Section 5.3 Isomorphism and invertible matrix
Proof.
Let \(V\) and \(W\) be finite-dimensional vector spaces over a field \(F\text{.}\) Suppose that \(\mathfrak{B}_V=(v_1,v_2,\ldots,v_n)\) and \(\mathfrak{B}_W=(w_1,w_2,\ldots,w_n)\) are ordered bases of \(V\) and \(W\text{,}\) respectively. Let \(T\colon V\to W\) be an \(F\)-linear isomorphism. Let the matrix of \(T\) relative to ordered bases \(\mathfrak{B}_V\) and \(\mathfrak{B}_W\) be
\begin{equation*}
[T]_{\mathfrak{B}_V}^{\mathfrak{B}_W}=\begin{pmatrix}\beta_{11}\amp\beta_{12}\amp\cdots\amp\beta_{1n}\\\beta_{21}\amp\beta_{22}\amp\cdots\amp\beta_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\beta_{n1}\amp\beta_{n2}\amp\cdots\amp\beta_{nn}\end{pmatrix}.
\end{equation*}
Recall from Subsection 4.7.2 that \(T(v_i)=\sum_k\beta_{ki}w_k\text{.}\)
\begin{equation*}
T^{-1}(w_j)=\sum_\ell\gamma_{\ell j}v_\ell.
\end{equation*}
The matrix of \(T^{-1}\) relative to ordered bases \(\mathfrak{B}_W\) and \(\mathfrak{B}_V\) is
\begin{equation*}
[T^{-1}]_{\mathfrak{B}_W}^{\mathfrak{B}_V}=\begin{pmatrix}\gamma_{11}\amp\gamma_{12}\amp\cdots\amp\gamma_{1n}\\\gamma_{21}\amp\gamma_{22}\amp\cdots\amp\gamma_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\gamma_{n1}\amp\gamma_{n2}\amp\cdots\amp\gamma_{nn}\end{pmatrix}.
\end{equation*}
By Remark 4.1.5, the composition \(T\circ T^{-1}=I_W\) is determined by its action on \(\{w_j\}\text{.}\) We have the following.
\begin{align*}
w_j=T\circ T^{-1}(w_j)\amp=T\big(\sum_\ell\gamma_{\ell j}v_\ell\big)\\
\amp=\sum_\ell\gamma_{\ell j}T(v_\ell)\\
\amp=\sum_\ell\gamma_{\ell j}\big(\sum_k\beta_{k\ell}w_k\big)\\
\amp=\sum_k\big(\sum_\ell\gamma_{\ell j}\beta_{k\ell}\big)w_k
\end{align*}
By Exercise 3.5.11, we must have
\begin{equation*}
\sum_\ell\gamma_{\ell j}\beta_{k\ell}=0\quad\text{for }k\neq j\quad\text{and}\quad\sum_\ell\gamma_{\ell j}\beta_{j\ell}=1.
\end{equation*}
We thus obtain that the \(j\)-th column of the following matrix
\begin{equation*}
\begin{pmatrix}\beta_{11}\amp\beta_{12}\amp\cdots\amp\beta_{1n}\\\beta_{21}\amp\beta_{22}\amp\cdots\amp\beta_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\beta_{n1}\amp\beta_{n2}\amp\cdots\amp\beta_{nn}\end{pmatrix}\cdot\begin{pmatrix}\gamma_{11}\amp\gamma_{12}\amp\cdots\amp\gamma_{1n}\\\gamma_{21}\amp\gamma_{22}\amp\cdots\amp\gamma_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\\gamma_{n1}\amp\gamma_{n2}\amp\cdots\amp\gamma_{nn}\end{pmatrix}
\end{equation*}
is \(\big(0,0,\ldots,0,1,0,\ldots,0\big)^t\text{.}\) Therefore, \([T]_{\mathfrak{B}_V}^{\mathfrak{B}_W}\cdot[T^{-1}]_{\mathfrak{B}_W}^{\mathfrak{B}_V}\) is the \(n\times n\) identity matrix.
Similar computations for \(T^{-1}\circ T=I_V\) yield \([T^{-1}]_{\mathfrak{B}_W}^{\mathfrak{B}_V}\cdot[T]_{\mathfrak{B}_V}^{\mathfrak{B}_W}\) is the \(n\times n\) identity matrix.
Lemma 5.3.2.
Every \(n\times n\) invertible matrix induces an \(F\)-linear isomorphism of \(M_{n\times 1}(F)\) onto itself.Proof.
\begin{equation*}
L_A\circ L_B(y)=A(By)=y\quad\text{and}\quad L_B\circ L_A(x)=B(Ax)=x.
\end{equation*}
Therefore, \(L_A\) as well as \(L_B\) are isomorphisms.Lemma 5.3.3.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) Suppose that \(\mathfrak{B}=\{u_1,\ldots,u_n\}\) and \(\mathfrak{B}^\prime=\{v_1,\ldots,v_n\}\) are bases of \(V\text{.}\) Then there exists an \(n\times n\) invertible matrix \(P\) such that the coordinates of any vector \(v\) with respect to \(\mathfrak{B}^\prime\) is the same as the coordinates of \((Pv)^t\) with respect to the basis \(\mathfrak{B}\text{.}\)Proof.
\begin{equation*}
v_i=\sum_j\alpha_{ji}u_j.
