Fundamental Theorem of Linear Algebra

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In the previous post, we saw that for two \(\mathbb{K}\)-vector spaces \(V,W\) of dimensions \(n\) and \(m\) respectively, \(\Hom(V,W)\) is an \(mn\)-dimensional \(\mathbb{K}\)-vector space. The space \(\Mat_{m\times n}(\mathbb{K})\) of \(m\times n\) matrices is also an \(mn\)-dimensional \(\mathbb{K}\)-vector space. Thus, by §Isomorphisms, ⁋Corollary 4, these two vector spaces are isomorphic.

The fundamental theorem of linear algebra that we prove in this post shows not only that they are isomorphic merely because they have the same dimension, but that a natural isomorphism exists between them, so that they are in fact the same space.

Fundamental Theorem: Euclidean Spaces

In §Space of Linear Maps, we agreed to understand a linear map \(L\) satisfying

\[\begin{aligned}L(x_1)&=\alpha_{11}y_1+\alpha_{21}y_2+\cdots+\alpha_{m1}y_m\\L(x_2)&=\alpha_{12}y_1+\alpha_{22}y_2+\cdots+\alpha_{m2}y_m\\&\phantom{a}\vdots\\L(x_n)&=\alpha_{1n}y_1+\alpha_{2n}y_2+\cdots+\alpha_{mn}y_m\end{aligned}\]

via the correspondence

\[v=\sum_{i=1}^n v_ix_i\quad\mapsto\quad \sum_{j=1}^m\left(\sum_{i=1}^n\alpha_{ji}v_i\right)y_j=L(v)\tag{1}\]

In particular, if \(V=\mathbb{K}^n\), \(W=\mathbb{K}^m\), and each is equipped with the standard basis \(\mathcal{E}_n=\{e_1,\ldots, e_n\},\mathcal{E}_m=\{e_1,\ldots,e_m\}\), then the above correspondence can be written as

\[\begin{pmatrix}v_1\\v_2\\\vdots\\v_n\end{pmatrix}\quad\mapsto\quad\begin{pmatrix}\sum_{i=1}^n\alpha_{1i}v_i\\\sum_{i=1}^n\alpha_{2i}v_i\\\vdots\\\sum_{i=1}^n\alpha_{mi}v_i\end{pmatrix}\]

But the right-hand side is exactly the matrix-vector product

\[\begin{pmatrix}\alpha_{11}&\alpha_{12}&\cdots&\alpha_{1n}\\\alpha_{21}&\alpha_{22}&\cdots&\alpha_{2n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m1}&\alpha_{m2}&\cdots&\alpha_{mn}\end{pmatrix}\begin{pmatrix}v_1\\v_2\\\vdots\\v_n\end{pmatrix}\tag{2}\]

We call the \(m\times n\) matrix in the above expression the matrix representation of \(L\) with respect to \(\mathcal{E}_n,\mathcal{E}_m\), and denote it by \([L]^{\mathcal{E}_n}_{\mathcal{E}_m}\).

Conversely, one can verify that an \(m\times n\) matrix specifies a linear map in exactly the same way.

Example 1 Consider the Euclidean \(n\)-space \(\mathbb{K}^n\) and a matrix \(A\in\Mat_{m\times n}(\mathbb{K})\). For any \(x\in\mathbb{K}^n\), define \(L_A(x)\) by the formula

\[L_A(x)=Ax\]

Then \(L_A\) is a linear map from \(\mathbb{K}^n\) to \(\mathbb{K}^m\).

Given any linear map \(L\) from \(\mathbb{K}^n\) to \(\mathbb{K}^m\), one can easily verify that \(L=L_{[L]^{\mathcal{E}_n}_{\mathcal{E}_m}}\). Therefore, the following correspondence exists.

\[\{\text{linear maps from $\mathbb{K}^n$ to $\mathbb{K}^m$}\}\longleftrightarrow\Mat_{m\times n}(\mathbb{K})\]

More precisely, the maps \(L\mapsto [L]^{\mathcal{E}_n}_{\mathcal{E}_m}\) and \(A\mapsto L_A\) (the definition from Example 1) are bijections that are inverses of each other.

Since the set on the left is precisely \(\Hom(\mathbb{K}^n, \mathbb{K}^m)\), we can ask whether this correspondence is a bijective linear map, that is, an isomorphism. The answer is yes, and together with Theorem 3 below, we call this result the fundamental theorem of linear algebra.

