Similar matrices

Linear mappings: Matrices and coordinate transformations

Similar matrices

Two square matrices $A$ and $B$ are similar if there exists an invertible matrix $T$ such that $B=T^{-1}A\,T$ .

In the previous section we have actually seen the following theorem.

Two matrices represent the same linear mapping if and only if the matrices are similar.

When we solve a problem about a linear mapping we often try to find a 'pretty' matrix mapping corresponding with the given linear transformation. 'Pretty' means for example 'in diagonal form'.

Let $A$ be a matrix mapping corresponding with a linear mapping $L$ . Then $L$ is called diagonalisable if and only if there exists an invertible matrix $T$ such that $T^{-1}A\,T$ is diagonal.

Not every linear mapping or matrix is diagonalisable. For example, $\matrix{1 & 1\\ 0 & 1}$ is not diagonalisable. The following theorem gives an example of a class of matrices that are diagonalisable.

Every real symmetric matrix is diagonalisable.

Finally, we note that some functions applied to similar matrices yield same value; the determinant of a matrix is such a function:

$\det(T^{-1}A\,T)=\det(A)$ for square matrices $A$ and invertible matrices $T$ of equal dimension as $A$ .

Another function with this property is the trace of a matrix, denoted as $\text{tr}(A)$ and defined by

$\text{tr}\matrix{a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \ldots & a_{nn}}=a_{11}+a_{22}+\cdots+a_{nn}=\sum_{i=1}^n a_{ii}$

For the trace of a matrix, the following applies:

$\text{tr}(T^{-1}A\,T)=\text{tr}(A)$ for square matrices $A$ and invertible matrices $T$ of equal dimension as $A$ .

For a math enthusiast we give a proof.

First, we show that for $n\times n$ matrices $A$ and $B$ holds $\text{tr}(A\,B)=\text{tr}(B\,A)$ .

$\text{tr}(A\,B)=\sum_{i=1}^n (AB)_{ii} = \sum_{i=1}^n\sum_{j=1}^n A_{ij}B_{ji}= \sum_{j=1}^n\sum_{i=1}^n B_{ji}A_{ij}= \sum_{j=1}^(BA)_{jj}= \text{tr}(BA)$

But then we also have:

$\text{tr}(T^{-1}A\,T)=\text{tr}(AT^{-1}T)=\text{tr}(A\,I)=\text{tr}(A)$