finding basis of 3x3 matrix

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for (a), I got

1 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 1
0 0 0
0 1 0
0 0 0
0 0 1
0 0 0
1 0 0
0 1 0
1 0 0
0 0 0
0 0 0
0 0 1
0 1 0
0 0 -1
0 0 0
1 0 0
0 1 0
-1 0 0
0 0 0
0 0 0
0 0 1
0 -1 0

Could you please check if this is correct?

I have no idea how to approach (b). Could you give me a hint?

Thanks,

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3 Answers

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For (b), use the standard basis (i.e. zeros everywhere and only a one in the i,j position) but add the identity matrix to all of them. This will give you invertible matrices (determinants are 1 or 2)and they form a basis.

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for (b), if you want to surprise your teacher with an unexpected (but rigorous) alternative solution, you can show that any random matrices work:

Pick 9 matrices with random coefficients in $[0,1]$ (with uniform probability for each coefficient), we can show this form a base.

First of all, being inversible is having a non-zero determinant. We want to show that $\det^{-1}(\{0\})$ has null measure in $M_{3,3}(\mathbb R)$, whose measure is the euclidean measure on $\mathbb R^9$. This is true because $\det$ is a polynomial, so the set of roots has measure $0$.

Therefore you have probability $0$ to pick a matrice with determinant zero, hence all your random matrices are inversible.

Secondly, you want to show that this is a base. Call $M_1,M_2,...,M_9$ your $9$ matrices, you just want to show that they are independant.

It is not hard to show that the measure of a sub-vectorial space is always $0$, therefore $span(M_1)$ has measure $0$ and $M_2$ has probability $0$ to be in it. Same goes for $span(M_1,M_2)$ so $M_3$ is not in it, and so on...

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To verify your answer to the first part indeed is a basis, we need to show the matrices are linearly independent (since this proves the matrices span a $9$-dimensional subspace of $M_{3 \times 3}(\mathbb{R})$, which is itself $9$-dimensional). If we write them as the rows of a $9 \times 9$ matrix, and compute its determinant, we find: $$\det\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 \\ \end{bmatrix}=-8 \neq 0$$ so they're linearly independent.

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