bhuvaneshnick
Junior Member
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- Dec 18, 2014
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we know that product of two orthogonal matrix is orthogonal
so the option is B that is two matrix are singular ,isn't it?
thank you
orthogonal or singular
- Thread starter bhuvaneshnick
- Start date
we know that product of two orthogonal matrix is orthogonal
so the option is B that is two matrix are singular ,isn't it?
thank you
Yes. The answer is B. A and B are singular. As an exercise, it would be good to be able to prove this. If, you don't care to work it out for your self, you could highlight the following [between the >>> and <<<] to see a proof.
>>>
Suppose it were true that neither A nor B were the zero matrix and
A B = 0.
If A (B) were non-singular then A-1 (B-1) would exist and we could multiply through by A-1 (B-1) to get B (A) equal to zero. This contradicts the given fact that neither A nor B was the zero matrix and thus A and B are singular.
<<<
>>>
Suppose it were true that neither A nor B were the zero matrix and
A B = 0.
If A (B) were non-singular then A-1 (B-1) would exist and we could multiply through by A-1 (B-1) to get B (A) equal to zero. This contradicts the given fact that neither A nor B was the zero matrix and thus A and B are singular.
<<<
HallsofIvy
Elite Member
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- Jan 27, 2012
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- 7,763
Ishuda is correct. I think I gave an incorrect answer on another board where you posted this question, thinking that is was only necessary for A or B to be singular.
I did not read the problem correctly and was thinking "AB singular" rather than "AB= 0". If, for example, A is not singular and AB= 0 then multiplying both sides of the equation, on the left, by \(\displaystyle A^{-1}\), B= 0, a contradiction. Conversely, if B is not singular and AB= 0, then multiplying both sides of the equation, on the right, by \(\displaystyle B^{-1}\) gives A= 0, a contradiction. In order to have AB= 0, with neither A nor B equal to 0, they must both be singular.
I did not read the problem correctly and was thinking "AB singular" rather than "AB= 0". If, for example, A is not singular and AB= 0 then multiplying both sides of the equation, on the left, by \(\displaystyle A^{-1}\), B= 0, a contradiction. Conversely, if B is not singular and AB= 0, then multiplying both sides of the equation, on the right, by \(\displaystyle B^{-1}\) gives A= 0, a contradiction. In order to have AB= 0, with neither A nor B equal to 0, they must both be singular.