Two square matrices are said to be similar if they represent the same linear operator under different bases. Two similar matrices have the same rank, trace, determinant and eigenvalues.
We start with a definition.
Proposition A matrix is said to be similar to another matrix if and only if there exists an invertible matrix such that
The transformation of into is called similarity transformation.
The matrix is called change-of-basis matrix.
In order to understand the relation between similar matrices and changes of bases, let us review the main things we learned in the lecture on the Change of basis.
Any vector can be represented as a linear combination of the basiswhere are scalar coefficients.
The coefficients of the linear combination form the so-called coordinate vector of with respect to , denoted by :
If we use a different basis , then the coordinates of any vector with respect to satisfywhere the matrix is called change-of-basis matrix and allows to convert coordinates with respect to into coordinates with respect to .
Remember that a linear operator can always be represented by a matrix such that, for any ,
In other words, if we pre-multiply the coordinates of with respect to by , we get the coordinates of as a result.
We have shown that the matrices of the linear operator under different bases are related to each other by the change-of-basis formula
Thus, is similar to . This result explains the characterization of similarity we have given in the introduction above: two similar matrices represent the same linear operator under different bases.
Similarity defines an equivalence relation between square matrices.
Proposition Matrix similarity is an equivalence relation, that is, given three matrices , and , the following properties hold:
Reflexivity: is similar to itself;
Symmetry: if is similar to , then is similar to ;
Transitivity: if is similar to and is similar to , then is similar to .
Similarity is reflexive becausewhere the identity matrix is the change-of-basis matrix. Symmetry holds because the equationimplieswhere is the change-of-basis matrix. Transitivity holds becauseimplywhere is the change-of-basis matrix.
The next proposition shows a first important property of similarity.
Proposition If two matrices are similar, then they have the same rank.
Let and be similar, so that , with invertible (hence full-rank). As proved in the lecture on matrix product and rank, because is full-rank and because is full-rank. Therefore, and have the same rank.
Trace is preserved by similarity transformations.
Proposition If two matrices are similar, then they have the same trace.
Let . Then, by an elementary property of the trace, we have that
The next property concerns the determinant.
Proposition If two matrices are similar, then they have the same determinant.
Let . We havewhere in steps and we have used two properties of the determinant: 1) the determinant of a product of two or more matrices is equal to the product of their determinants; 2) .
This is probably the most important property, as well as the reason why similarity transformations are so important in the theory of eigenvalues and eigenvectors.
Let and be similar, so that . Any eigenvalue of solves the characteristic equationwhile the eigenvalues of solve the equationwhere in steps and we have used two properties of the determinant: 1) the determinant of a product of two or more matrices is equal to the product of their determinants; 2) . Thus, solves the characteristic equation of if and only if it solves the characteristic equation of . Stated differently, is an eigenvalue of if and only if it is an eigenvalue of . Moreover, since and have the same characteristic equation, their eigenvalues have the same algebraic multiplicities. We still need to prove that the eigenvalues of and have the same geometric multiplicities. Note that
For a given eigenvalue , choose an eigenvector of associated to and denote it by . Then,If we post-multiply equation (1) by , we getorIn other words, is an eigenvector of associated to if and only if is an eigenvector of associated to . Suppose that , as an eigenvector of , has geometric multiplicity equal to . Choose a basis for the eigenspace of associated to (i.e., any eigenvector of associated to can be written as a linear combination of ). Let be the matrix obtained by adjoining the vectors of the basis:Thus, the eigenvectors of associated to satisfy the equationwhere is the vector of coefficients of the linear combination. If we pre-multiply both sides of the equation by , we getThus the eigenvectors of associated to are all the vectors that can be written as linear combinations of the columns of . But has the same rank of because is full-rank. Therefore, it has rank . So the geometric multiplicity of , as an eigenvector of is , the same it has as an eigenvector of .
Note from the previous proof that ifthen is an eigenvalue of if and only if it is an eigenvalue of , but is an eigenvector of associated to if and only if is an eigenvector of associated to .
In linear algebra we often use the term "unitarily similar".
Definition Two matrices and are said to be unitarily similar if and only if there exists a unitary matrix such that
Thus, two matrices are unitarily similar if they are similar and their change-of-basis matrix is unitary.
Since the inverse of a unitary matrix is equal to its conjugate transpose , the similarity transformation can be written as
When all the entries of the unitary matrix are real, then the matrix is orthogonal, and the similarity transformation becomes
The following proposition illustrates a simple but very useful property of similarity.
Proposition If two matrices and are similar, then their -th powers and are similar.
Let . We have
The proof also shows that the change-of-basis matrix employed in the similarity transformation of into is the same used in the similarity transformation of into .
Please cite as:
Taboga, Marco (2017). "Similar matrix", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/similar-matrix.
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