11: Canonical Forms
- Last updated
- Jul 25, 2023
- Save as PDF
( \newcommand{\kernel}{\mathrm{null}\,}\)
Given a matrix A, the effect of a sequence of row-operations on A is to produce UA where U is invertible. Under this “row-equivalence” operation the best that can be achieved is the reduced row-echelon form for A. If column operations are also allowed, the result is UAV where both U and V are invertible, and the best outcome under this “equivalence” operation is called the Smith canonical form of A (Theorem [thm:005369]). There are other kinds of operations on a matrix and, in many cases, there is a “canonical” best possible result.
If A is square, the most important operation of this sort is arguably “similarity” wherein A is carried to U−1AU where U is invertible. In this case we say that matrices A and B are similar, and write A∼B, when B=U−1AU for some invertible matrix U. Under similarity the canonical matrices, called Jordan canonical matrices, are block triangular with upper triangular “Jordan” blocks on the main diagonal. In this short chapter we are going to define these Jordan blocks and prove that every matrix is similar to a Jordan canonical matrix.
Here is the key to the method. Let T:V→V be an operator on an n-dimensional vector space V, and suppose that we can find an ordered basis B of B so that the matrix MB(T) is as simple as possible. Then, if B0 is any ordered basis of V, the matrices MB(T) and MB0(T) are similar; that is,
MB(T)=P−1MB0(T)Pfor some invertible matrix P
Moreover, P=PB0←B is easily computed from the bases B and D (Theorem [thm:028802]). This, combined with the invariant subspaces and direct sums studied in Section [sec:9_3], enables us to calculate the Jordan canonical form of any square matrix A. Along the way we derive an explicit construction of an invertible matrix P such that P−1AP is block triangular.
This technique is important in many ways. For example, if we want to diagonalize an n×n matrix A, let TA:Rn→Rn be the operator given by TA(x)=Ax or all x in Rn, and look for a basis B of Rn such that MB(TA) is diagonal. If B0=E is the standard basis of Rn, then ME(TA)=A, so
P−1AP=P−1ME(TA)P=MB(TA)
and we have diagonalized A. Thus the “algebraic” problem of finding an invertible matrix P such that P−1AP is diagonal is converted into the “geometric” problem of finding a basis B such that MB(TA) is diagonal. This change of perspective is one of the most important techniques in linear algebra.
