# Basis and Dimension of Vector Spaces

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• May 18th 2009, 03:47 PM
Maccaman
Basis and Dimension of Vector Spaces
Give a basis and the dimension of each of the following vector spaces...

(a) The space of 3 × 3 matrices which are invarient under a 90-degree clockwise rotation; that is, the matrices satisfying
$\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\end{pmatrix} = \begin{pmatrix} g&d&a\\h&e&b\\i&f&c\end{pmatrix}$

(b) The space of polynomials in $P_n(\mathbb{R}) (n\geq 2)$ which are divisible by $x^2 +1$. (i.e. they can be written as the product of $x^2 + 1$ with another polynomial).
• May 18th 2009, 06:24 PM
NonCommAlg
Quote:

Originally Posted by Maccaman
Give a basis and the dimension of each of the following vector spaces...

(a) The space of 3 × 3 matrices which are invarient under a 90-degree clockwise rotation; that is, the matrices satisfying
$\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\end{pmatrix} = \begin{pmatrix} g&d&a\\h&e&b\\i&f&c\end{pmatrix}$

so we have $a=c=g=i$ and $b=d=f=h.$ so an elment of your vector space is in the form: $aX + bY + eZ,$ where:

$X=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \ \ Y=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},$ and $Z=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$ so the dimension is 3 and a basis for the space is $\{X,Y,Z \}.$

Quote:

(b) The space of polynomials in $P_n(\mathbb{R}) (n\geq 2)$ which are divisible by $x^2 +1$. (i.e. they can be written as the product of $x^2 + 1$ with another polynomial).
an element of of your vector space here is in the form $(x^2+1)(c_{n-2}x^{n-2} + \cdots + c_1 x + c_0)= c_{n-2}(x^n + x^{n-2}) + \cdots + c_1(x^3+x) + c_0(x^2+1).$ it's clear now that the

dimension of your space is $n-1$ and a basis is $\{x^n + x^{n-2}, \cdots , x^3 + x , x^2 + 1 \}.$
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