Combining Aniket and Mark

Algebra Level 5

$A$ and $B$ are two square matrices of the same dimensions that satisfy

$AB = A, BA = B$

Which of the following statement need not be true?

Av = 0 \Leftrightarrow Bv = 0

A^2 = A, B^2 = B

AB = BA \Leftrightarrow A = B

Av = v \Leftrightarrow Bv = v

1 solution

Mark Hennings
Feb 19, 2017

Certainly $A^2 = ABA = AB = A$ and $B^2 = BAB = BA = B$ , so both $A$ and $B$ are idempotent.

If $Av=0$ , then $Bv = BAv = 0$ . If $Bv = 0$ then $Av = ABv = 0$ .

Since $[A,B] = A - B$ , it is clear that $A$ and $B$ commute precisely when they are equal.

Thus the first three statements are certainly true. The last need not be, since the matrices $A \; = \; \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) \hspace{2cm} B \; = \; \left(\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right)$ satisfy the conditions of the question, and $A\binom{1}{0} \; = \; \binom{1}{0} \hspace{2cm} B\binom{1}{0} \neq \binom{1}{0}$

(Edit: As Mark points out, statement 5 is not true. It is true only if the eigenspaces are orthogonal, but that need not be the case.)

The observation that you made could be simplified in the following way:

(Characteristic of idempotents) Since the minimal polynomial of these matrices is $x^2 - x = 0$ does not have repeated roots, we can split $V^n = A_1 \oplus A_0$ such that given $u_a \in A_1, v_a \in A_0$ we have $A(u_a + v_a) = u_a$ . This is the split into eigenspaces. Likewise, define $V^n = B_1 \oplus B_0$ .
Since $BA = B$ , hence $B v_a = BA v_a = B 0 = 0$ so $B_0 \subset A_0$ .
Since $AB =A$ , hence $A v_b = AB v_b = A 0 = 0$ so $A_0 \subset B_0$ .
Thus $A_0 = B_0$ .
So $A_1 = B_1$ , and thus these matrices are identical.

Calvin Lin Staff - 4 years, 3 months ago

Just read your report. Let me figure out what went wrong.

Calvin Lin Staff - 4 years, 3 months ago

The fact that $A_0=B_0$ does not force $A_1=B_1$ . The eigenspaces do not have to be orthogonal, for example.

I could have expressed my solution in terms of kernels, but thought that would make it less penetrable to the usual reader.

Mark Hennings - 4 years, 3 months ago

@Mark Hennings – Ah yes thanks!

I've updated the question. How does it look now?

Calvin Lin Staff - 4 years, 3 months ago

@Calvin Lin – That's fine. I will edit my solution to match...

Mark Hennings - 4 years, 3 months ago

Nice Solution! Thanks @Calvin Lin sir , for reposting it :)

Aniket Sanghi - 4 years, 3 months ago

Combining Aniket and Mark

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