf{B}) \quad \Rightarrow \quad \mathbf{w} \in \mathcal{N}(\mathbf{A}), $$ or equivalently, $$ \mathcal{R}(\mathbf{B}) \subseteq \mathcal{N}(\mathbf{A}). $$ The above implies that $$ \text{nullity}(\mathbf{A}) \ge \text{rank}(\mathbf{B})=\text{rank}(\mathbf{A}-\mathbf{I}). $$ But we are not quite done yet, since we want to show that the two sides are actually equal.

I am assuming that you should be familiar with the Rank-Nullity Theorem. How can you use that to show the other side of the inequality.

Edit 2: (This is the point where hints start looking a lot like an answer!) The Rank-Nullity Theorem theorem says that $$ n = \text{rank}(\mathbf{A}) + \text{nullity}(\mathbf{A}) \quad \Rightarrow \quad \text{nullity}(\mathbf{A}) = n - \text{rank}(\mathbf{A}). $$ What can you say about $\text{rank}(\mathbf{A} - \mathbf{I})$ with respect to $\text{rank}(\mathbf{A})$? How can you get it into the picture? (Remember! we are trying to show that $\text{nullity}(\mathbf{A}) \le \text{rank}(\mathbf{A} - \mathbf{I})$ because we have already shown the reverse inequality.)

Edit 3: We have already shown that $\text{nullity}(\mathbf{A}) \ge \text{rank}(\mathbf{A}-\mathbf{I})$. To show that the relation holds with equality, it suffices to show that $$ \text{nullity}(\mathbf{A}) \le \text{rank}(\mathbf{A}-\mathbf{I}). $$ We have already seen that $$ \text{nullity}(\mathbf{A}) = n - \text{rank}(\mathbf{A}). $$ Hence, to show the desired result, it suffices to show that $$ n - \text{rank}(\mathbf{A}) \le \text{rank}(\mathbf{A}−\mathbf{I}). $$ One way to argue about this, possibly not the best, is the following. Let $\mathbf{B} = \mathbf{A} - \mathbf{I}$. By the rank-nullity theorem, we know that $$ \text{rank}(\mathbf{B}) + \text{nullity}(\mathbf{B}) = n. $$ Now, the nullspace of $\mathbf{B}$, $\mathcal{N}(\mathbf{B})$, consists of all vectors $\mathbf{w}$ such that $\mathbf{B}\mathbf{w}=\mathbf{0}$, or equivalently: $$ \mathbf{w} \in \mathcal{N}(\mathbf{B}) \quad \Leftrightarrow \quad (\mathbf{A}-\mathbf{I})\mathbf{w} = \mathbf{0} \quad \Leftrightarrow \quad \mathbf{A}\mathbf{w} = \mathbf{w}. $$ These vectors $\mathbf{w}$ are clearly just a subset of the range of $\mathbf{A}$. In other words, $$ \mathbf{w} \in \mathcal{N}(\mathbf{B}) \quad \Rightarrow \quad \mathcal{R}(\mathbf{A}), $$ and in turn, $ \mathbf{w} \in \mathcal{N}(\mathbf{B}) \subseteq \mathcal{R}(\mathbf{A}). $ We conclude that $\text{nullity}(\mathbf{B}) \le \text{rank}(\mathbf{A})$ and finally, $$ \text{rank}(\mathbf{B}) = n-\text{nullity}(\mathbf{B}) \ge n-\text{rank}(\mathbf{A}), $$ which is the desired inequality $$ \text{rank}(\mathbf{A}-\mathbf{I}) \ge n-\text{rank}(\mathbf{A}). $$

  • $\begingroup$ thanks foranswer , i have almost got it , but how is that AB=O implies range of b lies in null spzce of A $\endgroup$ – godonichia Dec 22 '14 at 20:43
  • $\begingroup$ Let $\mathbf{b}_{1}, \dots, \mathbf{b}_{m}$ be the columns of $\mathbf{B}$. $\mathbf{A}\mathbf{B}=\mathbf{0}_{n \times n}$ implies that $\mathbf{A}\mathbf{b}_{i}= \mathbf{0}$, $\forall i$. So $\mathbf{b}_{i} \in \mathcal{N}(\mathbf{A})$. Do you agree? What about all vectors that are linear combinations of the $\mathbf{b}_{i}$s? $\endgroup$ – megas Dec 22 '14 at 20:46
  • $\begingroup$ Well still not certain about it $\endgroup$ – godonichia Dec 22 '14 at 20:49
  • $\begingroup$ Tell me what confuses you and we can take if from there. What is the part you feel uncertain about? $\endgroup$ – megas Dec 22 '14 at 20:53
  • $\begingroup$ For a LT, i know T (x) = Ax , for kernel setting Ax=0 and x lies in null space of A , Are you utilizing this fact but i am not certain $\endgroup$ – godonichia Dec 22 '14 at 20:56

here is what i think is a shorter proof of $$ A^2 = A \implies ker(A) = image(A-I) $$

proof: write $A^2 = A$ as $A(A-I) = 0$ this means that every column of $A-I$ is in $ker(A),$ therefore $image(A-I) \subset ker(A).$

to show the other direction, pick $x \in ker(A),$ so that $Ax = 0.$ then $x = x - Ax = (I-A)x$ that is $x \in image(A-I)$ that proves $\ker(A) \subset image(A-I)$

now the result $$rank(A-I) = dim(ker(A)) = nullity(A)$$ follows.

p.s.: this seems too good to be true. i hope i did not make any logical errors.

  • $\begingroup$ I think you meant every column of $A-I$ is in $ker(A)$. $\endgroup$ – megas Dec 23 '14 at 23:37
  • $\begingroup$ @m.a., thanks. yes that is a typo, fixed now. $\endgroup$ – abel Dec 23 '14 at 23:39

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