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For a complex $n\times n$ matrix $A$, $\mathrm{tr}(A^k)$ is the sum of the $k$th powers of its eigenvalues, counted with algebraic multiplicity. This does not require diagonalizability: a similar upper triangular matrix has the eigenvalues on its diagonal, and taking powers raises those diagonal entries to their corresponding powers. Newton's identities therefore recover its characteristic polynomial from the first $n$ such traces.

For the Jordan block $A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$, $\mathrm{tr}A=4$ and $\mathrm{tr}(A^2)=8$. Hence $e_1=4$ and $e_2=(4^2-8)/2=4$, giving $\det (zI-A)=z^2-4z+4=(z-2{)}^2$. The repeated eigenvalue is recovered even though this matrix is not diagonalizable.^[ref]

A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

For a matrix over a field $F$, diagonalizability requires an eigenbasis with entries in $F$ and eigenvalues in $F$. Passing to a larger field can supply missing roots of the characteristic polynomial. For example,is not a diagonalizable matrix over the rational numbers, since the quadratic factor has no rational root. It is a diagonalizable matrix over the real numbers and complex numbers, since its three eigenvalues are distinct. More generally, diagonalizability is equivalent to the minimal polynomial splitting into distinct linear factors over the specified field.

For an integer matrix, reduction modulo a prime number is a different operation from extension of its coefficient field, and can merge eigenvalues. In the same example modulo $5$, ${\chi }_A(t)=(t-1)(t-3{)}^2$, while $\ker (A-3I)$ is spanned by $(1,3,4{)}^T$. Thus the geometric multiplicity of $3$ is one and its algebraic multiplicity is two, preventing diagonalizability even though all eigenvalues belong to the finite field.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

If linear operators $A,B$ have a common eigenbasis $(v_i)$, then $A+B$ has the corresponding eigenvalues ${\alpha }_i+{\beta }_i$, counting algebraic multiplicity. This follows directly from linearity on each basis vector. The pairing is determined by the common eigenvectors; it is not the collection of all pairwise sums of arbitrary eigenvalues.

If a $k$-dimensional invariant subspace is mapped to the span of the first $k$ coordinate vectors by a similarity transformation, the transformed matrix has block formIts characteristic polynomial factors as $\det (\lambda I-B)\det (\lambda I-D)$, so its eigenvalues are the eigenvalues of the two diagonal blocks, counted with algebraic multiplicity.

Let $C$ be a connected component of the union of the row Gershgorin discs of an $n\times n$ matrix. If $C$ comprises $k$ of the indexed Gershgorin discs, then it contains exactly $k$ eigenvalues, counted with algebraic multiplicity. Coincident disks still count separately, once for each row. In particular, a Gershgorin disc disjoint from all the others contains one simple eigenvalue. More generally, any union of $k$ indexed Gershgorin discs disjoint from the remaining $n-k$ contains exactly $k$ eigenvalues, whether or not that union is connected.^[ref]

The separation is essential: overlapping Gershgorin discs do not individually promise one eigenvalue each. Discs that touch at a boundary point belong to the same connected component. For example, the matrix $\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 0\\ 0& 0& 8\end{array}\right)$ has Gershgorin discs of radii $1,0,0$ with centres $2,2,8$. The component around $2$ comprises two indexed disks and contains the eigenvalue $2$ with algebraic multiplicity two; the isolated point at $8$ contains the remaining eigenvalue.

For intuition, continuously turn on the off-diagonal entries using the diagonal matrix $E=\mathrm{diag}(a_{11},\dots ,a_{nn})$ and $A(t)=E+t(A-E)$, $0\le t\le 1$. At $t=0$ the eigenvalues are the disk centres; at every later time the Gershgorin discs stay inside their final union. The continuity of the eigenvalues as a collection means that no eigenvalue can cross a gap between final components. The count gives a useful certificate that an eigenvalue cluster is complete.^[ref]

