Hello, Richard!

If A is an n-by-n matrix, and AČ = 0, does this force A = 0? . no [size=3] Code:

We have: | a b | | a b | | 0 0 |
| | | | = | |
| c d | | c d | | 0 0 |
Then: [1] aČ + bc = 0 [2] ab + bd = 0
[3] ac + cd = 0 [4] bc + dČ = 0

From [2], we have: .b(a + d) .= .0

From [3], we have: .c(a + d) .= .0

If b,c ≠ 0, then: .d = -a

From [1], we have: .aČ .= .-bc

Since aČ is positive, *b* and *c* must have opposite signs.

And *a* is the mean proportional of |b| and |c|.

Some examples: Code:

| 4 -2 | | 3 0 -1 |
| 8 -4 | | 0 0 0 |
| 9 0 -3 |