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A block matrix inverse expresses the matrix inverse in terms of smaller matrices. Partition a square matrix over a field asIf $A$ is an invertible matrix, define its Schur complement $S=D-C{A}^{-1}B$. Then $M$ is an invertible matrix exactly when $S$ is an invertible matrix, andThe order of the matrix products matters. In particular, the lower-right block is $S^{-1}$, rather than generally $D^{-1}$: the coupling through $B$ and $C$ changes the response.^[ref]

For an original example, takeThe Schur complement is $S=5-1/2-4/3=19/6$. Substitution gives the concise answerMultiplying in either order gives the identity matrix; the lower-right entry is $6/19$, whereas $D^{-1}=1/5$. The difference records the feedback through the two eliminated variables.

Often the desired output is a solution of a linear system, rather than every entry of a matrix inverse. Block elimination solves with $A$ and then with $S$, recovering the eliminated variables by back substitution. Reusing an LU decomposition of $A$ avoids explicitly forming $A^{-1}$. A favourable block diagonal matrix or tridiagonal matrix block can make this useful; merely partitioning an unstructured dense matrix does not guarantee a faster solve.^[ref]

Every square matrix satisfies its own characteristic polynomial. More explicitly, for an $n$-by-$n$ matrix $A$ over a commutative ring with identity, formThenHere $0$ is the zero matrix, and the constant term multiplies the identity matrix. This is a polynomial evaluated at a matrix; it is stronger than saying that each eigenvalue is a root of a polynomial. Neither diagonalizability nor distinct eigenvalues is required.^[ref]

The practical payoff is that arbitrarily high matrix powers can be reduced to a combination of $I,A,\dots ,A^{n-1}$. These powers arise when repeatedly applying a linear map, solving a linear recurrence relation, or evolving a finite state space model. The characteristic polynomial turns that repeated action into a fixed-order scalar recurrence. The minimal polynomial may reduce the number of required powers further.

A cube root of a matrix $A$ is a square matrix $B$ of the same size over the specified field satisfying $B^3=A$. Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block $J_3(0)$ has none, even though its square is nonzero. If $B^3=J_3(0)$, then $B^9=0$, making $B$ a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives $B^3=0$, a contradiction.

If $A$ is an $m$-by-$n$ matrix and $B$ is an $n$-by-$m$ matrix over the same field, both products are square matrices andThe products can even have different sizes: for $A=(12)$ and $B=(34{)}^T$, $AB=(11)$ and $BA=\left(\begin{array}{cc}3& 6\\ 4& 8\end{array}\right)$, but both have trace $11$.^[ref]

Grouping all but one factor proves that $\mathrm{tr}(A_1{A}_2\cdots A_k)$ is unchanged by a cyclic permutation whenever the cyclic products are defined and square. In particular, every commutator has trace zero, since $\mathrm{tr}(AB-BA)=0$ by linearity.

Cyclic rotation does not allow arbitrary reordering. Forone has $ABC=A$, so $\mathrm{tr}(ABC)=1$, but $AC=0$, so $\mathrm{tr}(ACB)=0$. Rotating $ABC$ to $BCA$ or $CAB$ preserves its trace; swapping just $B$ and $C$ need not.

An $m\times n$ matrix has full row rank when its matrix rank is $m$, equivalently when its rows are linearly independent vectors. This requires $m\le n$. Its column space is all of ${\mathbb{F}}^m$, so the represented linear map is surjective and every linear system $Ax=b$ is solvable. If $n>m$, its nullity is $n-m>0$, so solutions are not unique.^[ref]

For example, the matrix transpose of the matrix in full column rank isIt has full row rank two. Every output $(p,q{)}^T$ has solutions $(p-t,q-t,t{)}^T$, with $t\in \mathbb{R}$. A square matrix of size $n$ has full row rank and full column rank together exactly when its matrix rank is $n$; then it is an invertible matrix. A nonsquare matrix cannot have both properties, and has no two-sided inverse: such an inverse would make its linear map a linear isomorphism between vector spaces of different dimensions.

The Gershgorin circle theorem locates every eigenvalue of a square matrix using only its entries. A diagonal entry supplies a centre in the complex plane, and the sum of the off-diagonal absolute values in its row supplies a radius. Small off-diagonal couplings therefore confine the eigenvalues near the diagonal entries. This turns a global eigenvalue problem into a picture built one row at a time.

