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Given $A,B$ are $n \times n$ matrices, $(A+B)^2=A+B,r(A+B)=r(A)+r(B)$.
Prove: $A^2=A,B^2=B$
I don't know how to use $r(A+B)=r(A)+r(B)$

Robert Shore
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Doggy
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 29 '22 at 06:45
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    To begin with, we have $r(A+B)\le r(A)+r(B)-\dim\left(im(A)\cap im(B)\right)$ because $im(A+B)\subseteq im(A)+im(B)$. The rank condition therefore implies that $im(A)\cap im(B)=0$. – user1551 Oct 29 '22 at 07:31

2 Answers2

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Following what user1551 said in the comments, we have: $\operatorname{Im}(A)\cap\operatorname{Im}(B) = \{0\}$.

(You did not specify the base field, so I'll just take any field $\mathbb{K}$.)

Denote the transpose of a matrix $M$ as $M^T$.
Since $r(M) = r(M^T)$ for all matrices $M \in \mathcal{M}_n(\mathbb{K})$ (one way to envision it is that the rank of a matrix $M$ is the maximum $k$ such that there exists an invertible minor $N$ of $M$ of size $k \times k$, and then see that for all minors $N$, $N^T$ is a minor of $M^T$, as well as recall that $\det(N) = \det(N^T)$), we also have: $$r\left(A^T + B^T\right) = r\left((A+B)^T\right) = r\left(A^T\right) + r\left(B^T\right)$$

Therefore, from the same arguments as before: $\operatorname{Im}\left(A^T\right)\cap\operatorname{Im}\left(B^T\right) = \{0\}$.

Now, let $x \in \mathbb{K}^n$. We get: $$Ax + Bx = \left(A+B\right)^2x = \left(A^2 + AB + BA + B^2\right)x = A\big((A + B)x\big) + B\big((A+B)x\big)$$ Thus, from $\operatorname{Im}(A)\cap\operatorname{Im}(B) = \{0\}$, we obtain: $$\begin{cases}Ax = A\big((A + B)x\big)\\ Bx = B\big((A+B)x\big)\end{cases} \quad\Rightarrow\quad \begin{cases}A^2x = (A - AB)x\\ B^2x = (B - BA)x\end{cases}$$ Since this holds for any $x \in \mathbb{K}^n$, we can conclude that: $$\begin{cases}A^2 = (A - AB)\\ B^2 = (B - BA)\end{cases}$$ By transposing everything, this grants: $$\begin{cases}(A^2)^T = (A - AB)^T\\ (B^2)^T = (B - BA)^T\end{cases} \quad\Rightarrow\quad \begin{cases}(A^T)^2 = A^T - B^TA^T\\ (B^T)^2 = B^T - A^TB^T\end{cases}$$ Again, let $y \in \mathbb{K}^n$. We have: $$\begin{cases}A^T(A^Ty) = A^Ty + B^T(-A^Ty)\\ B^T(B^Ty) = B^Ty + A^T(-B^Ty)\end{cases}$$ Hence, this time with $\operatorname{Im}\left(A^T\right)\cap\operatorname{Im}\left(B^T\right) = \{0\}$: $$\begin{cases}A^T(A^Ty) &= A^Ty \\ 0 &= B^T(-A^Ty)\\ B^T(B^Ty) &= B^Ty \\ 0 &= A^T(-B^Ty)\end{cases}$$ Due to this holding for all $y \in \mathbb{K}^n$, the first and third lines give us $(A^T)^2 = A^T$ and $(B^T)^2 = B^T$, therefore: $A^2 = A$ and $B^2 = B$, and we have finished. (The second and fourth lines give more precision to the result: the image of $A$ is a subspace of the kernel of $B$ and vice-versa).

Bruno B
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  • Thank you so much for catching that! – Bruno B Oct 29 '22 at 10:15
  • @EwanDelanoy this should be correct now? – Bruno B Oct 29 '22 at 13:32
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    It looks correct, but it can be shortened significantly. From $(A+B)^2=(A+B)$, we obtain $A(A+B-I)=-B(A+B-I)$. Hence both sides are zero, because $\operatorname{im}(A)\cap\operatorname{im}(B)=0$. That $A(A+B-I)=0$ means that $AB=A-A^2$ and in turn, $$ A^T(I-A^T)=B^TA^T $$ However, as row rank equals column rank, the rank condition also implies that $\operatorname{im}(A^T)\cap\operatorname{im}(B^T)=0$. Thus from the displayed equality above we infer that $B^TA^T=0$, i.e., $AB=0$. Hence from $A(A+B-I)=0$ we may deduce that $A^2=A$. – user1551 Oct 29 '22 at 13:43
  • @BrunoB I agree with user1551 : I think it's correct now, +1 – Ewan Delanoy Oct 29 '22 at 13:58
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It seems like you are working on your linear algebra worksheet(very similar to mine).@Burno B 's answer is beautiful,but there's also a tricky method just using the properties of the rank of matrix.

We know that the elementary row (or column) transformation doesn't change the rank of a matrix,so we have $$ \begin{pmatrix} A&0&0\\ 0&B&0\\ 0&0&I-A-B \end{pmatrix} \sim \begin{pmatrix} A&0&0\\ 0&B&0\\ A&B&I-A-B \end{pmatrix} \sim \begin{pmatrix} A&0&A\\ 0&B&B\\ A&B&I \end{pmatrix} $$

$$ \sim \begin{pmatrix} A-A^2&-AB&0\\ -BA&B-B^2&0\\ A&B&I \end{pmatrix} \sim \begin{pmatrix} A-A^2&-AB&0\\ -BA&B-B^2&0\\ 0&0&I \end{pmatrix} $$ and $$ r(A)+r(B)+r(I-A-B)=r\begin{pmatrix} A-A^2&-AB\\ -BA&B-B^2 \end{pmatrix}+r(I)=r\begin{pmatrix} A-A^2&-AB\\ -BA&B-B^2 \end{pmatrix}+n $$ Also, $$ (A+B)^2=(A+B)\implies r(A+B)+r(I-A-B)=n $$

$$ r(A+B)=r(A)+r(B)\implies r(A)+r(B)+r(I-A-B)=n $$ which implies $$ r\begin{pmatrix} A-A^2&-AB\\ -BA&B-B^2 \end{pmatrix}=0 $$ so $$ A^2=A,B^2=B,AB=BA=0 $$

GreenDuck
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