Tag Archives: linear transformation

Every degree n extension of a field F is embedded in the set of nxn matrices over F

Let F be a field, and let K be an extension of F of finite degree.

  1. Fix \alpha \in K. Prove that the mapping ‘multiplication by \alpha‘ is an F-linear transformation on K. (In fact an automorphism for \alpha \neq 0.)
  2. Deduce that K is isomorphically embedded in \mathsf{Mat}_n(F).

Let \varphi_\alpha(x) = \alpha x. Certainly then we have \varphi_\alpha(x+ry) = \alpha(x+ry) = \alpha x + r \alpha y = \varphi_\alpha(x) + r \varphi_\alpha(y) for all x,y \in K and r \in F; so \varphi_\alpha is an F-linear transformation. If \alpha \neq 0, then evidently \varphi_{\alpha^{-1}} \circ \varphi_\alpha = \varphi_\alpha \circ \varphi_{\alpha^{-1}} = 1.

Fix a basis for K over F; this yields a ring homomorphism \Psi : K \rightarrow \mathsf{Mat}_n(F) which takes \alpha and returns the matrix of \varphi_\alpha with respect to the chosen basis. Suppose \alpha \in \mathsf{ker}\ \Psi; then \varphi_\alpha(x) = 0 for all x \in K, and thus \alpha = 0. So \Psi is injective as desired.

Show that a given linear transformation has no eigenvectors

Let V = \mathbb{R}^\mathbb{N} be a countable-dimensional \mathbb{R} vector space, and define T : V \rightarrow V by T((a_i))_j = 0 if j = 0 and a_{j-i} otherwise. Show that T has no eigenvectors.


Suppose there exists a \in V, r \in \mathbb{R} nonzero, such that T(a) = ra. We claim that a_i = 0 for all i, and prove it by induction. For the base case i = 0, we have ra_0 = 0. Since r \neq 0, a_0 = 0. For the inductive step, if a_i = 0, then ra_{i+1} = a_i = 0. Again since r \neq 0, a_{i+1} = 0. So in fact a = 0.

So T has no eigenvectors.

Some properties of a QQ-vector space given the existence of a linear transformation with given properties

Let V be a finite dimensional vector space over \mathbb{Q} and suppose T is a nonsingular linear transformation on V such that T^{-1} = T^2 + T. Prove that the dimension of V is divisible by 3. If the dimension of V is precisely 3, prove that all such transformations are similar.


If T is such a transformation, then we have 1 = T^3 + T^2, and so T^3 + T^2 - 1 = 0. So the minimal polynomial of T divides p(x) = x^3 + x^2 - 1. Now p(x) is irreducible over \mathbb{Q} by the Rational Root Test (Prop. 11 on page 308 in D&F). So the minimal polynomial of T is precisely p(x). Now the characteristic polynomial of T divides some power of p(x) (Prop. 20 in D&F) and so has degree 3k for some k. But of course the degree of the characteristic polynomial is precisely the dimension of V, as desired.

Now if V has dimension 3, then the minimal polynomial of V is the characteristic polynomial, so that the invariant factors of T are simply p(x). In particular, all such T are similar.

Similar linear transformations have the same characteristic and minimal polynomials

Let V be an n-dimensional vector space over a field F. Prove that similar linear transformations on V have the same characteristic and the same minimal polynomial.


Suppose A and B are similar matrices (i.e. linear transformations) on V. Then we have an invertible matrix P such that B = P^{-1}AP.

Now \mathsf{char}(B) = \mathsf{det}(xI - B) = \mathsf{det}(xI - P^{-1}AP) = \mathsf{det}(P^{-1}xP - P^{-1}AP) = \mathsf{det}(P^{-1})\mathsf{det}(xI - A)\mathsf{det}(P) = \mathsf{det}(xI - P) = \mathsf{char}(A). So A and B have the same characteristic polynomial.

Recall that the minimal polynomial of a transformation T is the unique monomial generator of \mathsf{Ann}(V) in F[x] (under the usual action of F[x] on V). Thus, to show that A and B have the same minimal polynomial, it suffices to show that they have the same annihilator in F[x] under their respective actions on V.

To this end, suppose p(x) \in \mathsf{Ann}_A(V) (where \mathsf{Ann}_A denotes the annihilator induced by A). Then as a linear transformation, we have p(A) = 0. Say p(x) = \sum r_ix^i; then \sum r_iA^i = 0. But A^i = PB^iP^{-1}, so that \sum r_iPB^iP^{-1} = 0, and thus (left- and right-multiplying by P^{-1} and P, respectively) \sum r_iB^i = 0. So p(B) = 0 as a linear transformation, and we have p(x) \in \mathsf{Ann}_B(V). The reverse inclusion is similar, so that \mathsf{Ann}_A(V) = \mathsf{Ann}_B(V) as desired.

