## 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 $\mathsf{ker}\ \varphi$.