Linear Algebra – Vector Space with Matrices

linear algebramatricesvector-spaces

Given:
$ V_{A,B} = \{a_0A^4+a_1A^3+a_2A^2+a_3A+a_4B | a_i \in \mathbb{R}\} $

$A = \begin{pmatrix}-5 & 5\\\ 5 & -5\end{pmatrix}$, $B = \begin{pmatrix}-8 & 0\\\ 0 & 0\end{pmatrix}$

$M = \begin{pmatrix}10 & -10\\\ -10 & -6\end{pmatrix}$ , $N = \begin{pmatrix}2 & -10\\\ -10 & -6\end{pmatrix}$

1. Prove that $V_{A,B}$ is a vector space over $\mathbb{R}$.

2. Find $\dim V$ and 2 bases of $V_{A,B}$

3. If $M \in V_{A,B}$ / $N \in V_{A,B}$ find coordinates of $M$/$N$ relative to the bases found in 2.

My solution:

  1. (*)$ M_{2\times 2}(\Bbb R)$ is a vector space and $V_{A,B} \in M_{2\times 2}(\Bbb R)$.

    (**)

$u = (a_0A^4+a_1A^3+a_2A^2+a_3A+a_4B) \in V_{A,B}$, $v =
> (b_0A^4+b_1A^3+b_2A^2+b_3A+b_4B) \in V_{A,B}$, $$a_i, b_i \in \mathbb{R}$$ $$u + v \in V_{A,B}? $$
$$(a_0A^4+a_1A^3+a_2A^2+a_3A+a_4B+b_0A^4+b_1A^3+b_2A^2+b_3A+b_4B) \in V_{A,B}?$$
$$((a_0+b_0)A^4+(a_1+b_1)A^3+(a_2+b_2)A^2+(a_3+b_3)A+(a_4+b_4)B)\in
V_{A,B}$$

(***)

$\lambda \in \mathbb{R}$, $v = (b_0A^4+b_1A^3+b_2A^2+b_3A+b_4B) \in V_{A,B}$

$$\lambda v \in V_{A,B} ?$$

$$(\lambda (b_0A^4+b_1A^3+b_2A^2+b_3A+b_4B)) \in V_{A,B}?$$
$$((\lambda b_0) A^4 + (\lambda b_1) A^3 + (\lambda b_2) A^2 + (\lambda b_3) A + (\lambda b_4) B) \in V_{A,B}?$$ $$\lambda a_i \in \mathbb{R}, \lambda a_i = b_i, b_i \in \mathbb{R} => v = (b_0A^4+b_1A^3+b_2A^2+b_3A+b_4B) \in V_{A,B}$$

From (*), (**) and (***)$\implies$ $V_{A,B}$ is a vector space over $\Bbb R$.

  1. $\dim V = 2$ I think because the number of vectors in $V_{A,B}$ is 2 since the matrices are $2\times 2$.
    (I'm not sure if the vectors should be row or column vectors but I made them for row) The vectors should be:
    $$[(5000a_0 – 500a_1 – 5a_2 – 8a_3), (-5000a_0 + 500a_1 + 5a_2)]$$
    $$[(-5000a_0 + 500a_1+ 5a_2), (5000a_0 – 500a_1 – 5a_2)]$$.

I'm not sure how to solve the rest of 2. and 3. Any help is appreciated!

Best Answer

$2.$

$\dim V = 2$ I think because the number of vectors in $V_{A,B}$ is 2 since the matrices are 2x2.

Unfortunately, this does not make sense. The dimension of a finite-dimensional vector space is the number of elements of any basis of the vector space.

The set $\{A^4,A^3,A^2,A,B\}$ spans $V_{A,B}$, but is not a basis for $V_{A,B}$ because the set is not linearly independent. Indeed, $$A^2=-10A \ ; \ A^3=100A \ ; \ A^4=-1000A.$$ Substituting this into $ V_{A,B} = \{a_0A^4+a_1A^3+a_2A^2+a_3A+a_4B | a_i \in \mathbb{R}\} $, we have $$(*) \ \ \ \ \ V_{A,B} = \{(-1000a_0+100a_1-10a_2+a_3)A+a_4B | a_i \in \mathbb{R}\}=\{\alpha A+\beta B |\alpha,\beta \in \mathbb{R}\}.$$ Since $\{A,B\}$ is linearly independent and spans $V_{A,B}$, then the set $$\left\{A=\begin{pmatrix}-5&5\\5&-5\end{pmatrix},B=\begin{pmatrix}-8&0\\0&0\end{pmatrix}\right\}$$ is a basis for $V_{A,B}$. Since the number of elements of such basis is $2$, then $\dim V_{A,B}=2$. I will leave it up to you to find another basis for $V_{A,B}$.

$3.$ Note that $M\notin V_{A,B}$, because there is no $\alpha,\beta\in\Bbb R$ such that $$M=\alpha A+\beta B.$$ I will leave it up to you to solve the case of the matrix $N$.

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