Finding Basis And Dimension Of $U,W,U\cap W,U + W$

Question

Let there be two subspaces $U=\{1+x+x^3,x^2-1,1+x+x^2\}$ $W=\{p\in \mathbb{R}_{3}[x]: p(1)=p(2)\}$ Find Basis And Dimension Of $U,W,U\cap W,U +W$

$U$

$\begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 1\\ 0 & 1 & 1\\ 1 & 0 & 0 \end{pmatrix}\Rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}$

So $\{1+x+x^3,x^2-1,1+x+x^2\}$ is a basis for $U$ and $dim(U)=3$

$W$

$W=\{a+bx+cx^2+dx^3: a+b+c+d=a+2b+4c+8d\}\iff \{a+bx+cx^2+dx^3: b+3c+7d=0\}\iff \{a+bx+cx^2+dx^3: b=-3c-7d\}\iff \{a+bx+cx^2+dx^3:(-3c-7d)x+cx^2+dx^3\}=\{a+bx+cx^2+dx^3:c(-3x+x^2)+d(-7x+x^3)\}=\{a+bx+cx^2+dx^3:c(-3x+x^2)+d(-7x+x^3)\}$

So a basis for $W$ is $\{(-3x+x^2),(-7x+x^3)\}$ and $dim(W)=2$

$U\cap W$

For $U$

$ \left(\begin{array}{rrr|r} 1 & -1 & 1 & a \\ 1 & 0 & 1 & b\\ 0 & 1 & 1 &c\\ 1 & 0 & 0 & d \end{array}\right)\rightarrow \left(\begin{array}{rrr|r} 1 & 0 & 0 & d \\ 0 & -1 & 1 & a-d\\ 0 & 0 & 1 &b-d\\ 0 & 0 & 0 & a-2b+c+d \end{array}\right) $

For $W$:

$b+3c+7d=0$

So we have

$\begin{pmatrix} 1 & -2 & 1 & 1\\ 0 & 1 & 3 & 7\\ \end{pmatrix}$ which its null space is: $\{\begin{pmatrix} -7 \\ -3 \\ 1\\ 0 \end{pmatrix},\begin{pmatrix} -17 \\ -8 \\ 0\\ 1 \end{pmatrix}\}$

So basis for $U\cap W=\{-7-3x+x^2,-17 -8x +x^3\}$ and $dim(U\cap W)=2$

How to find $U+W$?

Answer 1

If $\begin{bmatrix} 0&1&3&7 \end {bmatrix}\mathbf u = 0$ then $\mathbf u$ is in $W$

let $u_1 = \begin{bmatrix} 1\\1\\0\\1 \end {bmatrix},u_2=\begin{bmatrix} -1\\0\\1\\0 \end {bmatrix},u_3=\begin{bmatrix} 1\\1\\1\\0 \end {bmatrix}$

$\begin{bmatrix} 0&1&3&7 \end {bmatrix}u_1 = 8, \begin{bmatrix} 0&1&3&7 \end {bmatrix}u_2 = 3,\begin{bmatrix} 0&1&3&7 \end {bmatrix}u_3 = 4$

$\begin{bmatrix} 0&1&3&7 \end {bmatrix}(u_1-2u_2) = 0$

We have found one vector in the intersection.

$(4u_2-3u_3)$ will also be in the intersection.

I will leave it to you to show that these are in fact independent.

Since there are vectors in $W$ that are not in $U\cap W$

$\text {rank} (U \cap W) < \text {rank} W$ and $\text {rank} W = 3$

$\text {rank} (U \cap W)\le 2$

We have found 2 independent vectors, so we must have a basis.