The discussion revolves around proving the uniqueness of the identity matrix in linear algebra. The original poster is exploring the implications of the identity matrix's properties, particularly focusing on the equation \( I_1 * A = A \) and \( I_2 * A = A \) for two identity matrices \( I_1 \) and \( I_2 \) acting on an \( n \times p \) matrix \( A \). There are concerns about how to conclude \( I_1 = I_2 \) without additional assumptions about the matrix \( A \).
The discussion is ongoing, with various participants exploring different angles on the problem. Some have offered insights into the implications of assuming \( A \) is invertible, while others emphasize the importance of defining the context for \( A \). There is no explicit consensus yet, but the conversation is delving into the foundational aspects of the problem.
Participants note that the uniqueness of the identity matrix may depend on the properties of the matrix \( A \), particularly in cases where \( A \) could be the zero matrix or not invertible. There is a recognition that additional assumptions may be necessary to clarify the proof's validity.
Ok, but just how are the two different? I'm not aware of any particular meaning of## [I_1- I_2].##.Mark44 said:I think you mean ##A(I_1 - I_2) = 0##.
WWGD said:Ok, but just how are the two different?
You wrote ##A(I_1 - A_2)## but I thought you meant ##A(I_1 - I_2)##.WWGD said:I'm not aware of any particular meaning of ##[I1−I2]##..
True, if we assume ##A## is invertible. Otherwise, its right kernel isn't trivial, i.e. , it's not just ##\{0\}##, so in that case of ##A## being singular, it doesn't follow that ##I_1=I_2##. But if ##A## is nonsingular, then you're right.Mark44 said:You wrote ##A(I_1 - A_2)## but I thought you meant ##A(I_1 - I_2)##.
If ##A(I_1 - I_2) = 0##, then using determinants you can deduce that ##I_1 = I_2##, which was the whole point in being able to say that the identity matrix must be unique.
From the beginning of this thread, it must be stated that the identities, ##I_1## and ##I_2##, work as identities for every possible ##A##. Otherwise, the conclusion that ##I_1=I_2## may be false.WWGD said:True, if we assume ##A## is invertible. Otherwise, its right kernel isn't trivial, i.e. , it's not just ##\{0\}##, so in that case of ##A## being singular, it doesn't follow that ##I_1=I_2##. But if ##A## is nonsingular, then you're right.
Ok, I guess I lost track of the " Initial Conditions". Using the Determinant alone ( assuming ##A## is square:FactChecker said:From the beginning of this thread, it must be stated that the identities, ##I_1## and ##I_2##, work as identities for every possible ##A##. Otherwise, the conclusion that ##I_1=I_2## may be false.
For the OP question it should not matter; uniqueness of the identity should be provable for any group (or at least any group that doesn't have some other weirdness like zero divisors). @FactChecker seems to me to have come up with the simplest derivation.PeroK said:I completely missed that we were talking about rectangular matrices!
The set of ##m \times n## matrices (where ##m \ne n##) do not form a multiplicative group. The usual matrix multiplication is not even well defined.PeterDonis said:For the OP question it should not matter; uniqueness of the identity should be provable for any group (or at least any group that doesn't have some other weirdness like zero divisors). @FactChecker seems to me to have come up with the simplest derivation.
This refers to multiplicative groups.PeroK said:The set of ##m \times n## matrices (where ##m \ne n##) do not form a multiplicative group. The usual matrix multiplication is not even well defined.
And in such cases there is no "identity" at all. So the OP question isn't even applicable.PeroK said:The set of ##m \times n## matrices (where ##m \ne n##) do not form a multiplicative group. The usual matrix multiplication is not even well defined.
An ##n \times n## matrix is a linear transformation on the set of ##n \times m## matrices (under left multiplication). The question related to the identity transformation in this context.PeterDonis said:And in such cases there is no "identity" at all. So the OP question isn't even applicable.