For complex invertible matrices $A,B$, $AB=-BA$ implies $B^{-1}AB=-A$ and $A^{-1}BA=-B$. Invariance of matrix trace under matrix similarity gives both zero traces. Invariance of determinant gives $\det A=(-1{)}^n\det A$, and invertibility forces $n$ even. Equivalently, eigenvalues occur in pairs $\lambda ,-\lambda$ with equal algebraic multiplicity, and zero is excluded. Both invertibility assumptions matter: $A=\mathrm{diag}(1,0,0)$ and $B=\mathrm{diag}(0,1,0)$ anticommute in dimension three but are singular and have nonzero traces.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

For an $n$-by-$n$ square matrix $A=(a_{ij})$ over a field, its matrix trace is the sum of its diagonal entries:For example, $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ has trace $2+1=3$. Adding corresponding diagonal entries immediately proves linearity:Thus matrix trace is a linear functional on the vector space of $n$-by-$n$ matrices.^[ref]

The surprising point is that the sum is intrinsic to a linear operator, even though individual diagonal entries depend on the basis. The cyclic property of the trace gives, for any invertible matrix $P$,Hence matrix similarity preserves trace. For the example above, choose the change-of-basis matrix $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. ThenThe diagonal entries have changed but their sum has not. This makes trace useful for extracting a feature of the underlying linear operator without finding a preferred basis.

Over the complex numbers, or over the real numbers after allowing complex eigenvalues, the trace equals their sum with algebraic multiplicity:No diagonalizable matrix hypothesis is needed.^[ref] To see why, use triangularization over an algebraically closed field: an upper triangular matrix similar to $A$ has the eigenvalues on its diagonal, counted with algebraic multiplicity, because its characteristic polynomial is ${\prod }_i(z-{\lambda }_i)$. Similarity preserves trace, so summing that diagonal proves the formula.

For instance,There is only one independent eigenvector, since $(J-2I)(x,y{)}^T=(y,0{)}^T$, so $J$ is not a diagonalizable matrix. Nevertheless its repeated eigenvalue contributes twice. For our earlier $A$, the characteristic polynomial is $z^2-3z+1$, its eigenvalues are $(3+\sqrt{5})/2$ and $(3-\sqrt{5})/2$, and their sum is $3$. An inexpensive diagonal sum therefore gives spectral information before solving the characteristic polynomial.

The trace of a real matrix does not give that linear map's volume factor: that factor is the absolute value of its determinant. For example, $2I_2$ has trace $4$ and area factor $4$, but $2I_3$ has trace $6$ and volume factor $8$. Their occasional numerical agreement in two dimensions is a coincidence. The connection between trace and local volume change gives a geometric interpretation near the identity matrix, and describes the volume rate of a flow map generated by a linear system of differential equations.

Let $P={\chi }_M$, and suppose $P(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$. The reciprocal identity gives, for $x\ne 0$,Near $x={\lambda }^{-1}$ the factor $x^{2m-d}R(1/x)$ is nonzero. Also $1-\lambda x=-\lambda (x-{\lambda }^{-1})$ has a simple zero there. HenceThis proves equality of the algebraic multiplicities, not merely that the inverse is another eigenvalue. Invertibility excludes zero eigenvalues throughout.

In the finite field ${\mathbb{F}}_5=\mathbb{Z}/5\mathbb{Z}$, the characteristic polynomial becomesbecause $(\lambda -3{)}^2={\lambda }^2-\lambda -1$ modulo $5$. Splitting into linear factors does not by itself guarantee diagonalizability: the repeated eigenvalue needs enough eigenvectors.

Solving $(A-3I)(x,y,z{)}^T=0$ givesso $y=3x$ and $z=4x$. Hencehas dimension one, although $3$ has algebraic multiplicity two. The geometric multiplicity is too small, and $A$ is not diagonalizable over ${\mathbb{F}}_5$. This illustrates diagonalizability depends on the field: characteristic zero gave three distinct eigenvalues, while reduction modulo $5$ merges two of them.

