Education: Exercise:Basis(Linear algebra)

Problem 1

Decide if each is a basis for

\mathbb{R}^3

$\langle \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \rangle$
$\langle \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \rangle$
$\langle \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \\ 0 \end{pmatrix} \rangle$
$\langle \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} \rangle$

Answer

By Theorem 1.12, each is a basis if and only if each vector in the space can be given in a unique way as a linear combination of the given vectors.

Yes this is a basis. The relation
$c_1\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} +c_2\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} +c_3\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
gives
$\left(\begin{array}{*{3}{c}|c} 1 &3 &0 &x \\ 2 &2 &0 &y \\ 3 &1 &1 &z \end{array}\right) \xrightarrow[-3\rho_1+\rho_3]{-2\rho_1+\rho_2} \;\xrightarrow[]{2\rho_2+\rho_3} \left(\begin{array}{*{3}{c}|c} 1 &3 &0 &x \\ 0 &-4 &0 &-2x+y \\ 0 &0 &1 &x-2y+z \end{array}\right)$
which has the unique solution $c_3=x-2y+z$ , $c_2=x/2-y/4$ , and $c_1=-x/2+3y/4$ .
This is not a basis. Setting it up as in the prior item
$c_1\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} +c_2\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} =\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
gives a linear system whose solution
$\left(\begin{array}{*{2}{c}|c} 1 &3 &x \\ 2 &2 &y \\ 3 &1 &z \end{array}\right) \xrightarrow[-3\rho_1+\rho_3]{-2\rho_1+\rho_2} \;\xrightarrow[]{2\rho_2+\rho_3} \left(\begin{array}{*{2}{c}|c} 1 &3 &x \\ 0 &-4 &-2x+y \\ 0 &0 &x-2y+z \end{array}\right)$
is possible if and only if the three-tall vector's components $x$ , $y$ , and $z$ satisfy $x-2y+z=0$ . For instance, we can find the coefficients $c_1$ and $c_2$ that work when $x=1$ , $y=1$ , and $z=1$ . However, there are no $c$ 's that work for $x=1$ , $y=1$ , and $z=2$ . Thus this is not a basis; it does not span the space.
Yes, this is a basis. Setting up the relationship leads to this reduction
$\left(\begin{array}{*{3}{c}|c} 0 &1 &2 &x \\ 2 &1 &5 &y \\ -1 &1 &0 &z \end{array}\right) \xrightarrow[]{rho_1\leftrightarrow\rho_3} \;\xrightarrow[]{\rho_1+\rho_2} \;\xrightarrow[]{(1/3)\rho_2+\rho_3} \left(\begin{array}{*{3}{c}|c} -1 &1 &0 &z \\ 0 &3 &5 &y+2z \\ 0 &0 &1/3 &x-y/3-2z/3 \end{array}\right)$
which has a unique solution for each triple of components $x$ , $y$ , and $z$ .
No, this is not a basis. The reduction
$\left(\begin{array}{*{3}{c}|c} 0 &1 &1 &x \\ 2 &1 &3 &y \\ -1 &1 &0 &z \end{array}\right) \xrightarrow[]{rho_1\leftrightarrow\rho_3} \;\xrightarrow[]{\rho_1+\rho_2} \xrightarrow[]{-1/3)\rho_2+\rho_3} \left(\begin{array}{*{3}{c}|c} -1 &1 &0 &z \\ 0 &3 &3 &y+2z \\ 0 &0 &0 &x-y/3-2z/3 \end{array}\right)$
which does not have a solution for each triple $x$ , $y$ , and $z$ . Instead, the span of the given set includes only those three-tall vectors where $x=y/3+2z/3$ .

Problem 2

Represent the vector with respect to the basis.

$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , $B=\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} -1 \\ 1 \end{pmatrix} \rangle \subseteq\mathbb{R}^2$
$x^2+x^3$ , $D=\langle 1,1+x,1+x+x^2,1+x+x^2+x^3 \rangle \subseteq\mathcal{P}_3$
$\begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}$ , $\mathcal{E}_4\subseteq\mathbb{R}^4$

