Eigenspace Decomposition

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Sums of Spaces

Consider an \(n\times n\) matrix \(A\) and an eigenvalue \(\lambda\). By definition, any vector \(v\) in the eigenspace \(E_\lambda\) corresponding to \(\lambda\) must satisfy the equation

\[Av=\lambda v\]

Therefore, when restricted to \(E_\lambda\), \(A\) becomes the very simple map \(v\mapsto \lambda v\).

More generally, think of \(A\) as a linear map from \(\mathbb{K}^n\) to \(\mathbb{K}^n\), and assume that the domain \(\mathbb{K}^n\) can be covered by the eigenspaces \(E_\lambda\). That is, suppose

\[\mathbb{K}^n=\span\left(\bigcup_{\lambda\in\sigma(A)}E_\lambda\right)\]

Then for any \(v\in\mathbb{K}^n\), there exist \(v_\lambda\in E_\lambda\) such that we can write

\[v=\sum_{\lambda\in\sigma(A)}v_\lambda\]

and therefore

\[Av=A\left(\sum_{\lambda\in\sigma(A)}v_\lambda\right)=\sum_{\lambda\in\sigma(A)}Av_\lambda\]

and by the argument above \(Av_\lambda=\lambda v_\lambda\), so we obtain the equation

\[Av=\sum_{\lambda\in\sigma(A)}\lambda v_\lambda\]

Of course, for this calculation to make sense, the way of expressing \(v\) as a sum of \(v_\lambda\) must be unique. We define this as follows.

Definition 1 A \(\mathbb{K}\)-vector space \(V\) is called the direct sum of its subspaces \((W_i)_{i\in I}\) if for every \(v\in V\) there exists a (necessarily unique) family \((v_i)_{i\in I}\) such that

\[v=\sum_{i\in I} v_i\]

holds.¹ We write this as \(V=\bigoplus_{i\in I}W_i\).

The easiest nontrivial case is when \(I\) is a two-element set.

Proposition 2 For two subspaces \(W_1,W_2\) of a \(\mathbb{K}\)-vector space \(V\), the condition \(V=W_1\oplus W_2\) is equivalent to \(V=W_1+W_2\) and \(W_1\cap W_2=\{0\}\).

Proof

First assume \(V=W_1\oplus W_2\). It is obvious from the definition that \(W_1+W_2\subseteq V\). Conversely, pick any \(v\in V\); then there exist \(w_i\in W_i\) such that \(v=w_1+w_2\), so \(V\subseteq W_1+W_2\) also holds. Hence we know \(V=W_1+W_2\). On the other hand, if \(W_1\cap W_2\neq \{0\}\), then for a nonzero \(w\in W_1+W_2\) we have

\[w=0+w=w+0\]

which contradicts the uniqueness in Definition 1.

Conversely, suppose \(V=W_1+W_2\) and \(W_1\cap W_2=\{0\}\). For any \(v\in V\), since \(V=W_1+W_2\), there must exist \(w_1\in W_i\) such that \(v=w_1+w_2\). Moreover, this expression is unique. If

\[v=w_1+w_2=w_1'+w_2'\]

then from

\[w_1-w_1'=w_2-w_2'\]

the left-hand side is an element of \(W_1\) and the right-hand side is an element of \(W_2\), so the condition \(W_1\cap W_2=\{0\}\) implies \(w_1-w_1'=w_2-w_2'=0\).

One direction of the above proposition still holds when \(I\) has three or more elements. That is, if \(V=\bigoplus_{i\in I}W_i\), then \(V=\sum_{i\in I}W_i\) and \(W_i\cap W_j=\{0\}\) whenever \(i\neq j\), and the proof is the same as above. However, the converse does not hold in general.

