We now prove: for every finite-dimensional complex vector space V and every linear map T : V V , there exists an ordered basis X of V such that [T]X is a Jordan canonical form. The proof is by induction on n = dim(V ). For n = 0 and n = 1, the result follows immediately. Now assume n > 1 and the result is already known for all complex vector spaces of smaller dimension than n. Pick a fixed eigenvalue c of the given linear map T, and let v 6= 0 be an associated eigenvector. Let U = T c IdV , where IdV denotes the identity map on V . Applying Fitting’s lemma to the linear map U, we get a direct sum V = Z W ; where Z and W are U-invariant (hence also T-invariant) subspaces of V such that U|Z is nilpotent and U|W is an isomorphism. On one hand, U|Z is nilpotent, so we know there is an ordered basis X1 of Z and a partition of the integer k = dim(Z) such that [U|Z]X1 = J(0; ). It follows that [T|Z]X1 = [U|Z + c IdZ]X1 = [U|Z]X1 + [c IdZ]X1 = J(0; ) + cIk = J(c; ). On the other hand, v cannot lie in W, since otherwise U|W(v) = U(v) = 0 = U|W(0) contradicts the fact that U|W is an isomorphism. So W 6= V , forcing dim(W) < dim(V ). By the induction hypothesis, there exists an ordered basis X2 of W such that [U|W]X2 is a Jordan canonical form matrix J1. By the same calculation used above, we see that [T|W]X2 is the matrix J2 obtained from J1 by adding c to every diagonal entry. This new matrix is also a Jordan canonical form. Finally, taking X to be the concatenation of X1 and X2, we know X is an ordered basis of V such that [T]X = blk-diag(J(c; ), J2). This matrix is a Jordan canonical form, so the induction proof is complete. The existence of Jordan canonical forms for linear maps implies the existence of Jordan canonical forms for matrices, as follows. Given A Mn(C), let T : Cn Cn be the linear map defined by T(v) = Av for all column vectors v Cn. Choose an ordered basis X of Cn such that J = [T]X is a Jordan canonical form. We know J = P1AP for some invertible P Mn(C), so A is similar to a Jordan canonical form. We should also point out that the only special feature of the field C needed in thisproof was that every linear map on an n-dimensional C-vector space (with n > 0) has an eigenvalue in C. This follows from the fact that all non-constant polynomials in C[x] split into products of linear factors. The Jordan canonical form theorem extends to any field F having the latter property (such fields are called algebraically closed). Solution We now prove: for every finite-dimensional complex vector space V and every linear map T : V V , there exists an ordered basis X of V such that [T]X is a Jordan canonical form. The proof is by induction on n = dim(V ). For n = 0 and n = 1, the result follows immediately. Now assume n > 1 and the result is already known for all complex vector spaces of smaller dimension than n. Pick a fixed eigenvalue c of the given linear map T, and let v 6= 0 be an associated eigenvector. Let U = T c IdV , where IdV denotes the .

1. We now prove: for every finite-dimensional complex vector space V and every linear
map T : V V , there exists an ordered basis X of V such that [T]X is a Jordan canonical
form.
The proof is by induction on n = dim(V ). For n = 0 and n = 1, the result follows
immediately. Now assume n > 1 and the result is already known for all complex vector
spaces of smaller dimension than n.
Pick a fixed eigenvalue c of the given linear map T, and let v 6= 0 be an associated
eigenvector. Let U = T c IdV , where IdV denotes the identity map on V .
Applying
Fitting’s lemma to the linear map U, we get a direct sum V = Z W ; where Z and W are
U-invariant (hence also T-invariant) subspaces of V such that U|Z is nilpotent and U|W is
an isomorphism. On one hand, U|Z is nilpotent, so we know there is an ordered basis X1 of
Z and a partition of the integer k = dim(Z) such that [U|Z]X1 = J(0; ).
It follows that
[T|Z]X1 = [U|Z + c IdZ]X1 = [U|Z]X1 + [c IdZ]X1 = J(0; ) + cIk = J(c; ).
