The concept of inverse of a matrix is a multidimensional generalization of the concept of reciprocal of a number:
the product between a number and its reciprocal is equal to 1;
the product between a square matrix and its inverse is equal to the identity matrix.
Let us start with a definition of inverse.
Definition Let be a matrix. Its inverse, if it exists, is the matrix that satisfieswhere is the identity matrix. If exists, then we say that is invertible.
When , then and which makes clear that the definition above generalizes the notion of reciprocal of a number.
Example Consider the matrix Then, we can verify thatby carrying out the multiplication between the two matrices:
Under what conditions is a square matrix invertible? The next proposition answers this question.
Proposition A matrix is invertible if and only if it is full-rank.
Let us first prove the "if" part (full-rank implies invertibility). Denote by the columns of the identity matrix . If is full-rank, then its columns are linearly independent. This implies that any -dimensional vector can be written as a linear combination of the columns of (see the lecture on standard bases for a proof). Therefore, for , we can write as a linear combination of the columns of :where are the coefficients of the linear combination. By using the results presented in the lecture on matrix multiplication and linear combinations, these coefficients can be stacked to form a vector such thatMoreover, the column vectors can be placed side by side to form a matrix such thatThus, is invertible and We are now going to prove the "only if" part (invertibility implies full-rank). If an inverse exists, thenPost-multiplying both sides of the equation by any vector , we getorThus, any vector can be written as a linear combination of the columns of , with coefficients taken from . In other words, the columns of span the space of all vectors. If they were not linearly independent, then we would be able to eliminate some of them and obtain a set of linearly independent vectors that 1) is a basis of the space of all vectors; 2) has cardinality less than . But this is not possible because any basis of has cardinality equal to (see the lecture on standard bases). Therefore, the columns of must be linearly independent, which means that has full rank.
Proposition If the inverse of a matrix exists, then it is unique.
In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column by finding the vectors that solvethat is, by writing the vectors of the canonical basis as linear combinations of the columns of . Since the representation of a vector in terms of a basis is unique, the vectors are unique. But the latter are the columns of . Therefore, also is unique.
An important fact is that gives the identity matrix not only when it is pre-multiplied, but also when it is post-multiplied by .
Proposition Let be a matrix. If its inverse exists, it satisfies not only the conditionbut also the conditionwhere is the identity matrix.
Post-multiply both sides of the equationby , and obtainor But we also have thatNow, it might seem intuitive that equations (1) and (2) imply that Nonetheless, it needs to be proved. The proof is as follows. Equation (1) says that the columns of , on the right-hand side of the equation, can be seen as linear combinations of the columns of itself, on the left-hand side, with coefficients taken from . Equation (2) says that the columns of , on the right-hand side of the equation, can be seen as linear combinations of the columns of itself, on the left-hand side, with coefficients taken from . Since is full-rank, its columns are a basis of the space of all vectors, and by the uniqueness of the representation in terms of a basis (see the lecture entitled Basis of a linear space), the coefficients of the linear combinations must be the same, that is,
The following proposition holds.
Proposition Let and be two matrices. Then, the product is invertible if and only if and are invertible. Furthermore,
The two matrices and are invertible if and only if they are full-rank (see above). If and are full-rank, then is full-rank (see the lecture on matrix products and rank). On the contrary, if at least one of the two matrices is not full-rank, then the rank of their product is less than becauseIn other words, the product is full-rank only if and are full-rank. Furthermore, it can be easily checked that satisfies the definition of inverse of :
The next proposition shows how to compute the inverse of the transpose of a matrix.
Proposition Let be a matrix and its transpose. If is invertible, then is invertible and
We have thatBy transposing both sides of the equation, we obtainbecause the identity matrix is equal to its transpose. By using the formula for the transposition of a product, we getSo, satisfies the definition of inverse of .
Below you can find some exercises with explained solutions.
We need to carry out the multiplication between the two matrices:Their product is equal to the identity matrix, so is indeed the inverse of .
Let , and be full-rank matrices. Express the inversein terms of the inverses of , and .
We have to repeatedly apply the formula for the inverse of a product:
By definition, the inverse of needs to satisfyButAs a consequence,
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