Theorem 2 \(\Hom(\mathbb{K}^n,\mathbb{K}^m)\cong\Mat_{m\times n}(\mathbb{K})\)

Proof

We need to show that the given map \(L\mapsto[L]^{\mathcal{E}_n}_{\mathcal{E}_m}\) is linear.

Let \(L_1,L_2\) both be elements of \(\Hom(\mathbb{K}^n,\mathbb{K}^m)\). Then for each \(e_i\in\mathcal{E}_n\), there exist families of scalars \((\alpha_{i,j})\) and \((\beta_{i,j})\) such that

\[\begin{aligned}L_1(e_1)&=\alpha_{1,1}e_1+\alpha_{2,1}e_2+\cdots+\alpha_{m,1}e_m\\L_1(e_2)&=\alpha_{1,2}e_1+\alpha_{2,2}e_2+\cdots+\alpha_{m,2}e_m\\&\vdots\\L_1(e_n)&=\alpha_{1,n}e_1+\alpha_{2,n}e_2+\cdots+\alpha_{m,n}e_m\end{aligned}\]

and

\[\begin{aligned}L_2(e_1)&=\beta_{1,1}e_1+\beta_{2,1}e_2+\cdots+\beta_{m,1}e_m\\L_2(e_2)&=\beta_{1,2}e_1+\beta_{2,2}e_2+\cdots+\beta_{m,2}e_m\\&\vdots\\L_2(e_n)&=\beta_{1,n}e_1+\beta_{2,n}e_2+\cdots+\beta_{m,n}e_m\end{aligned}\]

Now,

\[\begin{aligned}(L_1+L_2)(e_1)&=(\alpha_{1,1}+\beta_{1,1})e_1+(\alpha_{2,1}+\beta_{2,1})e_2+\cdots+(\alpha_{m,1}+\beta_{m,1})e_m\\(L_1+L_2)(e_2)&=(\alpha_{1,2}+\beta_{1,2})e_1+(\alpha_{2,2}+\beta_{2,2})e_2+\cdots+(\alpha_{m,2}+\beta_{m,2})e_m\\&\vdots\\(L_1+L_2)(e_n)&=(\alpha_{1,n}+\beta_{1,n})e_1+(\alpha_{2,n}+\beta_{2,n})e_2+\cdots+(\alpha_{m,n}+\beta_{m,n})e_m\end{aligned}\]

and therefore the matrix representation \([L_1+L_2]^{\mathcal{E}_n}_{\mathcal{E}_m}\) of \(L_1+L_2\) is exactly \([L_1]^{\mathcal{E}_n}_{\mathcal{E}_m}+[L_2]^{\mathcal{E}_n}_{\mathcal{E}_m}\). Similarly, the scalar multiplication property also holds.

Moreover, matrix multiplication also has a special meaning in \(\Hom(\mathbb{K}^n,\mathbb{K}^m)\).

Theorem 3 Let three Euclidean spaces \(\mathbb{K}^n,\mathbb{K}^m,\mathbb{K}^k\) be given. Then for any \(L_1:\mathbb{K}^n\rightarrow \mathbb{K}^m\), \(L_2:\mathbb{K}^m\rightarrow \mathbb{K}^k\), we always have

\[[L_2\circ L_1]^{\mathcal{E}_n}_{\mathcal{E}_k}=[L_2]^{\mathcal{E}_m}_{\mathcal{E}_k}[L_1]^{\mathcal{E}_n}_{\mathcal{E}_m}\]

That is, composition of linear maps corresponds to matrix multiplication.

Proof

To determine the left-hand side \([L_2\circ L_1]^{\mathcal{E}_n}_{\mathcal{E}_k}\), we only need to check where the elements \(e_i\) of \(\mathcal{E}_n\) are mapped by \(L_2\circ L_1\). Suppose \(L_1\) and \(L_2\) are given by

\[[L_1]^{\mathcal{E}_n}_{\mathcal{E}_m}=\begin{pmatrix}\alpha_{1,1}&\alpha_{1,2}&\cdots&\alpha_{1,n}\\\alpha_{2,1}&\alpha_{2,2}&\cdots&\alpha_{2,n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m,1}&\alpha_{m,2}&\cdots&\alpha_{m,n}\end{pmatrix},\quad[L_2]^{\mathcal{E}_m}_{\mathcal{E}_k}=\begin{pmatrix}\beta_{1,1}&\beta_{1,2}&\cdots&\beta_{1,m}\\\beta_{2,1}&\beta_{2,2}&\cdots&\beta_{2,m}\\\vdots&\vdots&\ddots&\vdots\\\beta_{k,1}&\beta_{k,2}&\cdots&\beta_{k,m}\end{pmatrix}\]