For example, let \(V=\mathbb{R}^2\) and take the two standard basis vectors \(e_1,e_2\) of \(V\). Set \(W_1=\mathbb{R}e_1\), \(W_2=\mathbb{R}e_2\), \(W_3=\mathbb{R}(e_1+e_2)\). Then \(V=W_1+W_2+W_3\), and \(W_i\cap W_j\) whenever \(i\neq j\), but \(V\neq W_1\oplus W_2\oplus W_3\).

\[e_1+e_2=e_1+e_2+0=0+0+(e_1+e_2)\]

because the way of representing \(e_1+e_2\in V\) is not unique, as above.

As another example, pick a basis \(\mathcal{B}=\{x_1,\ldots, x_n\}\) of \(V\). If we set \(W_i=\mathbb{K}x_i\), then the condition that \(\mathcal{B}\) is a basis coincides exactly with the condition that \(V\) is the direct sum of the \(W_i\). More generally, the following holds.

Proposition 3 For any \(\mathbb{K}\)-vector space \(V\) and subspaces \((W_i)_{i\in I}\), the condition \(V=\bigoplus_{i\in I} W_i\) is equivalent to the bases \(\mathcal{B}_i\) of \(W_i\) satisfying \(\mathcal{B}_i\cap\mathcal{B}_j=\emptyset\) whenever \(i\neq j\) and \(\bigcup_{i\in I}\mathcal{B}_i\) being a basis of \(V\).

Proof

First assume \(V=\bigoplus W_i\) and pick bases \(\mathcal{B}_i\) of the \(W_i\). If \(\mathcal{B}_i\cap\mathcal{B}_j\neq\emptyset\), then \(W_i\cap W_j\neq\emptyset\), contradicting the discussion after Proposition 2; hence we must have \(\mathcal{B}_i\cap\mathcal{B}_j=\emptyset\). For any \(v\in V\), since \(V=\bigoplus W_i\), there exist unique \(w_i\) satisfying the equation

\[v=\sum_{i\in I} w_i\]

Moreover, in each \(W_i\) the \(w_i\) can be uniquely expressed as a linear combination of elements of \(\mathcal{B}_i\). From this we see that \(\bigcup\mathcal{B}_i\) is a basis of \(V\).

Reversing this argument shows the converse as well.

Therefore we see that \(\dim V=\sum_{i\in I}\dim W_i\).

Diagonalization

Now we examine how to decompose \(\mathbb{K}^n\) into eigenspaces. From the preceding Proposition 3 we know that decomposing the vector space \(\mathbb{K}^n\) into eigenspaces \(E_\lambda\) is the same as collecting bases of the \(E_\lambda\) to obtain a basis of \(\mathbb{K}^n\). Also, if a nonzero vector \(x_1\) is an eigenvector corresponding to the eigenvalue \(\lambda_1\), then for another eigenvalue \(\lambda_2\) we have

\[Ax_1=\lambda_1x_1\neq\lambda_2 x_1\]

so we know \(x_1\not\in E_{\lambda_2}\). Therefore, however we choose bases of the \(E_\lambda\), the bases of \(E_{\lambda_1}\) and \(E_{\lambda_2}\) never overlap for distinct \(\lambda_1,\lambda_2\). Moreover, the following holds.

Proposition 4 For any matrix \(A\), suppose \(x_1,\ldots, x_m\) are eigenvectors corresponding to distinct eigenvalues \(\lambda_1,\ldots,\lambda_m\). Then the set \(\{x_1,\ldots,x_m\}\) is linearly independent.

Proof

Suppose for contradiction that the set \(\{x_1,x_2,\ldots, x_m\}\) is linearly dependent. That is, there exist scalars \(\alpha_i\), not all zero, satisfying the following equation

\[\alpha_1x_1+\alpha_2x_2+\cdots+\alpha_mx_m=0\tag{1}\]

Now among the families \((\alpha_i)_{1\leq i\leq m}\) satisfying this, choose one with minimal support, and denote it by \((\beta_i)_{1\leq i\leq m}\). In other words, if the number of indices \(i\) with \(\beta_i\neq0\) is \(k\), then no family \((\alpha_i)_{1\leq i\leq m}\) with fewer than \(k\) nonzero \(\alpha_i\) satisfies equation (1) above.