Answer

We solve
$c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} +c_2\begin{pmatrix} -1 \\ 1 \end{pmatrix} =\begin{pmatrix} 1 \\ 2 \end{pmatrix}$
with
$\left(\begin{array}{*{2}{c}|c} 1 &-1 &1 \\ 1 &1 &2 \end{array}\right) \xrightarrow[]{\rho_1+\rho_2} \left(\begin{array}{*{2}{c}|c} 1 &-1 &1 \\ 0 &2 &1 \end{array}\right)$
and conclude that $c_2=1/2$ and so $c_1=3/2$ . Thus, the representation is this.
${\rm Rep}_{B}(\begin{pmatrix} 1 \\ 2 \end{pmatrix})=\begin{pmatrix} 3/2 \\ 1/2 \end{pmatrix}_B$
The relationship $c_1\cdot(1)+c_2\cdot(1+x)+c_3\cdot(1+x+x^2)+c_4\cdot(1+x+x^2+x^3)=x^2+x^3$ is easily solved by eye to give that $c_4=1$ , $c_3=0$ , $c_2=-1$ , and $c_1=0$ .
${\rm Rep}_{D}(x^2+x^3)=\begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}_D$
${\rm Rep}_{\mathcal{E}_4}(\begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}) =\begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}_{\mathcal{E}_4}$

Problem 3

Find a basis for

\mathcal{P}_2

, the space of all quadratic polynomials. Must any such basis contain a polynomial of each degree:~degree zero, degree one, and degree two?

Answer

One basis is

\langle 1,x,x^2 \rangle

. There are bases for

\mathcal{P}_2

that do not contain any polynomials of degree one or degree zero. One is

\langle 1+x+x^2,x+x^2,x^2 \rangle

. (Every basis has at least one polynomial of degree two, though.)

Problem 4

Find a basis for the solution set of this system.

\begin{array}{*{4}{rc}r} x_1 &- &4x_2 &+ &3x_3 &- &x_4 &= &0 \\ 2x_1 &- &8x_2 &+ &6x_3 &- &2x_4 &= &0 \end{array}

Answer

The reduction

\left(\begin{array}{*{4}{c}|c} 1 &-4 &3 &-1 &0 \\ 2 &-8 &6 &-2 &0 \end{array}\right) \xrightarrow[]{2\rho_1+\rho_2} \left(\begin{array}{*{4}{c}|c} 1 &-4 &3 &-1 &0 \\ 0 &0 &0 &0 &0 \end{array}\right)

gives that the only condition is that

x_1=4x_2-3x_3+x_4

. The solution set is

\{\begin{pmatrix} 4x_2-3x_3+x_4 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \,\big|\, x_2,x_3,x_4\in\mathbb{R}\} =\{x_2\begin{pmatrix} 4 \\ 1 \\ 0 \\ 0 \end{pmatrix} +x_3\begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \end{pmatrix} +x_4\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \,\big|\, x_2,x_3,x_4\in\mathbb{R}\}

and so the obvious candidate for the basis is this.

\langle \begin{pmatrix} 4 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \rangle

We've shown that this spans the space, and showing it is also linearly independent is routine.

Problem 5

Find a basis for

\mathcal{M}_{2 \! \times \! 2}

, the space of

2 \! \times \! 2

matrices.

Answer

There are many bases. This is an easy one.

\langle \begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix}, \begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \rangle

Problem 6

Find a basis for each.

The subspace $\{a_2x^2+a_1x+a_0\,\big|\, a_2-2a_1=a_0\}$ of $\mathcal{P}_2$
The space of three-wide row vectors whose first and second components add to zero
This subspace of the $2 \! \times \! 2$ matrices
$\{\begin{pmatrix} a &b \\ 0 &c \end{pmatrix} \,\big|\, c-2b=0\}$

Answer

For each item, many answers are possible.

One way to proceed is to parametrize by expressing the $a_2$ as a combination of the other two $a_2=2a_1+a_0$ . Then $a_2x^2+a_1x+a_0$ is $(2a_1+a_0)x^2+a_1x+a_0$ and
$\{(2a_1+a_0)x^2+a_1x+a_0\,\big|\, a_1,a_0\in\mathbb{R}\} =\{a_1\cdot(2x^2+x)+a_0\cdot(x^2+1)\,\big|\, a_1,a_0\in\mathbb{R}\}$
suggests $\langle 2x^2+x,x^2+1 \rangle$ . This only shows that it spans, but checking that it is linearly independent is routine.
Parametrize $\{\begin{pmatrix} a &b &c \end{pmatrix}\,\big|\, a+b=0\}$ to get $\{\begin{pmatrix} -b &b &c \end{pmatrix}\,\big|\, b,c\in\mathbb{R}\}$ , which suggests using the sequence $\langle \begin{pmatrix} -1 &1 &0 \end{pmatrix},\begin{pmatrix} 0 &0 &1 \end{pmatrix} \rangle$ . We've shown that it spans, and checking that it is linearly independent is easy.
Rewriting
$\{\begin{pmatrix} a &b \\ 0 &2b \end{pmatrix}\,\big|\, a,b\in\mathbb{R}\} =\{a\cdot\begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix} +b\cdot\begin{pmatrix} 0 &1 \\ 0 &2 \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$
suggests this for the basis.
$\langle \begin{pmatrix} 1 &0 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 0 &1 \\ 0 &2 \end{pmatrix} \rangle$

Problem 7

Check Example 1.6.