If $c(\lambda )=\det (\lambda I-T)=0$, then $\lambda I-T$ has nonzero kernel of a linear map. A nonzero vector $v$ in that kernel of a linear map satisfies $Tv=\lambda v$, proving that $\lambda$ is an eigenvalue. Extend $v$ to a basis. Its first column is $(\lambda ,0,\dots ,0{)}^T$, so the matrix has block formHere $B$ represents the induced operator on $V/⟨v⟩$.

If the algebraic multiplicity is $r$, repeat this argument in that quotient: its characteristic polynomial still has $\lambda$ with multiplicity $r-1$. Induction and lifting a quotient basis gives a basis whose first $r$ columns have the prescribed upper-triangular form. With $U$ the span of those $r$ vectors, the full matrix isThus $U$ is invariant and is contained in $W_{\lambda }$. Conversely, if $(T-\lambda I{)}^{s}v=0$, project to $V/U$. The induced map $C-\lambda I$ is invertible, so its $s$th power cannot kill a nonzero class. Hence $v\in U$. Consequently $\boxed{W_{\lambda }=U,\dim W_{\lambda }=r}$; this proves simultaneously that it is a subspace and $T$-invariant. It also proves $(T-\lambda I{)}^r=0$ on this space, without using Cayley-Hamilton theorem in advance.

To prove independence of the different spaces, take a relation ${\sum }_{\lambda }{v}_{\lambda }=0$ with $v_{\lambda }\in W_{\lambda }$. Fix $\lambda$ and apply $P_{\lambda }(T)={\prod }_{\mu \ne \lambda }(T-\mu I{)}^{r_{\mu }}$. It kills all terms except $v_{\lambda }$. On $W_{\lambda }$, each factor $T-\mu I=(\lambda -\mu )I+N$ is invertible: its inverse is a finite geometric series in the nilpotent $N$. Hence $P_{\lambda }(T)v_{\lambda }=0$ implies $v_{\lambda }=0$. The sum is direct, and its dimension is $\sum r_{\lambda }=\dim V$, soFinally $c(T)={\prod }_{\lambda }(T-\lambda I{)}^{r_{\lambda }}$ kills each summand, hence every vector. This proves $\boxed{c(T)=0}$ without circularly assuming the theorem during the decomposition.

If $\varphi =T^{\prime }\psi$ and $v\in \ker T$, then $\varphi (v)=\psi (Tv)=0$. Thus $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^{\circ }$, using the annihilator of a vector subspace.

Conversely, let $\varphi$ vanish on the kernel of a linear map $\ker T$. Define a linear functional on $\mathrm{im}T$ by ${\psi }_0(Tv)=\varphi (v)$. This is well-defined: $Tv=Tw$ implies $v-w\in \ker T$, so $\varphi (v)=\varphi (w)$. Extend ${\psi }_0$ to a linear functional $\psi$ on $W$, using the extension argument from part (i). Then $T^{\prime }\psi =\varphi$, provingThe dimension formula from part (i) and the rank-nullity theorem now giveTo deduce the eigenvalue statement, apply this equality of rank of a linear map to $T-\lambda I$. Its dual map is $T^{\prime }-\lambda I$, so their kernel of a linear map dimensions agree. Thus the same $\lambda$ are eigenvalues, with the same geometric multiplicities.

For algebraic multiplicities as well, part (a)(iii) givesso equality of rank applies to every power. HenceAfter extending to a field where the characteristic polynomials split, take $r\ge \dim V$. These kernel of a linear map spaces are the generalized eigenspaces; the primary decomposition theorem for linear maps identifies their dimensions with the algebraic multiplicities. The same proof works over the original field for every eigenvalue in it. We obtainIn fact the argument also includes the zero eigenvalue.

Let $Hv=\lambda v$ with $v\ne 0$ and $H^{\ast }=H$. Then $v^{\ast }Hv=\lambda v^{\ast }v$, while taking the complex conjugate gives $v^{\ast }Hv=\overline{\lambda }{v}^{\ast }v$. Since $v^{\ast }v>0$, every eigenvalue of a Hermitian matrix is real.

