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To prove this result, we will start by proving a simpler one. Replacing A and C with the identity matrix I, we obtain another identity which is a bit simpler: (+) = (+). To recover the original equation from this reduced identity, replace by and by .
In linear algebra, the identity matrix of size is the square matrix with ones on the main diagonal and zeros elsewhere. It has unique properties, for example when the identity matrix represents a geometric transformation, the object remains unchanged by the transformation. In other contexts, it is analogous to multiplying by the number 1.
To prove that the backward direction + + is invertible with inverse given as above) is true, we verify the properties of the inverse. A matrix (in this case the right-hand side of the Sherman–Morrison formula) is the inverse of a matrix (in this case +) if and only if = =.
Proof. Laplace's formula for ... evaluated at the identity matrix, is equal to the trace. ... is a linear operator that maps an n × n matrix to a real number. Proof.
1 Proof. 2 Applications. 3 References. Toggle the table of contents. ... It is the determinant analogue of the Woodbury matrix identity for matrix inverses. Proof
The proof of this identity is the same as the standard power-series argument for the corresponding identity for the exponential of real numbers. That is to say, as long as X {\displaystyle X} and Y {\displaystyle Y} commute , it makes no difference to the argument whether X {\displaystyle X} and Y {\displaystyle Y} are numbers or matrices.
The fact that the Pauli matrices, along with the identity matrix I, form an orthogonal basis for the Hilbert space of all 2 × 2 complex matrices , over , means that we can express any 2 × 2 complex matrix M as = + where c is a complex number, and a is a 3-component, complex vector.
The following identity (Campbell 1897) leads to a special case of the Baker–Campbell–Hausdorff formula. Let G be a matrix Lie group and g its corresponding Lie algebra. Let ad X be the linear operator on g defined by ad X Y = [X,Y] = XY − YX for some fixed X ∈ g. (The adjoint endomorphism encountered above.)