Eigenspace Decomposition

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.

Direct Sums of Subspaces

Consider an $n\times n$ matrix $A$ and one of its eigenvalues $\lambda$. By definition, any vector $v$ in the eigenspace $E_\lambda$ corresponding to $\lambda$ must satisfy

\[Av=\lambda v\]

Thus, when restricted to $E_\lambda$, the matrix $A$ becomes the very simple map $v\mapsto \lambda v$.

More generally, regard $A$ as a linear map from $\mathbb{K}^n$ to $\mathbb{K}^n$, and assume that the domain $\mathbb{K}^n$ is spanned 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 vectors $v_\lambda\in E_\lambda$ such that

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

and therefore

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

By the argument above, since $Av_\lambda=\lambda v_\lambda$, we obtain

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

Of course, for this calculation to make sense, the representation of $v$ as a sum of the $v_\lambda$ 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 exist unique $(v_i)_{i\in I}$ 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 $I$ has two elements.

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 $V=W_1\oplus W_2$. By definition, $W_1+W_2\subseteq V$ is clear. Conversely, for any $v\in V$, there exist $w_i\in W_i$ such that $v=w_1+w_2$, so $V\subseteq W_1+W_2$ also holds. Hence $V=W_1+W_2$. On the other hand, if $W_1\cap W_2\neq \{0\}$, then for a nonzero $w\in W_1\cap W_2$,

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

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 exist $w_i\in W_i$ such that $v=w_1+w_2$. Moreover, this expression is unique: if

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

then

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

The left side lies in $W_1$ and the right side lies in $W_2$, 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 holds even 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 standard basis vectors $e_1,e_2$. If we set $W_1=\mathbb{R}e_1$, $W_2=\mathbb{R}e_2$, and $W_3=\mathbb{R}(e_1+e_2)$, then $V=W_1+W_2+W_3$ and $W_i\cap W_j=\{0\}$ whenever $i\neq j$, but $V\neq W_1\oplus W_2\oplus W_3$.

This is because the representation of $e_1+e_2\in V$ is not unique, as shown by

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

As another example, choose 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 is exactly 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 choose bases $\mathcal{B}_i$ of the $W_i$. If $\mathcal{B}_i\cap\mathcal{B}_j\neq\emptyset$, then $W_i\cap W_j\neq\emptyset$, which contradicts the discussion after Proposition 2, so we must have $\mathcal{B}_i\cap\mathcal{B}_j=\emptyset$. For any $v\in V$, from $V=\bigoplus W_i$ there exist unique $w_i$ satisfying

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

Also, in each $W_i$ the $w_i$ can be uniquely expressed as linear combinations 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.

Thus 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 Proposition 3 we know that decomposing the vector space $\mathbb{K}^n$ into the eigenspaces $E_\lambda$ is the same as collecting bases of the $E_\lambda$ to form a basis of $\mathbb{K}^n$. Also, if a nonzero vector $x_1$ is an eigenvector corresponding to an eigenvalue $\lambda_1$, then for another eigenvalue $\lambda_2$,

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

so $x_1\not\in E_{\lambda_2}$. 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 $\lambda_1,\ldots,\lambda_m$ respectively. 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

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

Among the $(\alpha_i)_{1\leq i\leq m}$ satisfying this, choose one with the smallest support and call it $(\beta_i)_{1\leq i\leq m}$. That is, if the number of $i$ with $\beta_i\neq0$ is $k$, then no $(\alpha_i)_{1\leq i\leq m}$ with fewer than $k$ nonzero entries satisfies equation (1).

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 $A$,

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

and since the $x_m$ are eigenvectors,

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

But 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 equation obtained above,

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

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

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

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 condition $\beta_i''=0$ is equivalent to $\beta_i=0$. Therefore, the number of $1\leq i\leq m-1$ with $\beta_i''\neq 0$ is $k-1$, 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, for any matrix $A$ with eigenvalues $\lambda\in\sigma(A)$, corresponding eigenspaces $E_\lambda$, and bases $\mathcal{B}_\lambda$, we know that $\mathcal{B}=\bigcup_{\lambda\in\sigma(A)}\mathcal{B}_\lambda$ is a linearly independent subset of $\mathbb{K}^n$. However, there is no reason for $\mathcal{B}$ to be a basis of $\mathbb{K}^n$ in general. For example, in §Characteristic Polynomial, ⁋Example 7, when $\mathbb{K}=\mathbb{R}$ we have $\sigma(J)=\emptyset$, so $\mathcal{B}=\emptyset$. Moreover, even if we assume that the characteristic polynomial of $A$ has exactly $n$ roots, a similar problem can arise; for instance, 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 the only eigenvector corresponding to 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 the geometric multiplicity is $1$.

The following 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$ never exceeds its algebraic multiplicity.

Proof

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

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

Since the columns of $X$ are linearly independent, $X^{-1}$ exists. Let the rows of $X^{-1}$ be $y_i$. From $X^{-1}X=XX^{-1}=I$,

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

holds. Therefore, setting $A'=X^{-1}AX$,

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

Thus, writing the characteristic polynomial of $A$ as $p_A(\mathbf{x})$, by §Characteristic Polynomial, ⁋Corollary 4 we have $p_A(\mathbf{x})=p_{A'}(\mathbf{x})$, and therefore

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

That is, the algebraic multiplicity of $\lambda$ in $p_A$ is at least $k$.

Given an $n\times n$ matrix $A$ with characteristic polynomial $p_A$, 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 equal $n$. Putting all this together, we obtain 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 a direct sum of the eigenspaces of $A$ is:

The characteristic polynomial of $A$ has $n$ roots counting multiplicity, and
For each eigenvalue, the geometric multiplicity equals the algebraic multiplicity.