Express a given linear transformation in terms of a dual basis

Let V \subseteq \mathbb{Q}[x] be the \mathbb{Q}-vector space consisting of those polynomials having degree at most 5. Recall that B = \{1,x,x^2,x^3,x^4,x^5\} is a basis for this vector space over \mathbb{Q}. For each of the following maps \varphi : V \rightarrow \mathbb{Q}, verify that \varphi is a linear transformation and express \varphi in terms of the dual basis B^\ast on V^\ast = \mathsf{Hom}_F(V,F).

  1. \varphi(p) = p(\alpha), where \alpha \in \mathbb{Q}.
  2. \varphi(p) = \int_0^1 p(t)\ dt
  3. \varphi(p) = \int_0^1 t^2p(t)\ dt
  4. \varphi(p) = p^\prime(\alpha), where \alpha \in \mathbb{Q}. (Prime denotes the usual derivative of a polynomial.)

Let v_i be the element of the dual basis B^\ast such that v_i(x^j) = 1 if i = j and 0 otherwise. I’m going to just assume that integration over an interval is linear.

  1. Note that \varphi(p+rq) = (p+rq)(\alpha) = p(\alpha) + rq(\alpha) = \varphi(p) + r \varphi(q); thus \varphi is indeed a linear transformation. Moreover, note that (\sum \alpha^i v_i)(\sum c_jx^j) = \sum \alpha^i v_i(\sum c_jx^j) = \sum \alpha^i \sum c_j v_i(x^j) = \sum \alpha^i c_i = (\sum c_ix^i)(\alpha). Thus \varphi = \sum \alpha^i v_i.
  2. Note that \varphi(\sum \alpha_i x^i) = \sum \frac{\alpha_i}{i+1}. Now (\sum \frac{1}{i+1} v_i)(\sum \alpha_j x^j) = \sum \frac{1}{i+1} v_i(\sum \alpha_j x^j) = \sum \frac{1}{i+1} \sum \alpha_j v_i(x^j) = \sum \frac{\alpha_i}{i+1} = \varphi(\sum \alpha_i x^i). So \varphi = \sum \frac{1}{i+1} v_i.
  3. Note that \varphi(\sum \alpha_i x^i) = \sum \frac{\alpha_i}{i+3}. Now (\sum \frac{1}{i+3} v_i)(\sum \alpha_j x^j) = \sum \frac{1}{i+3} v_i(\sum \alpha_j x^j) = \sum \frac{1}{i+3} \sum \alpha_j v_i(x^j) = \sum \frac{\alpha_i}{i+3} = \varphi(\sum \alpha_i x^i). Thus \varphi = \sum \frac{1}{i+3} v_i.
  4. Since differentiation (of polynomials) is linear and the evaluation map is linear, this \varphi is linear. Note that (\sum (i+1)\alpha^i v_{i+1})(\sum c_jx^j) = \sum (i+1)\alpha^i v_{i+1}(\sum c_jx^j) = \sum (i+1)\alpha^i \sum c_j v_{i+1}(x^j) = \sum (i+1)\alpha^i c_{i+1} = \varphi(\sum c_ix^i). Thus \varphi = \sum (i+1)\alpha^iv_{i+1}.

Find bases for the image and kernel of a given linear transformation over different fields

Let F be a field, and let V \subseteq F[x] be the 7-dimensional vector space over F consisting precisely of those polynomials having degree at most 6. Let \varphi : V \rightarrow V be the linear transformation given by \varphi(p) = p^\prime. (See this previous exercise about the derivative of a polynomial.) For each of the following concrete fields F, find bases for the image and kernel of \varphi.

  1. \mathbb{R}
  2. \mathbb{Z}/(2)
  3. \mathbb{Z}/(3)
  4. \mathbb{Z}/(5)

Note that the elements x^i, with 0 \leq i \leq 6, form a basis of V. We will now compute the matrix of \varphi with respect to this basis. To that end, note that \varphi(1) = 0 and \varphi(x^i) = ix^{i-1} for 1 \leq i \leq 6. These computations hold in any field (and indeed in any unital ring), as we regard i as the i-fold sum of 1.

  1. Over \mathbb{R}, the matrix of \varphi is
    A = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

    The reduced row echelon form of this matrix is

    A^\prime = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right],

    all of whose columns are pivotal except the first. Thus the set \{1,x,x^2,x^3,x^4,x^5\} is a basis of \mathsf{im}\ \varphi. The solutions of A^\prime X = 0 now have the form X(x_1) = [x_1\ 0\ 0\ 0\ 0\ 0\ 0]^\mathsf{T}, and thus \{1\} is a basis of \mathsf{ker}\ \varphi.