Answer

We will show that the second is a basis; the first is similar. We will show this straight from the definition of a basis, because this example appears before Theorem 1.12.

To see that it is linearly independent, we set up

c_1\cdot(\cos\theta-\sin\theta)+c_2\cdot(2\cos\theta+3\sin\theta)=0\cos\theta+0\sin\theta

. Taking

\theta=0

and

\theta=\pi/2

gives this system

\begin{array}{*{2}{rc}r} c_1\cdot 1 &+ &c_2\cdot 2 &= &0 \\ c_1\cdot (-1) &+ &c_2\cdot 3 &= &0 \end{array} \;\xrightarrow[]{\rho_1+\rho_2}\; \begin{array}{*{2}{rc}r} c_1 &+ &2c_2 &= &0 \\ &+ &5c_2 &= &0 \end{array}

which shows that

c_1=0

and

c_2=0

The calculation for span is also easy; for any

x,y\in\mathbb{R}

, we have that

c_1\cdot(\cos\theta-\sin\theta)+c_2\cdot(2\cos\theta+3\sin\theta)=x\cos\theta+y\sin\theta

gives that

c_2=x/5+y/5

and that

c_1=3x/5-2y/5

, and so the span is the entire space.

Problem 8

Find the span of each set and then find a basis for that span.

$\{1+x,1+2x\}$ in $\mathcal{P}_2$
$\{2-2x,3+4x^2\}$ in $\mathcal{P}_2$

Answer

Asking which $a_0+a_1x+a_2x^2$ can be expressed as $c_1\cdot (1+x)+c_2\cdot (1+2x)$ gives rise to three linear equations, describing the coefficients of $x^2$ , $x$ , and the constants.
$\begin{array}{*{2}{rc}r} c_1 &+ &c_2 &= &a_0 \\ c_1 &+ &2c_2 &= &a_1 \\ & &0 &= &a_2 \end{array}$
Gauss' method with back-substitution shows, provided that $a_2=0$ , that $c_2=-a_0+a_1$ and $c_1=2a_0-a_1$ . Thus, with $a_2=0$ , we can compute appropriate $c_1$ and $c_2$ for any $a_0$ and $a_1$ . So the span is the entire set of linear polynomials $\{a_0+a_1x\,\big|\, a_0,a_1\in\mathbb{R}\}$ . Parametrizing that set $\{a_0\cdot 1+a_1\cdot x\,\big|\, a_0,a_1\in\mathbb{R}\}$ suggests a basis $\langle 1,x \rangle$ (we've shown that it spans; checking linear independence is easy).
With
$a_0+a_1x+a_2x^2 =c_1\cdot(2-2x)+c_2\cdot(3+4x^2) =(2c_1+3c_2)+(-2c_1)x+(4c_2)x^2$
we get this system.
$\begin{array}{*{2}{rc}r} 2c_1 &+ &3c_2 &= &a_0 \\ -2c_1 & & &= &a_1 \\ & &4c_2 &= &a_2 \end{array} \;\xrightarrow[]{\rho_1+\rho_2} \;\xrightarrow[]{-4/3)\rho_2+\rho_3}\; \begin{array}{*{2}{rc}r} 2c_1 &+ &3c_2 &= &a_0 \\ & &3c_2 &= &a_0+a_1 \\ & &0 &= &(-4/3)a_0-(4/3)a_1+a_2 \end{array}$
Thus, the only quadratic polynomials $a_0+a_1x+a_2x^2$ with associated $c$ 's are the ones such that $0=(-4/3)a_0-(4/3)a_1+a_2$ . Hence the span is $\{(-a_1+(3/4)a_2)+a_1x+a_2x^2\,\big|\, a_1,a_2\in\mathbb{R}\}$ . Parametrizing gives $\{a_1\cdot (-1+x)+a_2\cdot ((3/4)+x^2)\,\big|\, a_1,a_2\in\mathbb{R}\}$ , which suggests $\langle -1+x,(3/4)+x^2 \rangle$ (checking that it is linearly independent is routine).