Then with a bit of calculation,

\[\begin{aligned}(L_2\circ L_1)(e_i)&=L_2(\alpha_{1,i}e_1+\cdots+\alpha_{m,i}e_m)\\&=\alpha_{1,i}L_2(e_1)+\alpha_{2,i}L_2(e_2)+\cdots+\alpha_{m,i}L(e_m)\\&=\alpha_{1,i}(\beta_{1,1}e_1+\beta_{2,1}e_2+\cdots+\beta_{k,1}e_k)\\&\phantom{==}+\alpha_{2,i}(\beta_{1,2}e_1+\beta_{2,2}e_2+\cdots+\beta_{k,2}e_k)\\&\phantom{===}+\cdots\\&\phantom{====}+\alpha_{m,i}(\beta_{1,m}e_1+\beta_{2,m}e_2+\cdots+\beta_{k,m}e_k)\end{aligned}\]

Now grouping the above expression by the basis elements \(e_1,\ldots, e_k\) of \(\mathbb{K}^k\), we get

\[(L_2\circ L_1)(e_i)=\left(\sum_{l=1}^m\alpha_{l,i}\beta_{1,l}\right)e_1+\cdots+\left(\sum_{l=1}^m\alpha_{l,i}\beta_{k,l}\right)e_k.\]

Since the \(i\)-th column of \([L_2\circ L_1]^{\mathcal{E}_n}_{\mathcal{E}_k}\) is the vector to which \(e_i\) is mapped by \(L_2\circ L_1\), the entry in row \(j\), column \(i\) of the matrix \([L_2\circ L_1]^{\mathcal{E}_n}_{\mathcal{E}_k}\) is the \(j\)-th component \(\sum_{l=1}^m\alpha_{l,i}\beta_{j,l}\) of this vector. Now from the calculation immediately following §Matrices, ⁋Definition 2, we know that this is the \((i,j)\)-entry of the product of the two matrices \([L_2]_{\mathcal{E}_k}^{\mathcal{E}_m}\) and \([L_1]_{\mathcal{E}_m}^{\mathcal{E}_n}\).

Fundamental Theorem: General Case

The fundamental theorem we proved above applies only to Euclidean spaces, but with a very small modification it also holds for general finite-dimensional \(\mathbb{K}\)-vector spaces. This process can be summarized simply by the following diagram.

For any finite-dimensional \(\mathbb{K}\)-vector space \(V\) and a basis \(\mathcal{B}=\{x_1,\ldots, x_n\}\), the coordinate representation is the following isomorphism

\[v=\sum_{i=1}^n v_ix_i\mapsto [v]_\mathcal{B}=\begin{pmatrix}v_1\\v_2\\\vdots\\v_n\end{pmatrix}\in\mathbb{K}^n\]

Similarly, let another finite-dimensional \(\mathbb{K}\)-vector space \(W\) and its basis \(\mathcal{C}=\{y_1,\ldots, y_m\}\) be given, and suppose a linear map \(L:V\rightarrow W\) is determined by the formula

\[\begin{aligned}L(x_1)&=\alpha_{1,1}y_1+\alpha_{2,1}y_2+\cdots+\alpha_{m,1}y_m\\L(x_2)&=\alpha_{1,2}y_1+\alpha_{2,2}y_2+\cdots+\alpha_{m,2}y_m\\&\vdots\\L(x_n)&=\alpha_{1,n}y_1+\alpha_{2,n}y_2+\cdots+\alpha_{m,n}y_m\end{aligned}\]

Then the matrix representation \([L]^\mathcal{B}_\mathcal{C}\) of \(L\) with respect to \(\mathcal{B},\mathcal{C}\) is this time defined by the formula

\[[L]^\mathcal{B}_\mathcal{C}=\begin{pmatrix}\alpha_{1,1}&\alpha_{1,2}&\cdots&\alpha_{1,n}\\\alpha_{2,1}&\alpha_{2,2}&\cdots&\alpha_{2,n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m,1}&\alpha_{m,2}&\cdots&\alpha_{m,n}\end{pmatrix}\]

Now comparing equations (2) and (1), one can verify that for any \(v\in V\), the coordinate representation of \(L(v)\) with respect to \(\mathcal{C}\) is given by

\[[L(v)]_\mathcal{C}=[L]^\mathcal{B}_\mathcal{C}[v]_\mathcal{B}\tag{3}\]

Then the general version of Theorem 2 is given by the following theorem.