Precisely, let $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$, with $n\ge 1$, and define the closed Gershgorin discsEvery eigenvalue $\lambda$ of $A$ satisfies $\lambda \in {\bigcup }_i{D}_i(A)$, including eigenvalues on the boundaries. No symmetry or diagonalizability is required. Despite the name, the theorem uses filled disks, not just their circular boundaries.^[ref]

The short reason is the triangle inequality. For a nonzero eigenvector $x$ with $Ax=\lambda x$, choose $i$ with $|x_i|={\max }_j|x_j|>0$. The $i$th row givesThus the largest eigenvector coordinate identifies a Gershgorin disc containing $\lambda$, even though that eigenvector need never be computed to use the bound.

This is useful when a full eigenvalue computation would be expensive: forming the Gershgorin discs needs one pass over the nonzero entries of a sparse matrix. If $0$ lies outside every Gershgorin disc, then $A$ is an invertible matrix, since zero is not an eigenvalue. The condition is exactly $|a_{ii}|>r_i$ in every row, namely strict diagonal dominance. This connects diagonal dominance to solvability of a linear system; the inverse infinity-norm bound from diagonal dominance adds a quantitative bound. It is a sufficient test: failure to exclude zero does not prove singularity.

For a concrete symmetric matrix, takeThe three Gershgorin discs have centres $2,5,9$ and are pairwise disjoint. The finite-dimensional spectral theorem makes the eigenvalues real, so their possible locations reduce to the intervals $[1.7,2.3]$, $[4.5,5.5]$, and $[8.6,9.4]$. Gershgorin eigenvalue counting gives exactly one eigenvalue in each. Independently, the characteristic polynomial iswhose numerical polynomial roots are $1.985811$, $4.989942$, and $9.024247$, rounded to six decimal places. All are positive, as Gershgorin bounds for symmetric matrices already guarantee without finding those polynomial roots.

S. Gerschgorin published the result in 1931 in Über die Abgrenzung der Eigenwerte einer Matrix, in the Bulletin de l'Académie des Sciences de l'URSS, issue 6, pages 749–754. The original paper's §2 states both eigenvalue inclusion and counting in separated groups of Gershgorin discs; §3 already discusses column bounds and transformations preserving the eigenvalues. Thus both the localization picture and ways to sharpen it were part of the original development.^[ref]

A row Gershgorin disc of a square matrix $A=(a_{ij})$ is the closed disk in the complex plane with centre $a_{ii}$ and radius $r_i={\sum }_{j\ne i}|a_{ij}|$. A zero radius gives a single point. The Gershgorin disc theorem places every eigenvalue in the union of these disks. For a Hermitian matrix, the real intervals cut out by the Gershgorin discs bound the eigenvalues, including energy levels when the matrix represents a Hamiltonian operator.

Applying the Gershgorin circle theorem to the transpose $A^T$ gives column Gershgorin discs with the same centres and radii $c_i={\sum }_{j\ne i}|a_{ji}|$. Since $A$ and $A^T$ have the same characteristic polynomial, every eigenvalue lies in the intersection of the row-disc union and the column-disc union. This is an intersection of two unions; one cannot generally replace each row radius by $\min (r_i,c_i)$. An isolated row Gershgorin disc contains exactly one eigenvalue by Gershgorin eigenvalue counting.^[ref]

An $n$-by-$n$ square matrix $A$ over a field is invertible when it has a two-sided matrix inverse. Equivalently, its linear map $x\mapsto Ax$ is a bijection: every output comes from exactly one input. For a square matrix, the following conditions are equivalent:Here matrix rank $n$ means both full column rank and full row rank. The square hypothesis matters: a rectangular matrix can have full column rank without reaching every output, or full row rank without determining every input.^[ref]

The equivalences explain why invertibility is a question about retaining information. A trivial kernel makes the linear map injective; the rank-nullity theorem then gives matrix rank $n$, so its columns form a basis and the linear map is also surjective. Assigning to each output its unique input gives a linear map: the preimage of $\alpha b+\beta c$ is $\alpha A^{-1}b+\beta A^{-1}c$. The determinant is nonzero precisely when the columns are linearly independent. Thus algebraic solvability and independent directions describe the same phenomenon; over the real numbers, nonzero oriented volume supplies its geometric interpretation.