Problem 9

Find a basis for each of these subspaces of the space

\mathcal{P}_3

of cubic polynomials.

The subspace of cubic polynomials $p(x)$ such that $p(7)=0$
The subspace of polynomials $p(x)$ such that $p(7)=0$ and $p(5)=0$
The subspace of polynomials $p(x)$ such that $p(7)=0$ , $p(5)=0$ , and~ $p(3)=0$
The space of polynomials $p(x)$ such that $p(7)=0$ , $p(5)=0$ , $p(3)=0$ , and~ $p(1)=0$

Answer

The subspace is $\{a_0+a_1x+a_2x^2+a_3x^3 \,\big|\, a_0+7a_1+49a_2+343a_3=0 \}$ . Rewriting $a_0=-7a_1-49a_2-343a_3$ gives $\{(-7a_1-49a_2-343a_3)+a_1x+a_2x^2+a_3x^3 \,\big|\, a_1,a_2,a_3\in\mathbb{R} \}$ , which, on breaking out the parameters, suggests $\langle -7+x,-49+x^2,-343+x^3 \rangle$ for the basis (it is easily verified).
The given subspace is the collection of cubics $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ such that $a_0+7a_1+49a_2+343a_3=0$ and $a_0+5a_1+25a_2+125a_3=0$ . Gauss' method
$\begin{array}{*{4}{rc}r} a_0 &+ &7a_1 &+ &49a_2 &+ &343a_3 &= &0 \\ a_0 &+ &5a_1 &+ &25a_2 &+ &125a_3 &= &0 \end{array} \;\xrightarrow[]{\rho_1+\rho_2}\; \begin{array}{*{4}{rc}r} a_0 &+ &7a_1 &+ &49a_2 &+ &343a_3 &= &0 \\ & &-2a_1 &- &24a_2 &- &218a_3 &= &0 \end{array}$
gives that $a_1=-12a_2-109a_3$ and that $a_0=35a_2+420a_3$ . Rewriting $(35a_2+420a_3)+(-12a_2-109a_3)x+a_2x^2+a_3x^3$ as $a_2\cdot(35-12x+x^2)+a_3\cdot(420-109x+x^3)$ suggests this for a basis $\langle 35-12x+x^2,420-109x+x^3 \rangle$ . The above shows that it spans the space. Checking it is linearly independent is routine. (Comment. A worthwhile check is to verify that both polynomials in the basis have both seven and five as roots.)
Here there are three conditions on the cubics, that $a_0+7a_1+49a_2+343a_3=0$ , that $a_0+5a_1+25a_2+125a_3=0$ ,and that $a_0+3a_1+9a_2+27a_3=0$ . Gauss' method
$\begin{array}{*{4}{rc}r} a_0 &+ &7a_1 &+ &49a_2 &+ &343a_3 &= &0 \\ a_0 &+ &5a_1 &+ &25a_2 &+ &125a_3 &= &0 \\ a_0 &+ &3a_1 &+ &9a_2 &+ &27a_3 &= &0 \end{array} \;\xrightarrow[-\rho_1+\rho_3]{-\rho_1+\rho_2} \;\xrightarrow[]{2\rho_2+\rho_3}\; \begin{array}{*{4}{rc}r} a_0 &+ &7a_1 &+ &49a_2 &+ &343a_3 &= &0 \\ & &-2a_1 &- &24a_2 &- &218a_3 &= &0 \\ & & & &8a_2 &+ &120a_3 &= &0 \end{array}$
yields the single free variable $a_3$ , with $a_2=-15a_3$ , $a_1=71a_3$ , and $a_0=-105a_3$ . The parametrization is this.
$\{(-105a_3)+(71a_3)x+(-15a_3)x^2+(a_3)x^3\,\big|\, a_3\in\mathbb{R}\}= \{a_3\cdot(-105+71x-15x^2+x^3)\,\big|\, a_3\in\mathbb{R}\}$
Therefore, a good candidate for the basis is $\langle -105+71x-15x^2+x^3 \rangle$ . It spans the space by the work above. It is clearly linearly independent because it is a one-element set (with that single element not the zero object of the space). Thus, any cubic through the three points $(7,0)$ , $(5,0)$ , and $(3,0)$ is a multiple of this one. (Comment. As in the prior question, a worthwhile check is to verify that plugging seven, five, and three into this polynomial yields zero each time.)
This is the trivial subspace of $\mathcal{P}_3$ . Thus, the basis is empty $\langle \rangle$ .

