Skip to content

12.9 Change of basis

Recall: Given a basis \(v_1,...,v_n\) of \(V\), we can write

  • \(v=a_1v_1+...+a_{n}v_{n}\rightarrow[v]_{\mathcal{B}}=\left(a_1,\ldots,a_{n}\right)\in\mathbb{F}^{n}\)
  • \(T\in\text{End}(V)\rightarrow [T]_\mathcal{B}\), where its j-column is \([T(v_j)]_\mathcal{B}\) (Here \([T]_{\mathcal{B}}=[T]_{\mathcal{BB}}\)​)

Example

imageWe define the linear map that let \(e_1\) become \(e_2\) and \(e_2\) become \(e_1\)

\(\mathcal{B}:e_1,e_2\) \(Te_1=e_2,Te_2=e_1\) Then \([T]_{\mathcal{B}}=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\)

\(\mathcal{B}^{\prime}:e_1+e_2,e_1-e_2\) \(T\left(e_1+e_2\right)=e_2+e_1,T\left(e_1-e_2\right)=e_2-e_1=-\left(e_1-e_2\right)\) Then \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\)

\(\mathcal{B}:v_1,...,v_n\) and \(\mathcal{B'}:w_1,...,w_n\)

\(w_{j}=\sum_{i=1}^{n}p_{ij}v_{i}\) Let \(P\) be the \(n\times m\) matrix whose \(ij\) coefficients us \(P_{ij}=p_{ij}\)

Now, let \(v\in V\) \(v=\beta_1w_1+\cdots+\beta_{n}w_{n}\), that is \([v]_{\mathcal{B^{\prime}}}=\left(\beta_1,\ldots,\beta_{n}\right)\)

Then \(v=\sum_{j=1}^{n}\beta_{j}w_{j}=\sum_{j=1}^{n}\beta_{j}\left(\sum_{i=1}^{n}p_{ij}v_{i}\right)=\sum_{j=1}^{n}\left(\sum_{i=1}^{n}\beta_{j}p_{ij}v_{i}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n}\beta_{j}p_{ij}\right)v_{i}=\sum_{i=1}^{n}\left(\sum_{j=1}^{n}p_{ij}\beta_{j}\right)v_{i}\Rightarrow\left\lbrack v\right\rbrack_{\mathcal{B}}=P\left\lbrack v\right\rbrack_{\mathcal{B}^{\prime}}\)

And \(\left(\sum_{j=1}^{n}p_{ij}\beta_{j}\right)\) is the i-th coordinates of \(P\quad\begin{pmatrix}\beta_1\\...\\\beta_n\end{pmatrix}\)

In particular \(P\) is invertible and \(P^{-1}[v]_{\mathcal{B}}=\left\lbrack v\right\rbrack_{\mathcal{B}^{\prime}}\)

‍

Notation

\(P\) is called the change of basis matrix from \(\mathcal{B}\) to \(\mathcal{B'}\)

Remark

\(P=[U]_{\mathcal{B}}\), where \(U\) is the linear map from \(V\) to \(V\) defined by \(v_i\mapsto w_i\)

\(w_{j}=\sum_{i=1}^{n}p_{ij}v_{i}\)​

Example

\(V=\R^2\) \(\mathcal{B}:e_1,e_2,\quad \mathcal{B'}:(1,-2),(2,3)\)

We have \(P=\begin{pmatrix}1 & 2\\ -2 & 3\end{pmatrix},P^{-1}=\frac17\begin{pmatrix}3 & -2\\ 2 & 1\end{pmatrix}\) and if \([v]_{\mathcal{B}}=\left(3,-2\right)\), then \([v]_{\mathcal{B^{\prime}}}=P^{-1}\begin{pmatrix}3\\ -2\end{pmatrix}=\begin{pmatrix}\frac{13}{7}\\ \frac47\end{pmatrix}\)

Then check yes

Recall: \([Tv]_{\mathcal{B}}=[T]_{\mathcal{B}}[v]_{\mathcal{B}}\)

