# Coordinates (Vectors and Similar Matrices)

Are you trying to understand why similar matrices are essential? Why are they used? This article covers coordinate vectors, coordinate space, and the matrix representation of a linear transformation. After that, I discuss similar matrices and their importance.

## Coordinate Vectors

Definition. Let $\mathcal{B}=(v_1,\ldots,v_n)$ be a basis of a subspace $V$ of $\mathbb{R}^n.$ For any $x \in V$ we can write $v= c_1 v_1 + \cdots +c_m v_m.$ The scalars $c_1,\ldots,c_m$ are called the $\mathcal{B}$-coordinates of $x$ and the vector $$\left [ x \right ]_{\mathcal{B}}:= \begin{bmatrix}c_1\ \vdots \ c_m\end{bmatrix}$$ is called the $\mathcal{B}$-coordinate vector of $x.$

For example the coordinates of $\begin{bmatrix}-2\\ 4\end{bmatrix}$ with respect the standard basis $$\mathcal{B}=(e_1,e_2)$$ is $\begin{bmatrix}-2\\ 4\end{bmatrix}$ since $\begin{bmatrix}-2\\ 4\end{bmatrix} =-2e_1+4e_2$ which written as $$\begin{bmatrix}-2\\ 4\end{bmatrix}{\mathcal{B}}=\begin{bmatrix}-2\\ 4\end{bmatrix}.$$ Notice the coordinates of $\begin{bmatrix}-2\\ 4\end{bmatrix}$ with respect the basis $$\mathcal{B}’ = \left(\begin{bmatrix}2\\ 0\end{bmatrix},\begin{bmatrix} 0\\ 2\end{bmatrix}\right)$$ is $\begin{bmatrix}-1\\ 2\end{bmatrix}$ since $\begin{bmatrix}-2\\ 4\end{bmatrix}=(-1)\begin{bmatrix}2\\ 0\end{bmatrix}+2\begin{bmatrix}0\\ 2\end{bmatrix}$ which written as $$\begin{bmatrix}-2\\ 4\end{bmatrix}_{\mathcal{B}’}=\begin{bmatrix}-1 \\ 2\end{bmatrix}.$$

Example. Consider the plane $2x_1-3x_2+4x_3=0$ with basis $$\mathcal{B}=\left(\begin{bmatrix}8 \\ 4 \\ -1\end{bmatrix} , \begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix}\right).$$

Solution. Let $[x]_\mathcal{B} = \begin{bmatrix}2 \\ -1\end{bmatrix}.$ Find $x.$ By definition of coordinates $$x= 2\begin{bmatrix}8 \\ 4 \\ -1 \end{bmatrix} +(-1)\begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix} = \begin{bmatrix}11 \\ 6 \\ -1\end{bmatrix}.$$ as desired.

## Coordinate Space

Lemma. If $\mathcal{B}=(v_1,\ldots,v_n)$ is a basis of a subspace $V$ of $\mathbb{R}^n$, then

$[x]{\mathcal{B}}+[y]_{\mathcal{B}}=[x+y]{\mathcal{B}}$, for all vectors $x,y$

$[kx]_{\mathcal{B}}=k[x]_{\mathcal{B}}$, for all vectors $x$, and all scalars $k.$

Proof. Let $[x]_{\mathcal{B}}=c_1 v_1+\cdots + c_n v_n$ and $$[y]_{\mathcal{B}}=d_1 v_1+\cdots + d_n v_n$$ be the representation of $x$ and $y$ with respect to $\mathcal{B}.$ Then $$[x]_{\mathcal{B}}+[y]_{\mathcal{B}} =\begin{bmatrix}c_1 \\ \vdots \\ c_3\end{bmatrix} + \begin{bmatrix}d_1 \\ \vdots \\ d_3\end{bmatrix} =\begin{bmatrix}c_1+d_1 \\ \vdots \\ c_n+d_n\end{bmatrix} =[x+y]{\mathcal{B}}$$ where the last equality holds since $$x+y=(c_1+d_1)v_1+\cdots (c_n+d_n)v_n.$$ We leave the proof of the second part for the reader.

