7.3 Basis and Dimension

Problem 1: Determine if $\mathcal{V}$ is a basis for $V$

a.

$\mathcal{V}=\left\{1+t,t+t^2,1-t,1-t^2\right\},V={\mathbf{P}}^2$

Solution:
${\mathbf{P}}^2$ has dimension 3. Since $\mathcal{V}$ contains 4 vectors, it is linearly dependent and cannot be a basis.

b.

$\mathcal{V}=\left\{1+t,t+t^2,1-t\right\},V={\mathbf{P}}^2$

Solution:
Check linear independence:

$$c_1(1+t)+c_2(t+t^2)+c_3(1-t)=0$$ $$(c_1+c_3)+(c_1+c_2-c_3)t+c_2{t}^2=0$$

This gives:

$$\{\begin{array}{l}{c}_1+c_3=0\\ c_1+c_2-c_3=0\\ c_2=0\end{array}$$

Only solution is $c_1=c_2=c_3=0$. The set is linearly independent and has 3 vectors, so it’s a basis.

c.

$$\mathcal{V}=\left\{\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\end{array}\right],\left[\begin{array}{ccc}0& 1& -1\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& -1\end{array}\right]\right\},V={\mathbf{P}}^{2\times 3}$$

Solution:
${\mathbf{P}}^{2\times 3}$ has dimension 6. $\mathcal{V}$ only has 3 vectors, so it cannot span the space and thus isn’t a basis.

d.

$$\mathcal{V}=\left\{\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right],\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{ll}1& 1\\ 1& 1\end{array}\right]\right\},V={\mathbf{R}}^{2\times 2}$$

Solution:
${\mathbf{R}}^{2\times 2}$ has dimension 4. $\mathcal{V}$ has only 3 vectors, so it cannot be a basis.

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Problem 2: Find a basis for the subspace of ${\mathbf{R}}^{2\times 2}$ where $a_{11}-a_{22}=0$

Solution:
General form:

$$\left[\begin{array}{cc}a& b\\ c& a\end{array}\right]=a\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+b\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]+c\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$$

Basis:

$$\left\{\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right\}$$

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Problem 3: Find a basis for polynomials in ${\mathbf{P}}^3$ with $p(-1)=p(1)=0$

Solution:
Conditions:

$$a_0-a_1+a_2-a_3=0$$ $$a_0+a_1+a_2+a_3=0$$

Solving gives $a_0=-a_2$ and $a_1=-a_3$. General form:

$$p(t)=-a_2+(-a_3)t+a_2{t}^2+a_3{t}^3=a_2(t^2-1)+a_3(t^3-t)$$

Basis:

$$\left\{t^2-1,t^3-t\right\}$$

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Problem 4: Remove vectors from $\mathcal{V}$ to yield a basis for ${\mathbf{P}}^2$

$$\mathcal{V}=\left\{1+t,1-t,1+2t,t-t^2,1+t+t^2\right\}$$

Solution:

1. Start with $\{1+t,1-t\}$ (linearly independent)
2. Add $1+t+t^2$ (not in span of first two)
3. We now have 3 linearly independent vectors, which is sufficient since $\dim ({\mathbf{P}}^2)=3$.

Final basis:

$$\left\{1+t,1-t,1+t+t^2\right\}$$

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Problem 5: Prove $\dim (T(m,n))=mn$

Proof:
$T(m,n)$ is the space of $m\times n$ matrices. The standard basis consists of all matrices $E_{ij}$ with 1 in the $(i,j)$-entry and 0 elsewhere. There are $mn$ such matrices, hence $\dim (T(m,n))=mn$.

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Problem 6: Prove scaled basis is still a basis

Proof:
Given basis $\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_k\}$, consider $\{{\mathbf{v}}_1,2{\mathbf{v}}_2,\dots ,k{\mathbf{v}}_k\}$.

1. Linear Independence:
   If ${\sum }_{i=1}^k{c}_i(i{\mathbf{v}}_i)=0$, then $\sum (i{c}_i){\mathbf{v}}_i=0$. Since $\{{\mathbf{v}}_i\}$ is linearly independent, $i{c}_i=0⟹c_i=0$.

2. Spanning:
   For any $\mathbf{v}\in V$, $\mathbf{v}=\sum a_i{\mathbf{v}}_i=\sum \left(\frac{a_i}{i}\right)(i{\mathbf{v}}_i)$. Thus, the scaled set spans $V$.

Therefore, the scaled set is also a basis.