An LU decomposition writes a square matrix as $A=LU$, with $L$ a lower triangular matrix and $U$ an upper triangular matrix. We use a unit triangular matrix $L$, meaning $l_{ii}=1$. The factors record Gaussian elimination: $U$ contains the reduced equations, while $L$ stores the elimination multipliers needed to reconstruct the original equations. This makes Reusing LU factors for multiple right-hand sides inexpensive.^[ref]

For an invertible matrix, unpivoted LU decomposition requires nonzero pivot elements at every elimination step; invertibility alone does not guarantee this. In general use Pivoted LU decomposition, with the explicit convention $PA=LU$ and a permutation matrix $P$. Once the factors exist, solving a linear system $Ax=b$ needs forward substitution followed by back substitution, rather than forming a matrix inverse.

When a square matrix satisfies $M^2=\kappa I$, its powers reduce to $M^{2n}={\kappa }^{n}I$ and $M^{2n+1}={\kappa }^{n}M$. Absolute convergence of the matrix exponential allows even and odd subseries to be separated. Thus $C(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n)!$ and $S(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n+1)!$. For real $\kappa >0$, these are $\cosh \sqrt{\kappa }$ and $\sinh \sqrt{\kappa }/\sqrt{\kappa }$. For $\kappa <0$, they are $\cos \sqrt{-\kappa }$ and $\sin \sqrt{-\kappa }/\sqrt{-\kappa }$. At $\kappa =0$, $e^M=I+M$, including nonzero nilpotent matrices of square zero. This also explains the limiting sine-over-argument coefficients without dividing by zero.

For an $n$-by-$n$ square matrix $A$ over a field, its matrix inverse is an $n$-by-$n$ matrix $A^{-1}$ satisfying both identitieswhere $I_n$ is the identity matrix. It exists exactly when $A$ is an invertible matrix, equivalently when its matrix rank is $n$ or its determinant is nonzero. It reverses the linear map: applying $A$ and then $A^{-1}$ recovers the input, while applying $A^{-1}$ and then $A$ recovers the output.^[ref]

The matrix inverse is unique. If $B$ and $C$ both satisfy the identities, then $B=B(AC)=(BA)C=C$. Although the definition is two-sided, for finite square matrices either identity alone suffices: $BA=I$ forces a trivial kernel, whereas $AB=I$ forces surjectivity; either makes $A$ an invertible matrix by the rank-nullity theorem. Rectangular matrices may have only one of these properties.

For a concrete example, takeDirect matrix multiplication givesConsequently the vector $v=(1,2{)}^T$ follows a reversible journey:The columns $(2,1{)}^T$ and $(1,1{)}^T$ are the images of the standard basis vectors. They turn the unit square into a parallelogram; the matrix inverse restores both the square and its surrounding grid. Since $\det A=1$, this example changes shape while preserving area, illustrating that invertibility does not require preserving lengths or angles.

For comparison, the singular matrixcannot have a matrix inverse: the output $(1,0{)}^T$ cannot specify which of the two inputs occurred. Their difference $(1,-1{)}^T$ belongs to its kernel. Indeed $S(x,y{)}^T=(x+y,0{)}^T$ collapses the plane to a line, its matrix rank is one, and its determinant is zero. No subsequent function of the output alone can reconstruct every input.

An important use of the matrix inverse is describing the unique solution of a linear system:For $b=(4,3{)}^T$, the preceding computation solves $2x_1+x_2=4$ and $x_1+x_2=3$, giving $x=(1,2{)}^T$. In computation, usually solve the linear system using Gaussian elimination and a factorization such as an LU decomposition, rather than forming every entry of the matrix inverse. The inverse formula specifies the answer; a factorization provides a reusable way to compute it for each right-hand side.^[ref]

Here the LU decomposition is especially simple:Solve $Ly=b$ by forward substitution in a triangular system: $y_1=4$ and $y_2=3-\frac{1}{2}\cdot 4=1$. Then solve $Ux=y$ by backward substitution in a triangular system: $x_2=2$ and $x_1=(4-2)/2=1$. The same factors work when $b$ changes. If the whole matrix inverse is actually needed, its $j$th column is the solution of $Ax=e_j$, where $e_j$ is the $j$th standard basis vector.

Reversing several operations means undoing the last one first. For invertible matrices of the same size,because $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=I$, and the product in the other order is also $I$. This is why a change-of-basis matrix uses its matrix inverse to reverse the direction of coordinate conversion.