If $Hv=\lambda v$ and $Hw=\mu w$, thenDistinct eigenvalues therefore force $v^{\ast }w=0$: their eigenvectors are orthogonal in the Hermitian inner product. A degenerate eigenvalue occurs more than once in the spectrum, counted with multiplicity. For a Hermitian matrix, its algebraic multiplicity equals its geometric multiplicity, so degeneracy gives an eigenspace of dimension greater than one; an orthonormal basis can be chosen within it.

The characteristic polynomial factors as $\det (\lambda I-A)=(\lambda -1)({\lambda }^2-1)$. Thus $\boxed{\lambda =1\text{ twice},\lambda =-1\text{ once}}$ in algebraic multiplicity. An orthonormal basis of corresponding eigenvectors isDirect multiplication verifies $A{v}_1=v_1$, $A{v}_2=v_2$, $A{v}_3=-v_3$. Their mutual Hermitian inner products vanish and each has norm one, which proves linear independence. For the repeated eigenvalue, Gaussian elimination givesHence $\boxed{\mathrm{rank}(A-I)=1,\dim \ker (A-I)=2}$. The second equality follows from the rank-nullity theorem and agrees with the two-dimensional eigenspace. A repeated eigenvalue alone would not imply a two-dimensional eigenspace for an arbitrary matrix; here the Hermitian matrix is diagonalizable.

Write the given matrix as $A$. Expanding its characteristic polynomial along the middle column givesThus the eigenvalues are $1,-2,-2$, counted with algebraic multiplicity.

For $\lambda =1$, solving $(A-I)v=0$ gives $z=2x$ and $y=-x$. For $\lambda =-2$, solving $(A+2I)v=0$ gives $z=3x$, with $y$ free. A convenient eigenbasis is thereforeThe last two eigenvectors span the two-dimensional eigenspace for $-2$. The three columns have determinant $-1$, so they are linearly independent and form a basis of ${\mathbb{R}}^3$. In particular the repeated eigenvalue causes no failure of diagonalizability here.

Use the Characteristic polynomials of AB and BA, which also account for the zero eigenvalue and algebraic multiplicities. The block determinantequals both $\det (I+tAB)$ and $\det (I+tBA)$, by taking the two Schur complements. For $\lambda \ne 0$, set $t=-1/\lambda$ and multiply by ${\lambda }^n$ to obtainBoth sides are polynomials, so agreement away from zero proves equality there as well. Their roots, with multiplicity, are the same. The familiar eigenvector argument $ABv=\mu v\Rightarrow BA(Bv)=\mu Bv$ works directly only when $\mu \ne 0$, because then $Bv\ne 0$; it should not be used to discard the zero case.

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

For $t\ne 0$, factor the matrix asThe determinant of a scalar multiple of an $n\times n$ matrix gains the scalar's $n$th power. ThusThe right side extends to a polynomial at $t=0$: multiplying ${\chi }_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A{)}^{-1}$, since the leading coefficient of ${\chi }_A(s)=\det (A-sI)$ is $(-1{)}^n$. This equals ${\chi }_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between eigenvalues and the characteristic polynomial,The first equivalence is the definition of an eigenvector; the others use the kernel of a linear map and the criterion for an invertible matrix to have nonzero determinant. Since ${\chi }_A$ has degree $n$, the fundamental theorem of algebra gives $n$ roots over $\mathbb{C}$ counted with multiplicity. The multiplicity of a root is its algebraic multiplicity, and the eigenspace is exactly $\ker (A-\lambda I)$; its dimension is the geometric multiplicity, which need not equal the algebraic multiplicity.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb{C}$ as $p(s)={\prod }_{j=1}^4(s-{\lambda }_j)$, including repeated roots when $\mu =0$. Substituting $A$ into this polynomial is valid because the resulting factors commute. If $\mu$ is an eigenvalue of $A^4$, thenAt least one factor must be zero, so its ${\lambda }_j$ is an eigenvalue of $A$ and satisfies ${\lambda }_j^4=\mu$. Thus the reverse direction of this instance of the spectral mapping theorem isThis argument also covers $\mu =0$, when the conclusion is that $A$ is singular and has eigenvalue zero.