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At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

The equation is solvable exactly when $b\in \mathrm{im}A$, the image of a linear map, equivalently when the rank of a matrix satisfies $\mathrm{rank}A=\mathrm{rank}[A\mid b]$. If it is solvable and $x_0$ is one solution, every solution is preciselyIndeed differences of solutions lie in the kernel of a linear map, and adding a kernel vector preserves the equation. Hence there are no solutions when $b\notin \mathrm{im}A$; one solution when $A$ is invertible; and infinitely many solutions when $b\in \mathrm{im}A$ and $A$ is singular. In the last case the rank-nullity theorem gives a nonzero kernel vector $v$, and $x_0+tv$ gives distinct solutions for every real $t$. This proves that a finite number greater than one is impossible.

Use the eigenvector basis $v_1=(-1,0,1{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$. The forcing has decomposition $b=-v_1+3v_3$. If $x=c_1{v}_1+c_2{v}_2+c_3{v}_3$, the linear system reduces toFor $\lambda =0$, all coefficients are determined, givingFor $\lambda =1$, the first equation would say $0=-1$, so there are no solutions. For $\lambda =2$, $c_1=1$, $c_3=3$, and $c_2$ is arbitrary. Thus all solutions areThese outcomes illustrate respectively an invertible matrix, incompatible forcing outside its image of a linear map, and a compatible forcing with a nontrivial kernel of a linear map.

For a linear map $\alpha :{\mathbb{R}}^m\to {\mathbb{R}}^n$, its kernel and image of a linear map areThe kernel is a subspace of the domain and the image a subspace of the codomain. Given the stated bases, define the matrix of a linear map byIts $j$th column is the coordinate vector of $\alpha (e_j)$; if $v={\sum }_j{v}_j{e}_j$, the output coordinates are $(\alpha v{)}_i={\sum }_j{A}_{ij}{v}_j$.

For the change of basis, write $e_j^{\prime }={\sum }_k{P}_{kj}{e}_k$ and $f_i^{\prime }={\sum }_{\ell }{Q}_{\ell i}{f}_{\ell }$. The matrices $P,Q$ are invertible. Input coordinates change by $v_{\mathrm{old}}=P{v}_{\mathrm{new}}$ and output coordinates by $w_{\mathrm{old}}=Q{w}_{\mathrm{new}}$, soThe domain and codomain basis changes need not be the same matrix.

For $\beta$, the two input vectors form a basis of ${\mathbb{R}}^2$, so its image is the span of $v=(1,2,3{)}^T$ and $w=(6,4,2{)}^T$. The vector $n=(1,-2,1{)}^T$ is orthogonal to both, since $n\cdot v=n\cdot w=0$. Every possible output therefore satisfies $n\cdot \beta x=0$, whereas the target itself is $n$ and $n\cdot n=6$. Hence there is no $x$ with the required output. This uses an image obstruction by an annihilating functional, rather than an inconsistent guessed input.

For $\gamma$, use the input basis $v_1=(1,2,0{)}^T$, $v_2=(0,1,1{)}^T$, $v_3=(0,1,0{)}^T$. Its determinant is $-1$, so it is genuinely a basis. Write $x=s{v}_1+t{v}_2+w{v}_3$. ThenSubstitution back into physical coordinates givesFor a direct check, its standard matrix is $\left(\begin{array}{ccc}1& 0& -2\\ 1& 1& 0\end{array}\right)$, which annihilates exactly that line.

For a linear map $T:V\to W$ with $V$ finite-dimensional, the rank-nullity theorem statesTo prove it, choose a basis $v_1,\dots ,v_k$ of the kernel of a linear map and use basis extension to obtain $v_1,\dots ,v_n$ as a basis of $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span the image of a linear map: applying $T$ to a basis expansion kills the first $k$ terms. They are also linearly independent. Indeed, if ${\sum }_{j>k}{a}_{j}T{v}_j=0$, then ${\sum }_{j>k}{a}_j{v}_j\in \ker T$, so it is a combination of $v_1,\dots ,v_k$. Independence of the full basis forces every $a_j=0$. Consequently $\dim \mathrm{im}T=n-k$, proving the theorem.

Here the image of a linear map has dimension $r\in \{0,1,2,3\}$, since it lies in ${\mathbb{R}}^3$. ThusEvery possibility occurs: for $0\le r\le 3$, map $(x_1,\dots ,x_5)$ to $(x_1,\dots ,x_r,0,\dots ,0)\in {\mathbb{R}}^3$. This map has rank $r$ and kernel dimension $5-r$.

Write a vector in the domain as $(u,v,w,t{)}^T$. The first and third equations defining the kernel of a linear map imply $v=2\alpha u-2t$. Substitution into the second equation then givesFor $\alpha \ne 1$, this forces $t=(\alpha +1)u$, $v=-2u$ and $w=3u$. HenceThe kernel of a linear map has dimension one, so the rank-nullity theorem gives rank three and proves the assertion about the image of a linear map. Alternatively, the minor formed from columns $2,3,4$ has determinant $2(\alpha -1)$, including a nonzero value at $\alpha =-1$.

