## 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 $i$th 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 $j$th 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 $\mathsf{ker}\ \varphi$.