Since at least two of the \(\beta_i\) are nonzero, we may assume without loss of generality that \(\beta_m\neq 0\). Then

\[x_m=\sum_{i=1}^{m-1}\left(-\frac{\beta_i}{\beta_m}\right)x_i\]

For convenience write this as \(x_m=\sum_{i=1}^{m-1}\beta'_ix_i\). Multiplying both sides by \(A\) gives

\[Ax_m=\sum_{i=1}^{m-1}\beta'_i(Ax_i)\]

and since these are eigenvectors,

\[\lambda_mx_m=\sum_{i=1}^{m-1}\beta'_i\lambda_i x_i\]

Multiplying both sides of \(x_m=\sum_{i=1}^{m-1}\beta'_ix_i\) by \(\lambda_m\) gives

\[\lambda_mx_m=\sum_{i=1}^{m-1}\beta_i'\lambda_mx_i\]

so combining this with the previously obtained equation yields

\[0=\sum_{i=1}^{m-1}\beta_i'(\lambda_i-\lambda_m)x_i\]

and since \(\beta_i'=-(\beta_i/\beta_m)\), multiplying both sides by \(\beta_m\) and simplifying gives

\[0=\sum_{i=1}^{m-1}\beta_i(\lambda_i-\lambda_m)x_i\]

If we define \((\beta''_i)_{1\leq i\leq n}\) by the following formula

\[\beta_i''=\begin{cases}\beta_i(\lambda_i-\lambda_m)&1\leq i\leq m-1\\0&i=m\end{cases}\]

then the above equation becomes

\[\beta_1''x_1+\beta_2''x_2+\cdots+\beta_m''x_m=0\]

By assumption \(\lambda_i-\lambda_m\neq 0\), so for \(1\leq i\leq m-1\) the conditions \(\beta_i''=0\) and \(\beta_i=0\) are equivalent. Hence there are \(k-1\) indices \(1\leq i\leq m-1\) with \(\beta_i''\neq 0\), and since \(\beta_m''=0\), the size of \(\supp(\beta_i'')_{1\leq i\leq m}\) is \(k-1\). This contradicts the minimality of \((\beta_i)_{1\leq i\leq m}\), so the set \(\{x_1,x_2,\ldots, x_m\}\) is linearly independent.

From this we know that for any matrix \(A\) with eigenvalues \(\lambda\in\sigma(A)\) and corresponding eigenspaces \(E_\lambda\), if we denote their bases by \(\mathcal{B}_\lambda\), then \(\mathcal{B}=\bigcup_{\lambda\in\sigma(A)}\mathcal{B}_\lambda\) is a linearly independent subset of \(\mathbb{K}^n\). However, in general there is no reason for \(\mathcal{B}\) to be a basis of \(\mathbb{K}^n\). For example, in §Characteristic Polynomial, ⁋Example 7, when \(\mathbb{K}=\mathbb{R}\) we have \(\sigma(J)=\emptyset\), so \(\mathcal{B}=\emptyset\).

Moreover, even assuming that the characteristic polynomial of \(A\) has exactly \(n\) roots counting multiplicity, a similar problem can arise: for example, for the matrix

\[A=\begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix}\]

the characteristic polynomial is \((\mathbf{x}-1)^3=0\), but one can check that the only eigenvector for the eigenvalue \(1\) is \((1,0,0)\). In the language introduced in the previous post, the algebraic multiplicity of the eigenvalue \(1\) is \(3\), and its geometric multiplicity is \(1\).

The next proposition shows that the geometric multiplicity of an eigenvalue of a matrix can never exceed its algebraic multiplicity.

Proposition 5 For an eigenvalue \(\lambda\in\mathbb{K}\) of an \(n\times n\) matrix \(A\), the geometric multiplicity of \(\lambda\) can never exceed the algebraic multiplicity of \(\lambda\).