For $\alpha =1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. ThusThe third row is the first row minus the second. The image of a linear map is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1{)}^T$ and $(1,0,1{)}^T$, are linearly independent, so the matrix rank is two and this containment is equality:This is the sole kernel jump in a parameter-dependent linear map; there is no further exceptional case at $\alpha =-1$.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

For a linear map $\alpha :U\to V$ with $U$ a finite-dimensional vector space, the rank-nullity theorem isChoose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_r$ of $U$. The vectors $\alpha (v_j)$ span the image of a linear map, because the kernel basis contributes zero. If ${\sum }_j{c}_j\alpha (v_j)=0$, then ${\sum }_j{c}_j{v}_j$ belongs to the kernel and is a combination of the $u_i$; independence of the extended basis makes every $c_j=0$. Thus the image basis has $r$ elements and $\dim U=k+r$, proving the theorem.

For a rank-two map on ${\mathbb{R}}^3$, the direct sum occurs for $\alpha (x,y,z)=(x,y,0)$: its kernel is $\mathbb{R}{e}_3$ and its image is $\mathrm{span}(e_1,e_2)$. It fails for $\beta (x,y,z)=(y,z,0)$: the kernel is $\mathbb{R}{e}_1$, contained in the two-dimensional image $\mathrm{span}(e_1,e_2)$. The rank-nullity theorem fixes the sum of the dimensions, but does not imply zero intersection.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

The homogeneous linear system has solution set $\ker A$, a vector subspace. If it contains only the zero vector, no two different solutions of the full system can exist: their difference would be a nonzero homogeneous solution.

A particular solution ${\mathbf{v}}_p$ to the full system exists exactly when $\mathbf{w}$ belongs to the image of a linear map $\mathrm{im}A$. If it exists, every solution isIndeed, subtracting $A{\mathbf{v}}_p=\mathbf{w}$ from $A\mathbf{v}=\mathbf{w}$ gives $A(\mathbf{v}-{\mathbf{v}}_p)=0$, and conversely adding a kernel vector preserves the right-hand side. Thus there are no solutions, one solution if the kernel is trivial, or infinitely many solutions if it is nontrivial. The last solution set is an affine subspace of dimension $\dim \ker A$, determined by the rank-nullity theorem. For a square nonsingular matrix there is one solution for every right-hand side.

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

The first three output components equal $x_1,x_2,x_3$, while the fourth is $x_4-{\sum }_{k=1}^4{x}_k=-x_1-x_2-x_3$. HenceIts outputs have component sum zero, and applying $P$ to any such output leaves it fixed. ThusThe first three columns give a basis of the image of a linear map:They are linearly independent because their first three coordinates are the standard basis. Their span is precisely the three-dimensional hyperplane $y_1+y_2+y_3+y_4=0$.

Apply the Gram-Schmidt process. First $q_1=v_1/\sqrt{2}$. Subtracting the $q_1$ projection from $v_2$ gives $(-1,2,0,-1{)}^T/2$, of length $\sqrt{3/2}$. Subtracting both preceding projections from $v_3$ gives $(-1,-1,3,-1{)}^T/3$, of length $2/\sqrt{3}$. Therefore an orthonormal basis isTheir pairwise dot products vanish and each has length one. Although this image admits an orthonormal basis, $P$ itself is an oblique projection: its kernel is the fourth coordinate axis, which is not perpendicular to the image, and $P\ne P^T$.

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

Let $Kx=a\phi (x)$ on a vector space, where $\phi$ is a linear functional. Its image of a linear map lies in the span of $a$. Applying $K$ twice gives $K^{2}x=a\phi (a)\phi (x)=\phi (a)Kx$. Mathematical induction then proves the displayed power formula. If $\phi (a)=0$, $K^2=0$, even when $K\ne 0$; this is a nilpotent operator. If $\phi (a)\ne 0$, then $K/\phi (a)$ is an idempotent linear map. In a real inner product space, the special case $\phi (x)=b\cdot x$ gives $\phi (a)=a\cdot b$.

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

The operator range is the image of a linear map $K$. It need not be closed in an infinite-dimensional Hilbert space; that failure makes the Moore–Penrose inverse of an operator unbounded on its domain.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

The rank of a linear map is the dimension of its image of a linear map. In finite-dimensional spaces it equals the rank of a matrix representing the map in any bases.

Let $T$ be a singular endomorphism of a finite-dimensional vector space. Choose $0\ne u\in \ker T$ and a nonzero linear functional $\lambda$ vanishing on $\mathrm{im}T$, possible because that image of a linear map is proper. Then $S(x)=\lambda (x)u$ is a nonzero rank-one endomorphism with $TS=ST=0$.