  2. Over \mathbb{Z}/(2), the matrix of \varphi is
    A = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

    The reduced row echelon form of this matrix is

    A^\prime = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right],

    whose 2nd, 4th, and 6th columns are pivotal. Thus \{1,x^2,x^4\} is a basis of \mathsf{im}\ \varphi. The solutions of A^\prime X = 0 now have the form X(x_1,x_3,x_5,x_7) = [x_1\ 0\ x_3\ 0\ x_5\ 0\ x_7]^\mathsf{T}. Choosing x_i appropriately, we see that \{1,x^2,x^4,x^6\} is a basis for \mathsf{ker}\ \varphi.

  3. Over \mathbb{Z}/(3), the matrix of \varphi is
    A = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

    The reduced row echelon form of this matrix is

    A^\prime = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right],

    whose 2nd, 3rd, 5th, and 6th columns are pivotal. Thus \{1,2x,x^3,2x^4\} is a basis for \mathsf{im}\ \varphi. Now the solutions of A^\prime X = 0 have the form X(x_1,x_4,x_7) = [x_1\ 0\ 0\ x_4\ 0\ 0\ x_7]^\mathsf{T}. Choosing the x_i appropriately, we see that \{1,x^3,x^6\} is a basis for \mathsf{ker}\ \varphi.

  4. Over \mathbb{Z}/(5), the matrix of \varphi is
    A = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

    The reduced row echelon form of this matrix is

    A^\prime = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right],

    all of whose columns are pivotal except the 1st and the 6th. Thus \{1,x,x^2,x^3,x^4\} is a basis for \mathsf{im}\ \varphi. The solutions of A^\prime X = 0 have the form X(x_1,x_7) = [x_1\ 0\ 0\ 0\ 0\ x_5\ 0]^\mathsf{T}; thus \{1,x^5\} is a basis for \mathsf{ker}\ \varphi.

Find bases for the image and kernel of a given linear transformation

Let V \subseteq \mathbb{Q}[x] be the 6 dimensional vector space over \mathbb{Q} consisting of the polynomials having degree at most 5. Let \varphi : V \rightarrow V be the map given by \varphi(p) = x^2p^{\prime\prime} - 6xp^\prime + 12p, where p^\prime and p^{\prime\prime} denote the first and second derivative of p with respect to x. (See this previous exercise.)

  1. Prove that \varphi is a linear transformation.
  2. We showed previously that the set \{1,x,x^2,x^3,x^4,x^5\} is a basis of V. Find bases for the image and kernel of \varphi with respect to this basis.

We begin with a lemma.

Lemma: Let R be a commutative unital ring. Then D : R[x] \rightarrow R[x] given by D(p) = p^\prime is a module homomorphism. Proof: Let p(x), q(x) \in R[x] and let r \in R. Then (p+rq)^\prime(x) = \sum (i+1)(p_i+rq_i)x^i = \sum(i+1)p_ix^i + r\sum (i+1)q_ix^i = p^\prime(x) + rq^\prime(x). Thus D(p+rq) = D(p) + rD(q), and so D is a module homomorphism. \square

Thus it is clear that \varphi is a linear transformation.

Note the following:

  1. \varphi(1) = 12
  2. \varphi(x) = 6x
  3. \varphi(x^2) = 2x^2
  4. \varphi(x^3) = 0
  5. \varphi(x^4) = 0
  6. \varphi(x^5) = 2x^5

Thus we see that the matrix of \varphi is

A = \left[ \begin{array}{cccccc} 12 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{array} \right].

The reduced row echelon form of A is the matrix

A^\prime = \left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right].

Since only the 1st, 2nd, 3rd, and 6th columns of A^\prime are pivotal, we see that the set \{12, 6x, 2x^2, 2x^5 \} forms a basis for \mathsf{im}\ \varphi. Next, the solutions of A^\prime X = 0 have the form X(x_4,x_5) = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ x_4 \\ x_5 \\ 0 \end{array} \right]. Letting (x_4,x_5) \in \{(1,0),(0,1)\}, we see that \{(0,0,0,1,0,0),(0,0,0,0,1,0)\} is a basis for \mathsf{ker}\ \varphi.

Find bases for the image and kernel of a given linear transformation

Let V = \mathsf{Mat}_{2,2}(\mathbb{R}) be the set of all 2 \times 2 matrices over \mathbb{R}, and consider V as an \mathbb{R} vector space in the usual way. Let \mathsf{tr} : V \rightarrow \mathbb{R} be defined by \mathsf{tr}\left( \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \right) = a+d.

  1. Show that E_{1,1} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], E_{1,2} = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], E_{2,1} = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right], and E_{2,2} = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] form a basis for V.
  2. Prove that \mathsf{tr} is a linear transformation and determine its matrix with respect to the basis given in part (1) and the basis \{1\} of \mathbb{R}. Find bases for the image and kernel of \mathsf{tr}.