Proof

Let the geometric multiplicity of \(\lambda\) be \(k\), and consider \(k\) linearly independent vectors \(x_1,\ldots, x_k\) spanning \(E_\lambda(A)\). To these we can add \((n-k)\) vectors \(x_{k+1},\ldots, x_k\) to form a new basis \(\{x_1,\ldots, x_n\}\) of \(\mathbb{K}^n\). Now if we define the matrix \(X\) as

\[X=(x_1|x_2|\cdots|x_n)\]

then the columns of \(X\) are linearly independent, so \(X^{-1}\) exists. Let the rows of \(X^{-1}\) be \(y_i\). From the equation \(X^{-1}X=XX^{-1}=I\) we obtain

\[y_i\cdot x_j=\begin{cases}1&i=j\\ 0&i\neq j\end{cases}\]

Therefore, setting \(A'=X^{-1}AX\), we obtain

\[\begin{aligned}A'&=X^{-1}(AX)=\begin{pmatrix}y_1\\ y_2\\ \vdots\\ y_n\end{pmatrix}(Ax_1|Ax_2|\cdots|Ax_n)\\ &=\begin{pmatrix}y_1\cdot Ax_1&y_1\cdot Ax_2&\cdots& y_1\cdot Ax_k&\cdots&y_1\cdot Ax_n\\ y_2\cdot Ax_1&y_2\cdot Ax_2&\cdots &y_2\cdot Ax_k&\cdots &y_2\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ y_k\cdot Ax_1&y_k\cdot Ax_2&\cdots&y_k\cdot Ax_k&\cdots&y_k\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ y_n\cdot Ax_1&y_n\cdot Ax_2&\cdots &y_n\cdot Ax_k&\cdots&y_n\cdot Ax_n \end{pmatrix}\\ &=\begin{pmatrix}y_1\cdot (\lambda x_1)&y_1\cdot (\lambda x_2)&\cdots& y_1\cdot (\lambda x_k)&\cdots&y_1\cdot Ax_n\\ y_2\cdot (\lambda x_1)&y_2\cdot (\lambda x_2)&\cdots &y_2\cdot (\lambda x_k)&\cdots &y_2\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ y_k\cdot (\lambda x_1)&y_k\cdot (\lambda x_2)&\cdots&y_k\cdot (\lambda x_k)&\cdots&y_k\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ y_n\cdot (\lambda x_1)&y_n\cdot (\lambda x_2)&\cdots &y_n\cdot (\lambda x_k)&\cdots&y_n\cdot Ax_n \end{pmatrix}\\ &=\begin{pmatrix}\lambda&0&\cdots& 0&\cdots&y_1\cdot Ax_n\\ 0&\lambda&\cdots &0&\cdots &y_2\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda&\cdots&y_k\cdot Ax_n\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&\cdots &0&\cdots&y_n\cdot Ax_n \end{pmatrix}\\ &=\begin{pmatrix}\lambda I_k&B\\ 0&C\end{pmatrix}\end{aligned}\]

Therefore, writing the characteristic polynomial of \(A\) as \(p_A(\mathbf{x})\), from §Characteristic Polynomial, ⁋Theorem 4 we have \(p_A(\mathbf{x})=p_{A'}(\mathbf{x})\), and hence we know

\[p_A(\mathbf{x}=p_{A'}(\mathbf{x})=\det(\mathbf{x}I-A')=(\mathbf{x}-\lambda)^k\det(\mathbf{x}I_{n-k}-C)\]

In other words, the algebraic multiplicity of \(\lambda\) in \(p_A\) is at least \(k\).

Given an \(n\times n\) matrix \(A\), let \(p_A\) be its characteristic polynomial. Then the sum of the algebraic multiplicities of the eigenvalues \(\lambda\) cannot exceed \(n\), the degree of \(p_A\). Also, for a fixed eigenvalue \(\lambda\), the above proposition shows that the geometric multiplicity of \(\lambda\) cannot exceed its algebraic multiplicity. Finally, from the argument after Proposition 4 we see that in order to decompose \(\mathbb{K}^n\) into eigenspaces, the sum of the geometric multiplicities of the \(\lambda\) must be equal to \(n\). Combining all of this yields the following proposition.

