The discussion centers on proving the uniqueness of the identity matrix in linear algebra, specifically addressing the relationship between two identity matrices, \(I_1\) and \(I_2\), acting on an \(n \times p\) matrix \(A\). The key argument presented is that if \(I_1 \cdot A = A\) and \(I_2 \cdot A = A\), then it follows that \(I_1 = I_2\) when \(A\) is invertible. The participants emphasize the importance of the conditions under which the proof holds, particularly noting that if \(A\) is the zero matrix, uniqueness fails. The conclusion drawn is that the identity matrix is unique under the assumption that \(A\) is not the zero matrix and is invertible.
PREREQUISITESStudents of linear algebra, mathematicians, and educators seeking to deepen their understanding of matrix theory and the uniqueness of identity elements in mathematical structures.
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.