Question

Show that the vector space ${\mathbf{R}}_{\mathbf{n}}\left[x\right]$of exercise 4 has dimension $\mathbf{n}\mathbf{+}\mathbf{1}$ over R.

Short answer

${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$has $\mathrm{n}+1$ dimension.

Step-by-step solution

The Vector space:

If the vectors are linearly independent and every vector in vector space is a linear combination of the given set, then every set of elements in vector space V is called a basis.

A linearly independent spanning set is a basis.

To show that the vector space Rn[x] has dimension n+1 over R.

Assume that the set of polynomials be ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$and the degree of all polynomials in ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$be $\le \mathrm{n}$.

Now, to show that ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$has $\mathrm{n}+1$dimension.

Proof:

Take the polynomial $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$$1,\mathrm{x},....{\mathrm{x}}^{\mathrm{n}}$.

Now, it is required to show that the polynomial $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$is a basis of ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$.

As per linear independence,

Let ${\mathrm{a}}_0,{\mathrm{a}}_2,....{\mathrm{a}}_{\mathrm{n}}\in \mathrm{R}$considering that role="math" localid="1659152966173" ${\mathrm{a}}_0,{\mathrm{a}}_1\mathrm{x},....{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0$.

Now, role="math" localid="1659152979648" ${\mathrm{a}}_0,{\mathrm{a}}_1\mathrm{x},....{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0$is possible if all the coefficients of variables are zero.

Which implies that ${{\mathrm{a}}_{\mathrm{i}}}^{\text{'}}\mathrm{s}=0$.

Hence, $\left\{1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}\right\}$is linearly independent.

Now, $\mathrm{w}\in {\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$is of the form $\mathrm{w}={\mathrm{a}}_0,{\mathrm{a}}_1\mathrm{x},....{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ as any arbitrary element of ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$is of the form ${\mathrm{a}}_0,{\mathrm{a}}_1\mathrm{x},....{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0$and also ${\mathrm{a}}_0,{\mathrm{a}}_2,....{\mathrm{a}}_{\mathrm{n}}\in \mathrm{R}$.

Hence, $\mathrm{w}\in {\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$is of the form $\mathrm{w}={\mathrm{a}}_0,{\mathrm{a}}_1,....{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$.

As a result of which every element can be written in the form of $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$.

Therefore, $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$spans ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$,

So, $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$is a basis of ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$.

Thus, ${\mathrm{R}}_{\mathrm{n}}\left[\mathrm{x}\right]$has $\mathrm{n}+1$dimension as $1,\mathrm{x}...{\mathrm{x}}^{\mathrm{n}}$has $\mathrm{n}+1$ number of elements.