Theorem 4 \(\Hom(V,W)\cong \Mat_{m\times n}(\mathbb{K})\).

Proof

Fix bases \(\mathcal{B}\) of \(V\) and \(\mathcal{C}\) of \(W\). We need to show that the map \(L\mapsto[L]^\mathcal{B}_\mathcal{C}\) is linear.

Let \(L_1,L_2\) both be elements of \(\Hom(V,W)\). Then for each \(x_i\in\mathcal{B}\), there exist families of scalars \((\alpha_{i,j})\) and \((\beta_{i,j})\) such that

\[\begin{aligned}L_1(x_1)&=\alpha_{1,1}y_1+\alpha_{2,1}y_2+\cdots+\alpha_{m,1}y_m\\L_1(x_2)&=\alpha_{1,2}y_1+\alpha_{2,2}y_2+\cdots+\alpha_{m,2}y_m\\&\vdots\\L_1(x_n)&=\alpha_{1,n}y_1+\alpha_{2,n}y_2+\cdots+\alpha_{m,n}y_m\end{aligned}\]

and

\[\begin{aligned}L_2(x_1)&=\beta_{1,1}y_1+\beta_{2,1}y_2+\cdots+\beta_{m,1}y_m\\L_2(x_2)&=\beta_{1,2}y_1+\beta_{2,2}y_2+\cdots+\beta_{m,2}y_m\\&\vdots\\L_2(x_n)&=\beta_{1,n}y_1+\beta_{2,n}y_2+\cdots+\beta_{m,n}y_m\end{aligned}\]

Now,

\[\begin{aligned}(L_1+L_2)(x_1)&=(\alpha_{1,1}+\beta_{1,1})y_1+(\alpha_{2,1}+\beta_{2,1})y_2+\cdots+(\alpha_{m,1}+\beta_{m,1})y_m\\(L_1+L_2)(x_2)&=(\alpha_{1,2}+\beta_{1,2})y_1+(\alpha_{2,2}+\beta_{2,2})y_2+\cdots+(\alpha_{m,2}+\beta_{m,2})y_m\\&\vdots\\(L_1+L_2)(x_n)&=(\alpha_{1,n}+\beta_{1,n})y_1+(\alpha_{2,n}+\beta_{2,n})y_2+\cdots+(\alpha_{m,n}+\beta_{m,n})y_m\end{aligned}\]

and therefore the matrix representation \([L_1+L_2]^\mathcal{B}_\mathcal{C}\) of \(L_1+L_2\) is exactly \([L_1]^\mathcal{B}_\mathcal{C}+[L_2]^\mathcal{B}_\mathcal{C}\). Similarly, the scalar multiplication property also holds.

Theorem 3 also admits a similar generalization.

Theorem 5 Let three \(\mathbb{K}\)-vector spaces \(V_1,V_2,V_3\) and bases \(\mathcal{B}_1=\{x_1,\ldots,x_n\}\), \(\mathcal{B}_2=\{y_1,\ldots, y_m\}\), \(\mathcal{B}_3=\{z_1,\ldots, z_k\}\) of each be given. Then for any \(L_1:V_1\rightarrow V_2\), \(L_2:V_2\rightarrow V_3\), we always have

\[[L_2\circ L_1]^{\mathcal{B}_1}_{\mathcal{B}_3}=[L_2]^{\mathcal{B}_2}_{\mathcal{B}_3}[L_1]^{\mathcal{B}_1}_{\mathcal{B}_2}\]

That is, composition of linear maps corresponds to matrix multiplication.