Now on the one hand: \([Tv]_{\mathcal{B}^{\prime}}=P^{-1}\left\lbrack Tv\right\rbrack_{\mathcal{B}}\) since this, then \([Tv]_{\mathcal{B}^{\prime}}=P^{-1}\left\lbrack Tv\right\rbrack_{\mathcal{B}}=P^{-1}\left\lbrack T\right\rbrack_{\mathcal{B}}\left\lbrack v\right\rbrack_{\mathcal{B}}\)

On the other hand: \([Tv]_{\mathcal{B^{\prime}}}=[T]_{\mathcal{B^{\prime}}}[v]_{\mathcal{B^{\prime}}}=[T]_{\mathcal{B^{\prime}}}P^{-1}[v]_{\mathcal{B}}\)

Therefore, \(P^{-1}\left\lbrack T\right\rbrack_{\mathcal{B}}=\left\lbrack T\right\rbrack_{\mathcal{B}^{\prime}}P^{-1}\) and hence \(\left\lbrack T\right\rbrack_{\mathcal{B}^{\prime}}=P^{-1}\left\lbrack T\right\rbrack_{\mathcal{B}}P\)

Theorem

Let \(V\) be a vector space and let \(\mathcal{B}:v_1,...,v_n\) and \(\mathcal{B'}:w_1,...,w_n\) be two bases of \(V\).

Given \(T\in \text{End}(V)\), then it holds that

\[ [T]_{\mathcal{B^{\prime}}}=P^{-1}\left\lbrack T\right\rbrack_{\mathcal{B}}P \]

where \(P=[U]_{\mathcal{B}}\) for \(U\) the linear map defiend by \(U:v_{i}\mapsto w_{i},i=1,\ldots,n\)

Example

Recall, now we have \(\mathcal{B}:e_1,e_2\) \([T]_{\mathcal{B}}=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\) \(\mathcal{B}^{\prime}:e_1+e_2,e_1-e_2\) \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\)

Then \(P=\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix},P^{-1}=-\frac12\begin{pmatrix}-1 & -1\\ -1 & 1\end{pmatrix}\). Compute \(P^{-1}\left\lbrack T\right\rbrack_{\mathcal{B}}P=P^{-1}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}P=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}=\left\lbrack T\right\rbrack_{\mathcal{B}^{\prime}}\)​

Example

\(T:\mathbb{R}^3\rightarrow\mathbb{R}^3,\left(x,y,z\right)\mapsto\left(x+y+z,-x+z,x+2y+z\right)\) \(\mathcal{B}:e_1,e_2,e_3\) \(\mathcal{B}^{\prime}:\left(1,0,1\right),\left(1,0,0\right),\left(1,-1,1\right)\)

First we can calculate \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 1 & 1\\ -1 & 0 & 1\\ 1 & 2 & 1\end{pmatrix}\)

Compute \(T(a,1,1-a)\) where \((a,1,1-a)=v\)

  1. By linear map formula: \((2,1-2a,3)\)​

  2. \(\left\lbrack Tv\right\rbrack_{\mathcal{B}}=[T]_{\mathcal{B}}[v]_{\mathcal{B}}=\begin{pmatrix}1 & 1 & 1\\ -1 & 0 & 1\\ 1 & 2 & 1\end{pmatrix}\begin{pmatrix}a\\ 1\\ 1-a\end{pmatrix}=\begin{pmatrix}2\\ 1-2a\\ 3\end{pmatrix}\)​

  3. \(\left\lbrack Tv\right\rbrack_{\mathcal{B}^{\prime}}=[T]_{\mathcal{B}^{\prime}}[v]_{\mathcal{B}^{\prime}}\)​

    Let \(P\) be the matrix of change of absis from \(\mathcal{B}\) to \(\mathcal{B'}\)​

    \(P=\begin{pmatrix}1 & 1 & 1\\ 0 & 0 & -1\\ 1 & 0 & 1\end{pmatrix},P^{-1}=\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & -1\\ 0 & -1 & 0\end{pmatrix}\)

    Then we have \([T]_{\mathcal{B^{\prime}}}=\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & -1\\ 0 & -1 & 0\end{pmatrix}\begin{pmatrix}1 & 1 & 1\\ -1 & 0 & 1\\ 1 & 2 & 1\end{pmatrix}\begin{pmatrix}1 & 1 & 1\\ 0 & 0 & -1\\ 1 & 0 & 1\end{pmatrix}=\begin{pmatrix}2 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}\)