Remark. The polynomial in the third item could alternatively have been derived by multiplying out

(x-7)(x-5)(x-3)

Problem 10

We've seen that it is possible for a basis to remain a basis when it is reordered. Must it remain a basis?

Answer

Yes. Linear independence and span are unchanged by reordering.

Problem 11

Can a basis contain a zero vector?

Answer

No linearly independent set contains a zero vector.

Problem 12

Let

\langle \vec{\beta}_1,\vec{\beta}_2,\vec{\beta}_3 \rangle

be a basis for a vector space.

Show that $\langle c_1\vec{\beta}_1,c_2\vec{\beta}_2,c_3\vec{\beta}_3 \rangle$ is a basis when $c_1, c_2, c_3\neq 0$ . What happens when at least one $c_i$ is $0$ ?
Prove that $\langle \vec{\alpha}_1,\vec{\alpha}_2,\vec{\alpha}_3 \rangle$ is a basis where $\vec{\alpha}_i=\vec{\beta}_1+\vec{\beta}_i$ .

Answer

To show that it is linearly independent, note that $d_1(c_1\vec{\beta}_1)+d_2(c_2\vec{\beta}_2)+d_3(c_3\vec{\beta}_3)=\vec{0}$ gives that $(d_1c_1)\vec{\beta}_1+(d_2c_2)\vec{\beta}_2+(d_3c_3)\vec{\beta}_3 =\vec{0}$ , which in turn implies that each $d_ic_i$ is zero. But with $c_i\neq 0$ that means that each $d_i$ is zero. Showing that it spans the space is much the same; because $\langle \vec{\beta}_1,\vec{\beta}_2,\vec{\beta}_3 \rangle$ is a basis, and so spans the space, we can for any $\vec{v}$ write $\vec{v}=d_1\vec{\beta}_1+d_2\vec{\beta}_2+d_3\vec{\beta}_3$ , and then $\vec{v}=(d_1/c_1)(c_1\vec{\beta}_1)+(d_2/c_2)(c_2\vec{\beta}_2)+(d_3/c_3)(c_3\vec{\beta}_3)$ . If any of the scalars are zero then the result is not a basis, because it is not linearly independent.
Showing that $\langle 2\vec{\beta}_1,\vec{\beta}_1+\vec{\beta}_2, \vec{\beta}_1+\vec{\beta}_3 \rangle$ is linearly independent is easy. To show that it spans the space, assume that $\vec{v}=d_1\vec{\beta}_1+d_2\vec{\beta}_2+d_3\vec{\beta}_3$ . Then, we can represent the same $\vec{v}$ with respect to $\langle 2\vec{\beta}_1,\vec{\beta}_1+\vec{\beta}_2,\vec{\beta}_1+\vec{\beta}_3 \rangle$ in this way $\vec{v}=(1/2)(d_1-d_2-d_3)(2\vec{\beta}_1)+d_2(\vec{\beta}_1+\vec{\beta}_2)+d_3(\vec{\beta}_1+\vec{\beta}_3)$ .

Problem 13

Find one vector

\vec{v}

that will make each into a basis for the space.

$\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\vec{v} \rangle$ in $\mathbb{R}^2$
$\langle \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\vec{v} \rangle$ in $\mathbb{R}^3$
$\langle x,1+x^2,\vec{v} \rangle$ in $\mathcal{P}_2$

Answer

Each forms a linearly independent set if

\vec{v}

is ommitted. To preserve linear independence, we must expand the span of each. That is, we must determine the span of each (leaving

\vec{v}

out), and then pick a

\vec{v}

lying outside of that span. Then to finish, we must check that the result spans the entire given space. Those checks are routine.

Any vector that is not a multiple of the given one, that is, any vector that is not on the line $y=x$ will do here. One is $\vec{v}=\vec{e}_1$ .
By inspection, we notice that the vector $\vec{e}_3$ is not in the span of the set of the two given vectors. The check that the resulting set is a basis for $\mathbb{R}^3$ is routine.
For any member of the span $\{c_1\cdot(x)+c_2\cdot(1+x^2)\,\big|\, c_1,c_2\in\mathbb{R}\}$ , the coefficient of $x^2$ equals the constant term. So we expand the span if we add a quadratic without this property, say, $\vec{v}=1-x^2$ . The check that the result is a basis for $\mathcal{P}_2$ is easy.