Proposition 6 For any \(n\times n\) matrix \(A\), the necessary and sufficient condition for \(\mathbb{K}^n\) to be expressible as the direct sum of the eigenspaces of \(A\) is

1. the characteristic polynomial of \(A\) has exactly \(n\) roots counting multiplicity, and
2. for each eigenvalue the geometric multiplicity equals the algebraic multiplicity.

In particular, if \(\mathbb{K}\) is an algebraically closed field then the first condition is always satisfied, so only the second condition needs to be considered.

Throughout our discussion of matrix diagonalization we assume that the field \(\mathbb{K}\) is algebraically closed. This is merely for convenience; if \(\mathbb{K}\) is not algebraically closed, one need only consider the field extension obtained by adjoining the roots of the characteristic polynomial of the matrix in question. ([Field Theory] §Algebraic Extensions) This is exactly the same as, for example, adjoining the imaginary root \(i\) of the equation \(\x^2+1=0\) to obtain \(\mathbb{C}\) from \(\mathbb{R}\).

Diagonalization of Matrices

Earlier we examined how to decompose \(\mathbb{R}^n\) through the eigenvalues and eigenspaces of \(A\), and from Proposition 6 we also learned when such a decomposition is possible. Let us revisit the proof of Proposition 5 that we used to show this. We added \(n-k\) arbitrary vectors to a basis \(x_1,\ldots, x_k\) of \(E_\lambda\), defined the matrix \(X=(x_1\mid\cdots\mid x_n)\) using these, and computed that the upper-left \(k\times k\) block of

\[XAX^{-1}=\begin{pmatrix}\lambda I_k&B\\0&C\end{pmatrix}\]

becomes the diagonal matrix \(\lambda I_k\). However, if \(A\) satisfies all the conditions of Proposition 6, instead of adding the \(n-k\) vectors \(x_{k+1},\ldots, x_n\) arbitrarily, we can choose the \(n\) vectors \(x_1,\ldots, x_n\) so that they all come from bases of the eigenspaces of \(A\). Then from

\[y_i\cdot x_j=\begin{cases}1&i=j\\0&i\neq j\end{cases}\]

in the proof of Proposition 5, we see that \(C\) also becomes a diagonal matrix and \(B\) becomes the zero matrix. Therefore the following holds.

Proposition 7 Consider an \(n\times n\) matrix \(A\) satisfying all the conditions of Proposition 6, and let \(x_1,\ldots, x_n\) be a basis of \(\mathbb{R}^n\) consisting of bases of the eigenspaces. Suppose \(Ax_i=\lambda_ix_i\) and let \(X=(x_1\mid\cdots\mid x_n)\). Then for the diagonal matrix

\[D=\begin{pmatrix}\lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&\lambda_n\end{pmatrix}\]

we have \(A=XDX^{-1}\).

Hence we can give a fitting name to a matrix \(A\) satisfying this condition.

Definition 8 An \(n\times n\) matrix \(A\) satisfying all the conditions of Proposition 5 is called diagonalizable.

Alternatively, since Proposition 6 gave a necessary and sufficient condition, there is no problem in calling a matrix similar to a diagonal matrix diagonalizable. In other words, any diagonalizable matrix is completely determined by its eigenvalues.

We have already seen sufficiently above why diagonalizable matrices are conceptually important. Not only that, diagonalizable matrices are also of great help in terms of computational convenience. For example, if a matrix \(A\) is diagonalizable so that \(A=XDX^{-1}\), then the powers of \(A\) are given by \(A^k=XD^kX^{-1}\), and since the power of a diagonal matrix is merely the diagonal matrix made from the powers of each entry, computing powers of \(A\) becomes a very easy task.

More generally, if matrices \(A_1,\ldots, A_k\) are diagonalizable by the same matrix \(X\), that is, if

\[A_i=XD_iX^{-1}\]

then

\[A_1A_2\cdots A_k =XD_1D_2\cdots D_kX^{-1}\]

and since the product of diagonal matrices is merely the diagonal matrix consisting of the products of the diagonal entries, computing \(A_1A_2\cdots A_k\) may also not be very difficult. We give such a case the following name.