On the other hand, v cannot lie in W, since otherwise U|W(v) = U(v) = 0 = U|W(0)
contradicts the fact that U|W is an isomorphism. So W 6= V , forcing dim(W) < dim(V ).
By the induction hypothesis, there exists an ordered basis X2 of W such that [U|W]X2 is a
Jordan canonical form matrix J1. By the same calculation used above, we see that [T|W]X2
is the matrix J2 obtained from J1 by adding c to every diagonal entry. This new matrix is
also a Jordan canonical form. Finally, taking X to be the concatenation of X1 and X2, we
know X is an ordered basis of V such that [T]X = blk-diag(J(c; ), J2). This matrix is a
Jordan canonical form, so the induction proof is complete.
The existence of Jordan canonical forms for linear maps implies the existence of Jordan
canonical forms for matrices, as follows. Given A Mn(C), let T : Cn Cn be the linear
map defined by T(v) = Av for all column vectors v Cn. Choose an ordered basis X of Cn
such that J = [T]X is a Jordan canonical form. We know J = P1AP for some invertible
P Mn(C), so A is similar to a Jordan canonical form.
We should also point out that the only special feature of the field C needed in thisproof was that
every linear map on an n-dimensional C-vector space (with n > 0) has an eigenvalue in C. This
follows from the fact that all non-constant polynomials in C[x] split into products of linear
factors. The Jordan canonical form theorem extends to any field F
having the latter property (such fields are called algebraically closed).
Solution

2. We now prove: for every finite-dimensional complex vector space V and every linear
map T : V V , there exists an ordered basis X of V such that [T]X is a Jordan canonical
form.
The proof is by induction on n = dim(V ). For n = 0 and n = 1, the result follows
immediately. Now assume n > 1 and the result is already known for all complex vector
spaces of smaller dimension than n.
Pick a fixed eigenvalue c of the given linear map T, and let v 6= 0 be an associated
eigenvector. Let U = T c IdV , where IdV denotes the identity map on V .
Applying
Fitting’s lemma to the linear map U, we get a direct sum V = Z W ; where Z and W are
U-invariant (hence also T-invariant) subspaces of V such that U|Z is nilpotent and U|W is
an isomorphism. On one hand, U|Z is nilpotent, so we know there is an ordered basis X1 of
Z and a partition of the integer k = dim(Z) such that [U|Z]X1 = J(0; ).
It follows that
[T|Z]X1 = [U|Z + c IdZ]X1 = [U|Z]X1 + [c IdZ]X1 = J(0; ) + cIk = J(c; ).
On the other hand, v cannot lie in W, since otherwise U|W(v) = U(v) = 0 = U|W(0)
contradicts the fact that U|W is an isomorphism. So W 6= V , forcing dim(W) < dim(V ).
By the induction hypothesis, there exists an ordered basis X2 of W such that [U|W]X2 is a
Jordan canonical form matrix J1. By the same calculation used above, we see that [T|W]X2
is the matrix J2 obtained from J1 by adding c to every diagonal entry. This new matrix is
also a Jordan canonical form. Finally, taking X to be the concatenation of X1 and X2, we
know X is an ordered basis of V such that [T]X = blk-diag(J(c; ), J2). This matrix is a
Jordan canonical form, so the induction proof is complete.
The existence of Jordan canonical forms for linear maps implies the existence of Jordan
canonical forms for matrices, as follows. Given A Mn(C), let T : Cn Cn be the linear
map defined by T(v) = Av for all column vectors v Cn. Choose an ordered basis X of Cn
such that J = [T]X is a Jordan canonical form. We know J = P1AP for some invertible
P Mn(C), so A is similar to a Jordan canonical form.
We should also point out that the only special feature of the field C needed in thisproof was that
every linear map on an n-dimensional C-vector space (with n > 0) has an eigenvalue in C. This
follows from the fact that all non-constant polynomials in C[x] split into products of linear
factors. The Jordan canonical form theorem extends to any field F
having the latter property (such fields are called algebraically closed).