Proof

To determine the left-hand side \([L_2\circ L_1]^{\mathcal{B}_1}_{\mathcal{B}_3}\), we only need to check where the elements of \(\mathcal{B}_1\) are mapped by \(L_2\circ L_1\). Suppose \(L_1\) and \(L_2\) are given by

\[[L_1]^{\mathcal{B}_1}_{\mathcal{B}_2}=\begin{pmatrix}\alpha_{1,1}&\alpha_{1,2}&\cdots&\alpha_{1,n}\\\alpha_{2,1}&\alpha_{2,2}&\cdots&\alpha_{2,n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m,1}&\alpha_{m,2}&\cdots&\alpha_{m,n}\end{pmatrix},\quad[L_2]^{\mathcal{B}_2}_{\mathcal{B}_3}=\begin{pmatrix}\beta_{1,1}&\beta_{1,2}&\cdots&\beta_{1,m}\\\beta_{2,1}&\beta_{2,2}&\cdots&\beta_{2,m}\\\vdots&\vdots&\ddots&\vdots\\\beta_{k,1}&\beta_{k,2}&\cdots&\beta_{k,m}\end{pmatrix}\]

Then with a bit of calculation,

\[\begin{aligned}(L_2\circ L_1)(x_i)&=L_2(\alpha_{1,i}y_1+\cdots+\alpha_{m,i}y_m)\\&=\alpha_{1,i}L_2(y_1)+\alpha_{2,i}L_2(y_2)+\cdots+\alpha_{m,i}L(y_m)\\&=\alpha_{1,i}(\beta_{1,1}z_1+\beta_{2,1}z_2+\cdots+\beta_{k,1}z_k)\\&\phantom{==}+\alpha_{2,i}(\beta_{1,2}z_1+\beta_{2,2}z_2+\cdots+\beta_{k,2}z_k)\\&\phantom{===}+\cdots\\&\phantom{====}+\alpha_{m,i}(\beta_{1,m}z_1+\beta_{2,m}z_2+\cdots+\beta_{k,m}z_k)\end{aligned}\]

Now, grouping the above expression by the \(z\)’s, we get

\[(L_2\circ L_1)(x_i)=\left(\sum_{l=1}^m\alpha_{l,i}\beta_{1,l}\right)z_1+\cdots+\left(\sum_{l=1}^m\alpha_{l,i}\beta_{k,l}\right)z_k\]

As we verified earlier, the \(i\)-th column of \([L_2\circ L_1]^{\mathcal{B}_1}_{\mathcal{B}_3}\) is exactly the coordinate representation in \(\mathcal{B}_3\) of the vector to which \(x_i\) is mapped by \(L_2\circ L_1\), so the entry in row \(j\), column \(i\) of the matrix \([L_2\circ L_1]^{\mathcal{B}_1}_{\mathcal{B}_3}\) is the \(j\)-th component \(\sum_{l=1}^m\alpha_{l,i}\beta_{j,l}\) of this vector. Just as in Theorem 3, this component is the \((i,j)\)-entry of the matrix product \([L_2]^{\mathcal{B}_2}_{\mathcal{B}_3}[L_1]^{\mathcal{B}_1}_{\mathcal{B}_2}\), so the proof is complete.

Theorem 4 above shows that once we choose bases for \(V,W\), we can treat \(\Hom(V,W)\) and \(\Mat_{m\times n}(\mathbb{K})\) as the same thing. For instance, the \(mn\) bases of \(\Mat_{m\times n}(\mathbb{K})\) correspond to the \(mn\) bases examined in §Space of Linear Maps, ⁋Proposition 5. The following corollary is also a consequence of the fundamental theorem.

Corollary 6 Let two \(n\)-dimensional \(\mathbb{K}\)-vector spaces \(V,W\) be given, and fix their bases \(\mathcal{B},\mathcal{C}\). Then for any \(L\in\Hom(V,W)\), the matrix representation \([L^{-1}]^{\mathcal{C}}_{\mathcal{B}}\) of \(L^{-1}\in\Hom(W,V)\) with respect to the bases \(\mathcal{C},\mathcal{B}\) is equal to the inverse matrix of \([L]^{\mathcal{B}}_\mathcal{C}\).

Proof

Obvious by the uniqueness of inverse matrices and inverse functions.

In this way, most of the concepts defined in §Matrices can be transferred to \(\Hom(V,W)\). One concept that cannot be immediately transferred is the transpose matrix \(A^t\); we will understand its meaning later when we examine dual spaces.

