Matrix

Suppose A is non-singular then A^(-1) existsA^2 = A => A^(-1) (A^2 ) = A^(-1) A => A = I which is a contradictionHence A must be singular

2011-10-30 11:33:09 補充：
Doraemonpaul: I do not understand your argument and solution.
You may find many many solutions for A (your second part). If for some of them, the solution is invertible then A^(-1) A A = A^(-1) A => A=I then this soultion must be the identity matrix. det(I) = 1 consistent with what you say.

2011-10-30 11:33:16 補充：
But the question already mentions A is not equal to I, therefore no matter how many other A exists that satisfy A^2 = A, they must all be singular. Isn't it right?

2011-10-30 11:37:13 補充：
By the way, a singular (奇異) matrix means an non-invertible (非可逆) matrix. A non-singular (非奇異) matrix means an invertible (可逆) matrix.

A^2 = A

det(A^2) = det(A)

det(A) det(A) = det(A)

det(A) = 0 or det A = 1

Since det(A) can be 1, we conclude that A is not necessary to be a singular matrix. For example A = I also satisfies A^2 = A.

自由自在，你這樣理解法實錯硬，你點可以私自限死A一定是可逆？

再者，這題是屬於二次矩陣多項式方程，其性質絕對不會是如你所說這樣的簡單，如果不是，例如http://tw.knowledge.yahoo.com/question/question?qid=1511101802650我都不需要煩惱了。

2011-10-30 11:04:47 補充：

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahooknowledge/matrixequation/quadraticmatrixequ9.jpg

參考資料：
my wisdom of maths + results from http://www.wolframalpha.com/input/?i=%28%280%2C0%2Cp%29%2C%280%2Cq%2C0%29%2C%28r%2C0%2C0%29%29%5E2 + http://www.wolframalpha.com/input/?i=%28%280%2C0%2Cp%29%2C%280%2C%28pr%29%5E%281%2F2%29%2C0%29%2C%28r%2C0%2C0%29%29%5E2 + http://www.wolframalpha.com/input/?i=%28%280%2C0%2Cp%29%2C%280%2C-%28pr%29%5E%281%2F2%29%2C0%29%2C%28r%2C0%2C0%29%29%5E2