Two square matrices $A,B$ over the same field are similar if $B=S^{-1}AS$ for an invertible matrix $S$. They can describe one linear operator using two different ordered bases, with the same basis used for input and output in each description. Fix the convention $x=Sy$: $x$ is the old coordinate vector, $y$ the new coordinate vector, and the columns of $S$ are the new basis vectors in old coordinates. This is the convention of the change-of-basis matrix.^[ref]

For a concrete linear operator on the physical plane, let $T$ double horizontal components and multiply vertical components by five. In the old standard basis $\mathcal{E}=(e_1,e_2)$, its matrix of a linear map is $A=\mathrm{diag}(2,5)$. Choose the new ordered basis $\mathcal{F}=(b_1,b_2)$ with $b_1=e_1$, $b_2=e_1+e_2$. ThenThe second column of $B$ says $T(b_2)=-3b_1+5b_2=2e_1+5e_2$. Its off-diagonal entry records how the chosen basis vectors describe the output; the physical linear operator still stretches along the old horizontal and vertical directions.

Take the physical vector $p=2e_1+e_2=b_1+b_2$. Its two coordinate vectors, and the two descriptions of $T(p)$, areReconstruction gives $T(p)=4e_1+5e_2=-b_1+5b_2$. The output coordinate vectors differ, but the endpoint in the physical plane is the same.

This is why matrix similarity is useful: a convenient basis can simplify calculations without changing the linear operator. A diagonalizable matrix can be replaced by a diagonal matrix, making repeated application a calculation with powers of its eigenvalues. In the example, $A$ makes the two independent stretch factors visible, while $B$ expresses the same action in directions chosen for another purpose. Matrix similarity invariants separate properties of the linear operator from features of its address system.

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.

For a square matrix over a field, the minimal polynomial $m_A$ is the unique monic polynomial of least degree satisfying $m_A(A)=0$. The Cayley-Hamilton theorem guarantees at least one such polynomial: the characteristic polynomial. The minimal polynomial divides every polynomial that annihilates $A$, so reducing matrix powers modulo $m_A$ can need fewer coefficients than reducing modulo the characteristic polynomial.

For example, $2I_2$ has minimal polynomial $t-2$, whereas the Jordan block $\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ has minimal polynomial $(t-2{)}^2$. Both have the same characteristic polynomial. Their different matrix powers are illustrated in repeated eigenvalues in the Cayley-Hamilton theorem.

For a real square matrix $A$ of odd size, if $\det A>0$, then $A$ has a positive real eigenvalue. Its characteristic polynomial $p(t)=\det (tI-A)$ is monic of odd degree, so $p(0)=-\det A<0$ and $p(t)\to +\infty$ as $t\to +\infty$. The intermediate value theorem supplies a positive root. This proves the claim without assuming that all eigenvalues are real. The odd-size hypothesis is essential: $-I$ in even dimension has positive determinant and only the negative eigenvalue $-1$.

Let $M$ be the square matrix whose $j$th column is $A_j$. The Leibniz formula for determinants is precisely the stated contraction:A square matrix has zero determinant exactly when its columns are linearly dependent, proving the required equivalence. One direction also follows directly from multilinearity and antisymmetry of a multilinear map: replacing one column by a linear combination of the others leaves only terms with repeated columns, each zero. Conversely, nonzero determinant gives an inverse matrix and hence independent columns.

An invertible matrix is a square matrix $X$ admitting a matrix inverse $X^{-1}$ with $X{X}^{-1}=X^{-1}X=I$, where $I$ is the identity matrix. Equivalently, its associated linear map is bijective.

Let $X\mathbf{v}=\lambda \mathbf{v}$ with eigenvector $\mathbf{v}\ne 0$. The eigenvalue cannot be zero: otherwise applying $X^{-1}$ gives $\mathbf{v}=0$. Apply $X^{-1}$ to the eigenvector equation:Thus $1/\lambda$ is an eigenvalue of $X^{-1}$ with the same eigenvector.

For an invertible matrix $P$ whose columns give the new basis in old coordinates, the new matrix is $B=P^{-1}AP$. Multiplicativity of the determinant givessince $\det (P^{-1})\det P=1$.

To justify the cyclic property of the trace directly, for two square matrices $R,S$,Apply it with $R=P^{-1}A$ and $S=P$:Thus both the determinant and trace are invariant under change of basis.

The definition of an eigenvalue and the kernel criterion for a singular square matrix giveFor the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

A two-sided matrix inverse requires a square matrix, so $\boxed{n=3}$ is necessary. It exists only if that square matrix is also an invertible matrix, equivalently $\det M\ne 0$; choosing the dimension alone does not ensure invertibility.