Change-of-Basis Matrices

To summarize Theorem 4 in one sentence: a linear map from an \(n\)-dimensional vector space \(V\) to an \(m\)-dimensional vector space \(W\) can be represented as an \(m\times n\) matrix once we fix bases \(\mathcal{B},\mathcal{C}\) for each, and conversely any \(m\times n\) matrix can be understood as a linear map. Then one natural question is what happens when we change the basis, and in fact the answer is already given in Theorem 5.

Definition 7 For any finite-dimensional \(\mathbb{K}\)-vector space \(V\) and two bases \(\mathcal{B},\mathcal{B}'\) of \(V\), the change-of-basis matrix from \(\mathcal{B}\) to \(\mathcal{B}'\) is

\[[\id_V]_{\mathcal{B}'}^\mathcal{B}\]

That this matrix must be square is obvious from the fact that the dimension of a vector space is well-defined. Also, from the equation

\[I=[\id_V]^{\mathcal{B}}_{\mathcal{B}}=[\id_V]_{\mathcal{B}}^{\mathcal{B}'}[\id_V]^\mathcal{B}_{\mathcal{B}'}\]

we see that such a matrix is always invertible.

To see how a change-of-basis matrix operates, fix a finite-dimensional \(\mathbb{K}\)-vector space \(V\) and let two bases \(\mathcal{B},\mathcal{B}'\) defined on \(V\) be given. The fundamental theorem of linear algebra means that the following diagram commutes.

Here the two vertical maps are \(v\mapsto [v]_\mathcal{B}\) and \(v\mapsto[v]_{\mathcal{B}'}\) respectively. Therefore, we can think of the change-of-basis matrix as the matrix that takes the coordinate representation of \(v\in V\) with respect to \(\mathcal{B}\) and converts it to the coordinate representation with respect to \(\mathcal{B}'\). More generally, given any linear map \(L:V\rightarrow W\) and bases \(\mathcal{B},\mathcal{C}\) of \(V,W\) as well as another pair of bases \(\mathcal{B}',\mathcal{C}'\), the fundamental theorem of linear algebra gives us the equation

\[[L]_{\mathcal{C}'}^{\mathcal{B}'}=[\id_W]_{\mathcal{C}'}^\mathcal{C}[L]_{\mathcal{C}}^\mathcal{B}[\id_V]^{\mathcal{B}'}_{\mathcal{B}}\]

Let two \(m\times n\) matrices \(A,B\) be given. Then from the above equation, if there exist suitable invertible matrices \(P,Q\) satisfying

\[B=PAQ\]

one is tempted to treat \(A\) and \(B\) as the same thing. This means that given a fixed linear map \(L\), we consider all matrix representations obtained by choosing bases for the domain and codomain of \(L\) to be the same.

However, despite this plausible motivation, the result is not very good. If we can change bases for both the domain and codomain of \(L\), then by choosing an arbitrary basis \(\{x_1,\ldots, x_n\}\) for the domain and then selecting the linearly independent elements \(L(x_1),\ldots, L(x_k)\) from among \(L(x_1),\ldots, L(x_n)\) and extending them to a basis for the codomain using §Dimension of Vector Spaces, ⁋Proposition 6, this linear map can always be represented in block matrix form

\[\begin{pmatrix}I&O\\O&O\end{pmatrix}\]

That is, if we classify matrix representations of \(L\) in this way, the only thing that matters is the rank of \(L\).

Therefore, we need to define a finer relation than this equivalence relation.

Definition 8 Let arbitrary \(n\times n\) matrices \(A,B\) be given. Then \(A\) and \(B\) are called similar matrices if there exists an invertible matrix \(P\) such that \(A=PBP^{-1}\).

That two matrices \(A,B\) are similar means that if we think of \(A\) as the matrix representation of a linear transformation \(L:V\rightarrow V\) with respect to a basis \(\mathcal{B}\) of a fixed vector space \(V\), then there exists a suitable basis \(\mathcal{C}\) such that we can think of \(B\) as the matrix representation of \(L\) with respect to the basis \(\mathcal{C}\). Then in this case

\[A=[L]_{\mathcal{B}}^\mathcal{B}=[\id_V]^\mathcal{B}_\mathcal{C}[L]^\mathcal{C}_\mathcal{C}[\id_V]^\mathcal{C}_\mathcal{B}=PBP^{-1}\]

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References

[Lee] 이인석, 선형대수와 군, 서울대학교 출판문화원, 2005.

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