Note that

\alpha E_{1,1} + \beta E_{1,2} + \gamma E_{2,1} + \delta E_{2,2} = \left[ \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right].

In particular, this linear combination is zero if and only if each coefficient is zero. Thus the E_{i,j} are linearly independent. Moreover, they certainly generate V, and thus form a basis.

It is clear that \mathsf{tr} is a linear transformation. Note that \mathsf{tr}(E_{1,1}) = 1, \mathsf{tr}(E_{1,2}) = 0, \mathsf{tr}(E_{2,1}) = 0, and \mathsf{tr}(E_{2,2}) = 1. Thus the matrix of \mathsf{tr}, with respect to this basis, is A = [1\ 0\ 1\ 0]. Note that A is in reduced row echelon form, and only its first column is pivotal. Thus \{1\} is a basis for \mathsf{im}\ \mathsf{tr}. The solutions of AX = 0 have the form X(x_2,x_3,x_4) = \left[ \begin{array}{c} -x_3 \\ x_2 \\ x_3 \\ x_4 \end{array} \right]. Letting (x_2,x_3,x_4) \in \{(1,0,0),(0,1,0),(0,0,1)\}, we see that the set \{(0,1,0,0),(-1,0,1,0),(0,0,0,1)\} is a basis for \mathsf{ker}\ \mathsf{tr}.

Find bases for the image and kernel of a given linear transformation

Consider \mathbb{R}^4 and \mathbb{R}^2 as \mathbb{R}-vector spaces in the usual way. Let \varphi : \mathbb{R}^4 \rightarrow \mathbb{R}^2 be the linear transformation such that \varphi(1,0,0,0) = (1,-1), \varphi(1,-1,0,0) = (0,0), \varphi(1,-1,1,0) = (1,-1), and \varphi(1,-1,1,-1) = (0,0). Find bases for the image and kernel of \varphi.


We will use the strategy given in this previous exercise.

Letting f_1 = (1,0,0,0), f_2 = (1,-1,0,0), f_3 = (1,-1,1,0), and f_4 = (1,-1,1,-1) and letting e_i denote the standard basis vectors, we evidently have e_1 = f_1, e_2 = f_1 - f_2, e_3 = f_3 - f_2, and e_4 = f_3 - f_4. Thus \varphi(e_1) = (1,-1), \varphi(e_2) = (1,-1), \varphi(e_3) = (1,-1), and \varphi(e_4) = (1,-1).

Thus the matrix of \varphi with respect to the standard bases is A = \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ \text{-}1 & \text{-}1 & \text{-}1 & \text{-}1 \end{array} \right]. The reduced row echelon form A is then A^\prime = \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]. Since only the first column of A^\prime is pivotal, the image of \varphi is spanned by (1,-1), which is certainly a basis.

The solutions of A^\prime X = 0 have the form X(x_2,x_3,x_4) = \left[ \begin{array}{c} -x_2-x_3-x_4 \\ x_2 \\ x_3 \\ x_4 \end{array} \right]. Letting (x_2,x_3,x_4) \in \{(1,0,0),(0,1,0),(0,0,1)\}, we see that the set \{(-1,1,0,0),(-1,0,1,0),(-1,0,0,1)\} is a basis for \mathsf{ker}\ \varphi.

Find bases for the image and kernel of a given linear transformation

Let \varphi be the linear transformation from the \mathbb{R} vector space \mathbb{R}^4 (with the standard basis) to itself is given by the matrix

A = \left[ \begin{array}{cccc} 1 & \text{-}1 & 0 & 3 \\ \text{-}1 & 2 & 1 & \text{-}1 \\ \text{-}1 & 1 & 0 & \text{-}3 \\ 1 & \text{-}2 & \text{-}1 & 1 \end{array} \right].

Find bases for the image and kernel of \varphi.


We will use the strategy given in this previous exercise.

Evidently the reduced row echelon form of A is the matrix

A^\prime = \left[ \begin{array}{cccc} 1 & 0 & 1 & 5 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].

Since only the first two columns of A^\prime are pivotal, the set \{(1,-1,-1,1), (-1,2,1,-2)\} is a basis for the image of \varphi.

Moreover, the solutions of A^\prime X = 0 have the form X(x_3,x_4) = \left[ \begin{array}{c} -x_3-5x_4 \\ -x_3-2x_4 \\ x_3 \\ x_4 \end{array} \right]. Letting (x_3,x_4) \in \{(1,0), (0,1)\}, we see that the set \{-1,-1,1,0), (-5,-2,0,1)\} is a basis for \mathsf{ker}\ \varphi.