Example. Determine whether the vector $x = \begin{bmatrix}1 \\ -2 \\ -2\end{bmatrix}$ is in $\operatorname{span} V$ of the vectors $v_1=\begin{bmatrix}8 \\ 4 \\ -1\end{bmatrix}$ and $v_2=\begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix}$ and if so write the coordinates of $x$ with respect to this basis of $V.$

Solution. We need to find scalars $c_1$ and $c_2$ such that $x=c_1 v_1+c_2 v_2.$ Solving this system we find $c_1=-3$ and $c_2=5.$ Therefore we find, $[x]_{\mathcal{B}}=\begin{bmatrix}-3 \\ 5\end{bmatrix}$ where $\mathcal{B}=(v_1, v_2).$

Example. Consider the plane $x+2y+z=0.$ Find a basis of this plane. Find another basis $\mathcal{B}$ of this plane such that $[x]_\mathcal{B}=\begin{bmatrix}2 \\ -1\end{bmatrix}$ for $x = \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}.$

Solution. We find a basis by letting $z=t$ and $y=s$ be free variables. Then $x=-t-2s.$ All solutions to the equation $x+2y+z=0$ are $$\begin{bmatrix}x\\ y\\ z\end{bmatrix} =\begin{bmatrix}-t-2s\\ s\\ t\end{bmatrix} =t\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}+s\begin{bmatrix}-2\\ 1\\ 0\end{bmatrix}.$$ So a basis for the plane is $\mathcal{B} =\left(\begin{bmatrix}-1\\ 0\1\end{bmatrix},\begin{bmatrix}-2\\ 1\\ 0\end{bmatrix}\right).$ Notice that $$\begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = (1)\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}+(-1)\begin{bmatrix}-2\\ 1 \\ 0\end{bmatrix}$$ and thus $\begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}_{\mathcal{B}}=\begin{bmatrix}1 \\ -1\end{bmatrix}.$ This is not the basis we seek. However, notice that $$\begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix} =2 \begin{bmatrix} x_1 \\ 0 \\ -x_1\end{bmatrix} + (-1)\begin{bmatrix}-2y_2 \\ y_2 \\ 0 \end{bmatrix}$$ holds when $y_2=1$ and $x_1=-1/2.$ Also notice the vectors $\begin{bmatrix}-1/2 \\ 0 \\ 1/2\end{bmatrix}$ and $\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}$ span the plane and are linearly independent. Therefore we have the basis we seek, namely$\mathcal{B}=\left(\begin{bmatrix}-1/2 \\ 0 \\ 1/2\end{bmatrix},\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}\right).$

## Matrix Representation

Notice the matrix representation of a linear transformation consists of columns of coordinate vectors.

Lemma. Let $T$ be a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^n$ and $\mathcal{B}=(v_1,\ldots,v_n)$ a basis of $\mathbb{R}^n.$ The $n\times n$ matrix $B$ that transforms $[x]_{\mathcal{B}}$ into $[T(x)]_{\mathcal{B}}$ is called the $\mathcal{B}$-matrix of $T$, written as $[T(x)]_{\mathcal{B}}=B [x]_{\mathcal{B}}$ for all $x$ in $\mathbb{R}^n$ and $$B= \begin{bmatrix} [T(v_1)]_\mathcal{B} & \cdots & [T(v_n)]_\mathcal{B} \end{bmatrix}.$$

Proof. Since $\mathcal{B}$ is a basis of $\mathbb{R}^n$, there exists scalars $c_1, \ldots, c_n$ such that $x=c_1 v_1+c_2v_2+\cdots + c_n v_n.$ Using the linearity of $T$ we find $$T(x)=c_1 T(v_1)+c_2 T(v_2)+\cdots +c_n T(v_n).$$ So we find \begin{align} [T(x)]_{\mathcal{B}} & =c_1 [T(v1)]_{\mathcal{B}}+c_2 [T(v_2)]_{\mathcal{B}}+\cdots +c_n [T(v_n)_{\mathcal{B}} \\ & = \begin{bmatrix} [T(v_1)]_\mathcal{B} & \cdots & [T(v_n)]_\mathcal{B} \end{bmatrix} [x]_\mathcal{B} \end{align} as desired.

Example. Find the matrix $\mathcal{B}$ of the linear transformation $T(x)=A x$ where $$A=\begin{bmatrix} 5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8 \end{bmatrix}$$ with respect to the basis $$\mathcal{B}=\left ( \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ -2\end{bmatrix} \right ).$$ This $T$ with respect to $\mathcal{B}$ has the following matrix $B.$\begin{align*} & \begin{bmatrix} \left[T\begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}\right]_{\mathcal{B}} & \left[T\begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}\right]_{\mathcal{B}} & \left[T\begin{bmatrix}0 \\ 1 \\ -2\end{bmatrix}\right]_{\mathcal{B}} \end{bmatrix} \\ & = \begin{bmatrix} \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}{_\mathcal{B}} & \begin{bmatrix} 9 \\ -9 \\ 0\end{bmatrix}{_\mathcal{B}} & \begin{bmatrix}0 \\ 9 \\ -18\end{bmatrix}{_\mathcal{B}} \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}. \end{align*} as desired.

