Tag Archives: basis

Exhibit a quadratic field as a field of matrices

Let K = \mathbb{Q}(\sqrt{D}), where D is a squarefree integer. Let \alpha = a+b\sqrt{D} be in K, and consider the basis B = \{1,\sqrt{D}\} of K over \mathbb{Q}. Compute the matrix of the \mathbb{Q}-linear transformation ‘multiplication by \alpha‘ (described previously) with respect to B. Give an explicit embedding of \mathbb{Q}(\sqrt{D}) in the ring \mathsf{Mat}_2(\mathbb{Q}).


We have \varphi_\alpha(1) = a+b\sqrt{D} and \varphi(\alpha)(\sqrt{D}) = bD + a\sqrt{D}. Making these the columns of a matrix M_\alpha, we have M_\alpha = \begin{bmatrix} a & bD \\ b & a \end{bmatrix}, and this is the matrix of \varphi_\alpha with respect to B. As we showed in the exercise linked above, \alpha \mapsto M_\alpha is an embedding of K in \mathsf{Mat}_2(\mathbb{Q}).

Compare to this previous exercise about \mathbb{Z}[\sqrt{D}].

The columns of a square matrix over a field are linearly independent if and only if the determiniant of the matrix is nonzero

Let F be a field and let A = [A_1 | \cdots | A_n] be a square matrix of dimension n \times n over F. Prove that the set \{A_i\}_{i=1}^n is linearly independent if and only if \mathsf{det}\ A \neq 0.


Let B be the reduced row echelon form of A, and let P be invertible such that PA = B.

Suppose the columns of A are linearly independent. Now B has column rank n. In particular, B = I. Now 1 = \mathsf{det}(PA) = \mathsf{det}(P) \mathsf{det}(A); so \mathsf{det}(A) \neq 0.

We prove the converse contrapositively. Suppose the columns of A are linearly dependent; then the column rank of B is strictly less than n, so that B has a row of all zeros. Using the cofactor expansion formula, \mathsf{det}(B) = 0. Since P is invertible, its determinant is nonzero; thus \mathsf{det}(A) = 0. Thus if \mathsf{det}(A) \neq 0, then the columns of A are linearly independent.

The dual basis of an infinite dimensional vector space does not span the dual space

Let F be a field and let V be an infinite dimensional vector space over F; say V has basis B = \{v_i\}_I. Prove that the dual basis \{\widehat{v}_i\}_I does not span the dual space \widehat{V} = \mathsf{Hom}_F(V,F).


Define a linear transformation \varphi on V by taking v_i to 1 for all i \in I. Note that for all i \in I, \varphi(v_i) \neq 0. Suppose now that \varphi = \sum_{i \in K} \alpha_i\widehat{v}_i where K \subseteq I is finite; for any j \notin K, we have (\sum \alpha_i \widehat{v}_i)(v_j) = 0, while \varphi(v_j) = 1. Thus \varphi is not in \mathsf{span}\ \{\widehat{v}_i\}_I.

So the dual basis does not span \widehat{V}.

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.

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

Let V = \mathbb{R}^5 and W = \mathbb{R}^4 be considered as vector spaces over \mathbb{R} with the standard bases. Let \varphi : V \rightarrow W be the linear transformation given by \varphi(x,y,z,u,v) = (x+2y+3z+4u+4v, -2x-4y+2v, x+2y+u-2v, x+2y-v). Find bases for the image and kernel of \varphi.


We will follow the strategy given in this previous exercise. To wit, we compute the matrix A of \varphi as well as the reduced row echelon form A^\prime of A. Then the columns of A corresponding to the pivotal columns of A^\prime give a basis for the image of \varphi. We construct a basis for the space of solutions to the equation A^\prime X = 0.

Let e_i denote the ith standard basis vector. Evidently then we have \varphi(e_1) = (1,\text{-}2,1,1), \varphi(e_2) = (2,\text{-}4,2,2), \varphi(e_3) = (3,0,0,0), \varphi(e_4) = (4,0,1,0), and \varphi(e_5) = (4,2,\text{-}2,\text{-}1). Thus the matrix of \varphi is A = \left[ \begin{array}{ccccc} 1 & 2 & 3 & 4 & 4 \\ \text{-}2 & \text{-}4 & 0 & 0 & 2 \\ 1 & 2 & 0 & 1 & \text{-}2 \\ 1 & 2 & 0 & 0 & \text{-}1 \end{array} \right]. Evidently, the reduced row echelon form of A is A^\prime = \left[ \begin{array}{ccccc} 1 & 2 & 0 & 0 & \text{-}1 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & \text{-}1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right].

Since only the 1st, 3rd, and 4th columns of A^\prime are pivotal, the set

\left\{ \left[ \begin{array}{c} 1 \\ \text{-}2 \\ 1 \\ 1 \end{array} \right], \left[ \begin{array}{c} 3 \\ 0 \\ 0 \\ 0 \end{array} \right], \left[ \begin{array}{c} 4 \\ 0 \\ 1 \\ 0 \end{array} \right] \right\}

is a basis for \mathsf{im}\ \varphi.

Moreover, evidently the solutions of A^\prime X = 0 are of the form X(y,v) = \left[ \begin{array}{c} v - 2y \\ y \\ -3v \\ v \\ v \end{array} \right]. Letting (y,v) \in \{ (1,0), (0,1) \}, we see that the set

\left\{ \left[ \begin{array}{c} \text{-}2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array} \right], \left[ \begin{array}{c} 1 \\ 0 \\ \text{-}3 \\ 1 \\ 1 \end{array} \right] \right\}

is a basis for \mathsf{ker}\ \varphi.

Computing the image and kernel of a linear transformation

Let V be an n-dimensional vector space with basis B = \{e_i\}_{i=1}^n and let W be an m-dimensional vector space with basis E = \{f_i\}_{i=1}^m. Let \varphi : V \rightarrow W be a linear transformation, and let A = M^E_B(\varphi). (That is, the jth column of A is the coordinates of e_j in terms of the ordered basis E. Let A^\prime be the reduced row echelon form of A.

  1. Prove that the dimension of \mathsf{im}\ \varphi is the row rank (i.e. number of nonzero rows) of A^\prime. Prove moreover that the columns of A corresponding to the pivotal columns of A^\prime form a basis of \mathsf{im}\ \varphi.
  2. Note that the kernel of \varphi is precisely the solution of the equation A^\prime X = 0. We saw how to solve such equations in this previous exercise. Prove that \varphi is injective if and only if A^\prime has row rank n. If \varphi is not injective, construct a basis for \mathsf{ker}\ \varphi.

  1. Write A = [A_1 | \cdots | A_n] as a column matrix. Then A[x_1 | \cdots | x_n]^\mathsf{T} = \sum A_i x_i. In particular, the image of \varphi is the span of the columns of A. In particular, as we showed previously, the columns of A corresponding to the pivotal columns of A^\prime form a basis of \mathsf{im}\ \varphi. In particular, the dimension of this image is the row rank of A^\prime.
  2. Note that \varphi is injective if and only if \mathsf{im}\ \varphi has dimension n, if and only if (by part 1) A^\prime has row rank n. We showed previously how to construct the solutions of A^\prime X = 0; specifically, for each nonpivotal column of A^\prime, we fix one corresponding entry in X to be 1 and the other (nonpivotal) entries 0. The other entries of X are then determined uniquely. These vectors are certainly linearly independent, and generate \mathsf{ker}\ \varphi.