Definition 9 A family \(\{A_i\}\) of matrices is called simultaneously diagonalizable if there exists an invertible matrix \(X\) such that \(X^{-1}A_iX\) is a diagonal matrix for every \(i\).

If two matrices \(A,B\) are simultaneously diagonalizable by a fixed matrix \(X\), then from the following equation

\[AB=XD_AX^{-1}XD_BX^{-1}=XD_AD_BX^{-1}=XD_BD_AX^{-1}=BA\]

we know that the two matrices \(A, B\) commute. The next proposition shows that the converse also holds (for diagonalizable matrices).

Proposition 10 If two diagonalizable matrices \(A,B\) satisfy \(AB=BA\), then \(A, B\) are simultaneously diagonalizable.

Proof

Essentially, it suffices to show that the two matrices \(A,B\) give the same eigenspace decomposition. Consider the eigenspace decomposition using \(A\):

\[V=\bigoplus_{\lambda}E_\lambda(A)\]

Then for any \(v\in E_\lambda(A)\), from the following equation

\[A(Bv)=ABv=BAv=B(\lambda v)=\lambda(Bv)\]

we know that \(Bv\in E_\lambda(A)\). Now regarding \(B\) as a linear operator on the vector space \(E_\lambda(A)\), since the original linear operator \(B\) was diagonalizable, \(B\) is also diagonalizable on \(E_\lambda(A)\), and hence there exists a basis of \(E_\lambda(A)\) consisting of eigenvectors of \(B\). Now every element of \(E_\lambda(A)\) is an eigenvector of \(A\) (corresponding to eigenvalue \(\lambda\)), so these are also eigenvectors of \(A\).

Eigenspace Decomposition of Linear Operators

So far we have examined the process of diagonalizing a given matrix, and basically this is the same as decomposing a vector space into eigenspaces when a (diagonalizable) linear operator is given. For the proof of this we actively used bases of the eigenspaces. Describing this without choosing bases will be helpful when we examine the Jordan canonical form, which will be covered in the next post.

For a finite-dimensional vector space \(V\) and a linear operator \(L:V\rightarrow V\), we have seen that the following equation

\[\rank L +\nullity L=\dim V\]

holds. (§Isomorphisms, ⁋Theorem 7) Here \(\rank L=\dim\im L\) and \(\nullity L=\dim\ker L\). However, this does not mean that we can immediately write \(V\) as the direct sum of \(\im L\) and \(\ker L\). For example, for the matrix \(A\) that was an example of a non-diagonalizable matrix after Proposition 4, considering the operator defined by

\[A-I=\begin{pmatrix}0&1&1\\0&0&1\\0&0&0\end{pmatrix}\]

we have \(\ker (A-I)\cap \im(A-I)\neq \{0\}\). Yet if only \(\ker L\cap \im L=\{0\}\) holds, then from §Dimension, ⁋Example 8 and Proposition 2 we know that necessarily \(V=\ker L\oplus \im L\). The following lemma gives a condition equivalent to this one.

Lemma 10 In the above situation, \(\ker L\cap \im L=\{0\}\) is equivalent to \(\ker L^2=\ker L\).

Proof

With a little thought, we see that \(\ker L^2=\ker L\) is equivalent to \(\ker L^2\subset \ker L\). Therefore what we must show is the following equivalence

\[\ker L\cap \im L=\{0\}\iff \ker L^2\subset\ker L\]

First suppose \(\ker L\cap \im L=\{0\}\) and let \(v\in\ker L^2\). Then \(0=L^2 v=L(Lv)\), so \(Lv\in\ker L\), and hence by assumption we must have \(Lv=0\). That is, \(v\in\ker L\). Conversely, assume \(\ker L^2\subset \ker L\) and let \(v\in \ker L\cap \im L\). Then since \(v\in \im L\), there exists \(w\in V\) such that \(v=Lw\). But since \(v\in\ker L\) as well,

\[0=Lv=L(Lw)=L^2w\implies w\in\ker(L^2)\subset \ker L\]

so \(w\in \ker L\). Hence \(v=Lw=0\).