Theorem. Let $T$ be a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^n$ with standard matrix $A$ and let $B$ be the $\mathcal{B}$-matrix of $T$ where $\mathcal{B}= (v_1,\ldots,v_n)$ then $A S= S B$ where $S= \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} .$

## Similar Matrices and Their Importance

Similar matrices share many properties such as rank, determinant, trace, eigenvalues, minimal polynomials, and much more. The columns of the change of basis matrix are just the basis vectors of according to their coordinates.

Definition. Two $n\times n$ matrices $A$ and $B$ are called similar if there exists an invertible matrix $S$ such that $A S= S B.$

Example. Determine whether the following two matrices are similar. $$A=\begin{bmatrix} 1 & 2 \\ 4 &3 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix} 5 & 0 \\ 0 & -1 \end{bmatrix}$$ We are looking for a matrix $S= \begin{bmatrix} x & y \\ z & t\end{bmatrix}$ such that $AS=SB.$ Writing out we find $$\begin{bmatrix} 1 & 2 \\ 4 &3 \end{bmatrix} \begin{bmatrix} x & y \\ z & t\end{bmatrix} \begin{bmatrix} x & y \\ z & t\end{bmatrix} \begin{bmatrix} 5 & 0 \\ 0 & -1 \end{bmatrix}$$ which implies that $$\begin{bmatrix} x+2z & y+2t \\ 4x+3z & 4y+3t \end{bmatrix} \begin{bmatrix} 5x & -y \\ 5z & -t \end{bmatrix}.$$ This leads to $z=2x$ and $t=-y$ so that $S$ is any invertible matrix of the form $$\label{simform} \begin{bmatrix}x & y \\ 2x & -y\end{bmatrix}.$$ Since $S= \begin{bmatrix}1 & 1 \\ 2 & -1 \end{bmatrix}$ is invertible and of the form in \eqref{simform}, we can say, yes $A$ is similar to $B.$

Theorem. If $A$ is similar to $B$ then $A^k$ is similar to $B^k$ for any positive integer $k.$

Proof. If $A$ is similar to $B$ then there exists an invertible matrix $S$ such that $B=S^{-1}A S.$ Then $$B^k=(S^{-1}AS)(S^{-1}AS)\cdots (S^{-1}AS)=S^{-1}A^k S$$ which shows that $B^k$ is similar to $A^k.$

Theorem. Two $n\times n$ matrices $A$, $A’$ are similar if and only if they are matrices of the same linear transformation $T$ from $\mathbb{R}^n$ to $\mathbb{R}^n$ with respect to two bases for $\mathbb{R}^n$.

The idea of change of coordinates gives rise to similar matrices.

Theorem. Similar matrices have the same rank.

Proof. Let $A$ and $B$ be similar matrices. We know that $A$ and $B$ represent the same linear transformation. Thus they must have the same rank.

Example. Are the following similar matrices? $$A= \begin{bmatrix} -1 & 2 & 1 \\ 2 & 0 & 1 \\ 2 & 2 & 2 \end{bmatrix} \qquad \text{and}\qquad B= \begin{bmatrix} -1 & 2 & 1 \\ 2 & 0 & 1 \\ 1 & 2 & 2 \end{bmatrix}$$ Since $\mathop{rank}(A)=3$ and $\mathop{rank}(B)=2$, these are not similar matrices.

Theorem. Similarity of matrices is an equivalence relation.

Proof. To show transitivity, assume $A$ is similar to $B$ and $B$ is similar to $C$, then there exists invertible matrices $E$ and $F$ such that $AE=EB$ and $BF=FC.$ Then $$A=EBE^{-1} =E(FCF^{-1})E^{-1} =(EF)C(F^{-1}E^{-1}) =(EF)C(EF)^{-1}$$ which shows that $A$ is similar to $C.$ The proof of the reflexive property and the symmetric property are left for the reader. A linear transformation represents a whole class of (similar) matrices. That is the collection of $n\times n$ matrices is partitioned into non overlapping sets of matrices, with all matrices in any such set being similar and representing the same linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^n.$

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.