Need a 2 x 2 matrix who's square is zero

Irks23
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Homework Statement



If the product of two matrices is zero, it is not necessary that either one be zero. In particular, show that a 2 x 2 matrix who's square is zero may be written in two parameters a and b, and find the general form of the matrix. Verify that it's determinant is 0.

The Attempt at a Solution



So I tried to find a 2 x 2 matrix who's square is 0 and determinant is 0. The best I could come up with was:
A iA
iA A
which squares to:
0 2iA^2
2iA^2 0
but only the real part is 0, not the imaginary part, and the determinant is not zero. I'm not seeing at this point how any matrix's square could be zero except the 0 matrix, am I interpreting the question wrong?
 
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If A2=0, what can you say about its eigenvalues?

Then if it's, say, upper triangular...
 
A B
C D
if its upper triangular, in ABCD format above with C being 0, when you square the matrix, you get A^2+BC=A^2 for the first element, which only is 0 if A is zero...

I really don't know what A^2 = 0 tells about the eigenvalues...
 
I can find 3 families, but I don't think you can all write them in one general form

since a=0 didn't work, try b= or c=0 to get two of the families

If the matrix has all nonzero entries, you can do something with
ab + bd = 0 and ac + cd = 0
 
got it, thank you
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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