Let [math]A[/math] and [math]B[/math] be n x n matrices. [math]A[/math] [b]is similar to[/b] [math]B[/math] if there exists an invertible n x n matrix [math]P[/math] such that [math]P^{-1}AP=B[/math]. Obviously, this relation is symmetric i.e. [math]A[/math] is similar to [math]B[/math] implies [math]B[/math] is also similar to [math]A[/math] (since [math]Q^{-1}BQ=A[/math] when [math]Q=P^{-1}[/math]). Therefore, sometimes we just say [math]A[/math] and [math]B[/math] [b]are similar[/b]. [br][br]As we know, the square matrix [math]A[/math] of a linear transformation [math]T:\mathbb{R}^n\rightarrow\mathbb{R}^n[/math] means [math]T\left(v\right)=Av[/math] for any column vector [math]v[/math], whose entries are (by default) coordinates using [u]the standard coordinate system[/u] in [math]\mathbb{R}^n[/math]. However, if we use another coordinate system defined by a new basis for [math]\mathbb{R}^n[/math] and express the same linear transformation [math]T[/math] as a matrix i.e. [math]T\left(v\right)=Bv[/math], where now both [math]v[/math] and [math]T(v)[/math] are column vectors whose entries are coordinates using [u]the new coordinate system[/u], it can be shown that [math]B=P^{-1}AP[/math], where [math]P[/math] is the n x n matrix formed by the column vectors of the new basis in the standard coordinate system. In other word, [math]A[/math] and [math]B[/math] are similar. P is usually called the [b]change-of-coordinates matrix[/b].[br][br]In the applet below, you can see the effect on the matrix representation of a linear transformation under the change of coordinates. First, you can define the linear transformation [math]T[/math] by setting the vectors [math]T(\hat{\mathbf{i}}), T(\hat{\mathbf{j}})[/math] in the "Range". Then you can set the two vectors [math]p_1,p_2[/math] that define the new coordinate system, which is represented by the dotted green gridlines (when [math]\{p_1,p_2\}[/math] are linearly independent). Then you can freely drag the vector [math]v[/math] and see how [math]v[/math] and [math]T(v)[/math] are related using two different coordinate systems.[br][br]

Suppose A and B are similar. The following theorem shows that they have a lot in common:[br][br][u]Theorem[/u]: Let [math]A[/math] and [math]B[/math] be n x n matrices such that [math]A[/math] and [math]B[/math] are similar. Then we have the following:[br][list=1][*]They have the same determinant.[br][/*][*]They have the same characteristic equation and hence the same eigenvalues (with the same multiplicities).[/*][*]They have the same rank i.e.[math]\dim(\text{Col} \ A)=\dim(\text{Col} \ B)[/math][/*][*][math]\dim(\text{Nul} \ A)=\dim(\text{Nul} \ B), \dim(\text{Row} \ A)=\dim(\text{Row} \ B)[/math][br][/*][/list][br]Proof: By definition, there exists an invertible matrix [math]P[/math] such that [math]P^{-1}AP=B[/math].[br]1. [math]\det(B)=\det(P^{-1}AP)=\det(P^{-1})\det(A)\det(P)=\frac1{\det(P)}\cdot \det(A)\det(P)=\det(A)[/math][br][br]2. [math]\det(B-\lambda I)=\det(P^{-1}AP-P^{-1}(\lambda I)P)=\det(P^{-1}(A-\lambda I)P)=\det(P^{-1})\det(A-\lambda I)\det(P)=\det(A-\lambda I)[/math][br][br]3. Let [math]S,T,U:\mathbb{R}^n\rightarrow\mathbb{R}^n[/math] be the linear transformations corresponding to the matrices [math]A,B,P[/math] respectively. Then [math]\text{Col} \ A=\text{Im}(S)[/math] and [math]\text{Col} \ B =\text{Im}(T)=\text{Im}(U^{-1}\circ S\circ U)[/math]. Since [math]U[/math] is bijective, [math]\text{Im}(U^{-1}\circ S\circ U)=\text{Im}(U^{-1}\circ S)=U^{-1}(\text{Im}(S))[/math]i.e. [math]U^{-1}[/math] maps [math]\text{Col} \ A[/math] to [math]\text{Col} \ B[/math] and thus [math]\dim(\text{Col} \ A)=\dim(\text{Col} \ B)[/math].[br][br]4. [math]\dim(\text{Nul} \ A)=\dim(\text{Nul} \ B)[/math] follows from (3) and rank-nullity theorem. [math]\dim(\text{Row} \ A)=\dim(\text{Row} \ B)[/math] follows from (3) and the fact that any column space and row space of a matrix have the same dimension.[br]