\end{equation*}
We thus have the following.
\begin{align*}
v\amp=\sum_ia_iv_i\\
\amp=\sum_ia_i\sum_j\alpha_{ji}u_j\\
\amp=\sum_j\big(\sum_ia_i\alpha_{ji}\big)u_j
\end{align*}
By Exercise 3.5.11, the coordinates of \(v\) with respect to the basis \(\mathfrak{B}\) are
\begin{equation*}
(\sum_ia_i\alpha_{1i},\ldots,\sum_ia_i\alpha_{ni}).
\end{equation*}
Put \(P=(\alpha_{ij})\) Using (5.2.1) we therefore have the following.
\begin{equation*}
L_P\circ{}_nT\circ T_{\mathfrak{B}^\prime}(v)={}_nT\circ T_{\mathfrak{B}}(v).
\end{equation*}
Since, \({}_nT\) and \(T_{\mathfrak{B}^\prime}\) are \(F\)-isomorphisms (see Proposition 5.1.9 and Checkpoint 5.1.8), the \(F\)-linear map \(L_P\circ{}_nT\circ T_{\mathfrak{B}^\prime}\) is also an \(F\)-isomorphism.. Moreover the matrix of \(L_P\circ{}_nT\circ T_{\mathfrak{B}^\prime}\) with respect to \(\mathfrak{B}\text{,}\) \([L_P\circ{}_nT\circ T_{\mathfrak{B}^\prime}]_{\mathfrak{B}}^{\mathfrak{B}}=P\text{.}\) By Proposition 5.3.1, \(P\) is invertible and thus we get the result.Remark 5.3.4.
Note that the matrix \(P\) in Lemma 5.3.3 is the matrix of the identity map \(\unit_V\) with respect to bases \(\mathfrak{B}^\prime\) and \(\mathfrak{B}\) in that order, i.e.,
\begin{equation*}
[\unit_V]_{\mathfrak{B}^\prime}^{\mathfrak{B}}=P.
\end{equation*}
Theorem 5.3.5.
Let \(V,W\) be finite-dimensional vector spaces over a field \(F\text{.}\) Let \(T\colon V\to W\) be an \(F\)-linear map. Let \(\mathfrak{B}_i\) be bases of \(V\) and \(\mathfrak{C}_i\) be bases of \(W\text{.}\) Then there exists invertible matrices \(P,Q\) such that
\begin{equation*}
[T]_{\mathfrak{B}_1}^{\mathfrak{C}_1}=P\cdot [T]_{\mathfrak{B}_2}^{\mathfrak{C}_2}\cdot Q.
\end{equation*}
Proof.
\begin{equation*}
(V,\mathfrak{B}_1)\xrightarrow{\unit_V}(V,\mathfrak{B}_2)\xrightarrow{T}(W,\mathfrak{C}_2)\xrightarrow{\unit_W}(W,\mathfrak{C}_1)
\end{equation*}
We have \(T=\unit_W\circ T\circ\unit_V\text{.}\) By Exercise 4.8.5 we get the result.As a special case of the above theorem we obtain the following result.
Corollary 5.3.6.
Let \(V\) be finite-dimensional vector space over a field \(F\text{.}\) Suppose that \(\mathfrak{B}_{1}=(v_1,\ldots,v_n),\mathfrak{B}_{2}=(x_1,\ldots,x_n)\) are ordered bases of \(V\text{.}\) Let \(T\colon V\to V\) be an \(F\)-linear map. Let \([T]_{\mathfrak{B}_1}\) and \([T]_{\mathfrak{B}_2}\) be matrices of \(T\) with respect to bases \(\mathfrak{B}_1\) and \(\mathfrak{B}_2\text{,}\) respectively. Then, there exists an invertible \(n\times n\) matrix \(C\) such that
\begin{equation*}
[T]_{\mathfrak{B}_2}=C^{-1}\,[T]_{\mathfrak{B}_1}\,C.
\end{equation*}
Theorem 5.3.7.
Let \(V, W\) be finite-dimensional vector spaces over a field \(F\) and let \(T\colon V\to W\) be an \(F\)-linear map. Then there exists bases in which the matrix of \(T\) has a block-diagonal form
\begin{equation*}
\begin{pmatrix}I_r\amp 0_1\\0_2\amp 0_3\end{pmatrix},
\end{equation*}
where \(r=\dim_F\Im(T)\) and \(I_r\) is \(r\times r\) identity matrix, and \(0_i\) are zero matrices of appropriate size.Proof.
\begin{equation}
V=\ker(T)\bigoplus\langle u_1,\ldots,u_r\rangle.\tag{5.3.1}
\end{equation}
We claim that \(\{T(u_1),\ldots,T(u_r)\}\) is a basis of \(\Im(T)\text{.}\) Indeed, suppose that \(\sum_i\alpha_iT(u_i)=0\text{.}\) Hence, \(\sum\alpha_iu_i\in\ker(T)\) . By eq. (5.3.1) and linear independence of \(\{u_1,\ldots,u_r\}\text{,}\) we have \(\alpha_i=0\) for every \(i\text{.}\)
By Proposition 3.7.6, we write
\begin{equation*}
W=\langle T(u_1),\ldots,T(u_r)\rangle\bigoplus W^\prime.
\end{equation*}
Let \(\{w_1,\ldots,w_{m-r}\}\) be a basis of \(W^\prime\text{.}\)
Consider the ordered bases \((u_1,\ldots,u_r,v_1,\ldots,v_{n-r})\) and \(\big(T(u_1),\ldots,T(u_r),w_1,\ldots,w_{m-r}\big)\) of \(V\) and \(W\text{,}\) respectively. The matrix of \(T\) with respect to these bases is
\begin{equation*}
\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}.
\end{equation*}
Hence the theorem is proved.