Returning to the original story, we are especially curious about the case where \(L\) is of the form \(A-\lambda I\) for some linear operator and its eigenvalue. The following proposition characterizes diagonalizability concisely using Lemma 10.

Proposition 11 A linear operator \(A:V\rightarrow V\) is diagonalizable if and only if for every eigenvalue \(\lambda\in\sigma(A)\) we have

\[\ker(A-\lambda I)^2=\ker(A-\lambda I)\]

Proof

First suppose \(A\) is diagonalizable. Then \(V=\bigoplus_{\mu\in\sigma(A)} E_\mu(A)\). Take any \(v\in\ker(A-\lambda I)^2\); then we can write \(v=\sum_{\mu\in\sigma(A)}v_\mu\) uniquely, and

\[(A-\lambda I)^2v=\sum_{\mu\in\sigma(A)}(A-\lambda I)^2v_\mu=\sum_{\mu\in\sigma(A)}(\mu-\lambda)^2v_\mu=0\]

By the uniqueness of the eigenspace decomposition, \((\mu-\lambda)^2v_\mu=0\) must hold for all \(\mu\), and when \(\mu\neq\lambda\) we have \(v_\mu=0\), so

\[v=v_\lambda\in E_\lambda(A)=\ker(A-\lambda I)\]

Conversely, suppose \(\ker(A-\lambda I)^2=\ker(A-\lambda I)\) for every eigenvalue \(\lambda\). From Lemma 10 we have

\[\ker(A-\lambda I)\cap\im(A-\lambda I)=\{0\}\]

for each \(\lambda\), and by §Isomorphisms, ⁋Theorem 7

\[\dim\ker(A-\lambda I)+\dim\im(A-\lambda I)=\dim V\]

so \(V=\ker(A-\lambda I)\oplus\im(A-\lambda I)\). For convenience write \(\ker(A-\lambda I)=E_\lambda(A)\) and \(\im (A-\lambda I)=W_\lambda(A)\). First, for any \(v\in W_\lambda(A)\), writing \(v=(A-\lambda I)w\), from

\[Av=A(A-\lambda I)w=(A-\lambda I)Aw\in W_\lambda(A)\]

we see that \(W_\lambda(A)\) is an \(A\)-invariant subspace. That is,

\[A\vert_{W_\lambda(A)}: W_\lambda(A) \rightarrow W_\lambda(A)\]

is well defined. Then from Proposition 4, if \(w\in W_\lambda(A)\) is an eigenvector of \(A\vert_{W_\lambda(A)}\) with eigenvalue \(\mu\), then regarding \(w\) as an element of \(V\) it is also an eigenvector of \(A\) (corresponding to eigenvalue \(\mu\)); and conversely, if an eigenvalue \(\mu\neq \lambda\) of \(A\) and a corresponding eigenvector are given, then this can be viewed as an eigenvalue–eigenvector pair of \(A\vert_{W_\lambda(A)}\). Moreover, for any eigenvalue \(\mu\) of \(A\vert_{W_\lambda(A)}\),

\[\ker (A_{W_\lambda(A)}-\mu I)=\ker (A_{W_\lambda(A)}-\mu I)^2\]

also holds on \(W_\lambda(A)\) for a similar reason. That is, we can repeat this process inductively. On the other hand, since we are assuming that \(\mathbb{K}\) is algebraically closed, we know that any linear operator \(W \rightarrow W\) always has an eigenvalue provided \(W\) is not \(0\)-dimensional, and from this we know that this induction exactly yields the eigenspace decomposition of \(A\).

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[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
[Lee] 이인석, 선형대수와 군, 서울대학교 출판문화원, 2005.

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1. Of course, as always, we assume that this sum is in fact a finite sum. That is, we assume that \((v_i)_{i\in I}\) is finitely supported. ↩