    This is a nice basis change since image​

    \([v]_{\mathcal{B}^{\prime}}=P^{-1}\left\lbrack v\right\rbrack_{\mathcal{B}}=\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & -1\\ 0 & -1 & 0\end{pmatrix}\begin{pmatrix}a\\ 1\\ 1-a\end{pmatrix}=\begin{pmatrix}2-a\\ 2a-1\\ -1\end{pmatrix}\)

    Thus \(\left\lbrack Tv\right\rbrack_{\mathcal{B}^{\prime}}=[T]_{\mathcal{B}^{\prime}}[v]_{\mathcal{B}^{\prime}}=\begin{pmatrix}4-2a\\ -1\\ 2a-1\end{pmatrix}\)

    Check: \([Tv]_{\mathcal{B}}=\begin{pmatrix}4-2a\\ -1\\ 2a-1\end{pmatrix}\rightarrow Tv=\left(4-2a\right)\left(1,0,1\right)+\left(-1\right)\left(1,0,0\right)+\left(2a-1\right)\left(1,-1,1\right)=\left(2,1-2a,3\right)\)​

Example

Consider the rotation image \([R]=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\)

Question: Is there a basis \(\mathcal{B'}\) of \(\R^2\) such that \([R]_{\mathcal{B^{\prime}}}=\begin{pmatrix}a&0\\0&b\end{pmatrix}\)?

Assume yes: \(\mathcal{B^{\prime}}:v_1,v_2\). Thus \(R(v_1)=av_1,R(v_2)=bv_2\) Impossible

However if \(V=\mathbb{C}^2\), \(\mathcal{B}:(1,0),(0,1)\). Let \(R\) such that \([R]_{\mathcal{B}}=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\)

Consider \(P=\begin{pmatrix}i & i\\ 1 & -1\end{pmatrix},P^{-1}=\begin{pmatrix}-\frac{i}{2} & \frac12\\ -\frac{i}{2} & -\frac12\end{pmatrix}\). Then \(P^{-1}\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}P=\begin{pmatrix}1 & 0\\ 0 & -i\end{pmatrix}\)​

Now \(\mathcal{B^{\prime}}=P\left(\mathcal{B}\right),\mathcal{B^{\prime}}:v_1=\left(i,1\right),v_2=\left(i,-1\right)\)

Check: \([Rv_1]_{\mathcal{B}}=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\begin{pmatrix}i\\ 1\end{pmatrix}=\begin{pmatrix}-1\\ i\end{pmatrix}=i\begin{pmatrix}i\\ 1\end{pmatrix}=iv_1\)

\([Rv_2]_{\mathcal{B}}=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\begin{pmatrix}i\\ -1\end{pmatrix}=\begin{pmatrix}1\\ i\end{pmatrix}=-i\begin{pmatrix}i\\ -1\end{pmatrix}=-iv_2\)

Definition

Let \(A,B\in M_{n\times n}\left(\mathbb{F}\right)\). We say that \(B\) is similar to \(A\) if there is an invertible \(P\) such that \(B=P^{-1}AP\)

Notation

\(B\) is conjugate of \(A\) by \(P\)

Remark

If \(B\) is similar to \(A\) (\(B\sim A\)), then \(A\sim B\) (\(B=P^{-1}AP\Rightarrow PBP^{-1}=A\Rightarrow\left(\left(P\right)^{-1}\right)^{-1}BP^{-1}=A\))

Moreover, \(\sim\) is an equivalence relation

Notice that \([T]_\mathcal{B}\) and \([T]_\mathcal{B'}\) are similar

Properties of similar matrices

If \(B\sim A\Rightarrow trB=trA\) [\(tr(xy)=tr(yx)\)]

\(B\sim A\Rightarrow B=P^{-1}AP\Rightarrow tr\left(B\right)=tr\left(\left(P^{-1}A)P\right)\right.=tr\left(P\left(P^{-1}A\right)\right)=tr\left(A\right)\)​

If \(B\sim A\Rightarrow\dim\text{Null}B=\dim\text{Null}A\) and \(\text{Rank}B=\text{Rank}A\)

Remark

\(\text{End}(V)\;\iff\;M_{n\times n}\left(V\right)/\sim\)​

‍

‍

‍