In particular, if $\mathbb{K}$ is algebraically closed, the first condition is always satisfied, so only the second condition need be considered.

We assume that the field $\mathbb{K}$ is algebraically closed while dealing with diagonalization of matrices. This is purely for convenience; if $\mathbb{K}$ is not algebraically closed, we simply consider a field extension obtained by adjoining the roots of the characteristic polynomial of the matrix of interest. (§Algebraic Extensions) This is exactly the same as obtaining $\mathbb{C}$ from $\mathbb{R}$ by adjoining the imaginary root $i$ of the equation $\x^2+1=0$, for example.

Diagonalization of Matrices

We have previously examined how to decompose $\mathbb{R}^n$ through the eigenvalues and eigenspaces of an arbitrary $n\times n$ matrix $A$, and from Proposition 6 we have also learned when such a decomposition is possible. Let us look again at the proof of Proposition 5 that we used to establish this. We added $n-k$ arbitrary vectors to a basis $x_1,\ldots, x_k$ of $E_\lambda$, then defined the matrix $X=(x_1\mid\cdots\mid x_n)$ through these, and showed by calculation that

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

has the upper left $k\times k$ block equal to 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 them so that all $n$ vectors $x_1,\ldots, x_n$ form a basis of 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 eigenvectors of $A$. Let $Ax_i=\lambda_ix_i$ and $X=(x_1\mid\cdots\mid x_n)$. 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 $A=XDX^{-1}$.

Thus we can give a suitable name to a matrix $A$ satisfying this condition.

Definition 8 An $n\times n$ matrix $A$ satisfying all the conditions of Proposition 6 is called diagonalizable.

Alternatively, since Proposition 6 gives 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.

That diagonalizable matrices are conceptually important has been sufficiently examined above. Moreover, diagonalizable matrices are also of great computational utility. For example, if a matrix $A$ is diagonalizable with $A=XDX^{-1}$, 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 formed from the powers of each diagonal entry, computing powers of $A$ becomes a very easy task.

More generally, if matrices $A_1,\ldots, A_k$ are diagonalizable through 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 of matrices $\{A_i\}$ 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 through a fixed matrix $X$, then from

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

we know that the two matrices $A, B$ commute. The following 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 $A,B$ admit 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

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

we know that $Bv\in E_\lambda(A)$. Now viewing $B$ 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 therefore there exists a basis of $E_\lambda(A)$ consisting of eigenvectors of $B$. Any 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 fundamentally this is the same as decomposing a vector space into eigenspaces when a (diagonalizable) linear operator is given. To prove this we actively used bases of eigenspaces. Describing this without the choice of a basis will be helpful when we examine the Jordan canonical form in the next post.

For a finite-dimensional vector space $V$ and a linear operator $L:V\rightarrow V$, we have seen that

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

holds. (§Isomorphisms, ⁋Theorem 7 (Rank-nullity theorem)) Here $\rank L=\dim\im L$ and $\nullity L=\dim\ker L$. However, this does not mean that $V$ can be expressed as a direct sum of $\im L$ and $\ker L$. For example, for the matrix $A$ that was the non-diagonalizable example after Proposition 4,

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

the operator defined by this satisfies $\ker (A-I)\cap \im(A-I)\neq \{0\}$. However, if $\ker L\cap \im L=\{0\}$ holds, then from §Dimension of Vector Spaces, ⁋Example 8 and Proposition 2 we know that necessarily $V=\ker L\oplus \im L$. The following lemma gives a condition equivalent to this.

Lemma 11 In the above situation, the condition $\ker L\cap \im L=\{0\}$ 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 need to show is the equivalence

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

First, assume $\ker L\cap \im L=\{0\}$ and let $v\in\ker L^2$. Then $0=L^2 v=L(Lv)$, so $Lv\in\ker L$, and therefore 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 $w\in \ker L$. That is, $v=Lw=0$.

Returning to the original story, we are particularly interested in the case where $L$ is of the form $A-\lambda I$ for some linear operator and its eigenvalue. The following proposition uses Lemma 11 to characterize diagonalizability concisely.

Proposition 12 A linear operator $A:V\rightarrow V$ is diagonalizable if and only if for every eigenvalue $\lambda\in\sigma(A)$,

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

holds.

Proof

First, suppose $A$ is diagonalizable. Then $V=\bigoplus_{\mu\in\sigma(A)} E_\mu(A)$. For any $v\in\ker(A-\lambda I)^2$, we can uniquely write $v=\sum_{\mu\in\sigma(A)}v_\mu$, 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 that for every eigenvalue $\lambda$, we have $\ker(A-\lambda I)^2=\ker(A-\lambda I)$. From Lemma 11, for each $\lambda$,

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

and by §Isomorphisms, ⁋Theorem 7 (Rank-nullity theorem),

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

so $V=\ker(A-\lambda I)\oplus\im(A-\lambda I)$. For convenience, write $\ker(A-\lambda I)=E_\lambda(A)$ and $\im (A-\lambda I)=W_\lambda(A)$. We first know that for any $v\in W_\lambda(A)$, if we write $v=(A-\lambda I)w$, then 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 viewing $w$ 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 its corresponding eigenvector are given, this can be viewed as an eigenvalue-eigenvector pair of $A\vert_{W_\lambda(A)}$. Also, 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 as long as $W$ is not $0$-dimensional, and from this we know that this induction exactly gives the eigenspace decomposition of $A$.

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

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

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Direct Sums of Subspaces

Diagonalization

Diagonalization of Matrices

Eigenspace Decomposition of Linear Operators

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