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We observe the subspace known as $U\subset \mathbb{R}^{3\times 3}$ that is spanned by the following vectors:

$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} $, $ \begin{bmatrix} 0 & -3 & 0 \\ 0 & 2 & 0 \\ 0 & -1 & 0 \\ \end{bmatrix} $, $ \begin{bmatrix} 0 & 0 & 1 \\ 0 & -2 & 0 \\ 3 & 0 & 0 \\ \end{bmatrix} $, $ \begin{bmatrix} 0 & 0 & 0 \\ -1 & 2 & -3 \\ 0 & 0 & 0 \\ \end{bmatrix} $.

Choose a basis for $U$ and show that $ \begin{bmatrix} 2 & -3 & -2 \\ -3 & 8 & -9 \\ -6 & -1 & 6 \\ \end{bmatrix} \in U$.

Determine the coordinate vector for this vector with respect to the chosen basis for $U$.


Hints to how I solve this?

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We have:

$$ (1) \qquad a \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} +b \begin{bmatrix} 0 & -3 & 0 \\ 0 & 2 & 0 \\ 0 & -1 & 0 \\ \end{bmatrix} +c \begin{bmatrix} 0 & 0 & 1 \\ 0 & -2 & 0 \\ 3 & 0 & 0 \\ \end{bmatrix} +d \begin{bmatrix} 0 & 0 & 0 \\ -1 & 2 & -3 \\ 0 & 0 & 0 \\ \end{bmatrix} =\begin{bmatrix} a & -3b & c \\ -d & -2(a-b+c-d) & -3d \\ 3c & -b & 3a \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ iff $ a=b=c=d=0$. So these matrices are linearly independent and they are a basis $\mathcal{U}$ fo $U$.

And, for $(a,b,c,d)=(2,1,-2,3)$ the linear combination $(1)$ gives the matrix $$M= \begin{bmatrix} 2 & -3 & -2 \\ -3 & 8 & -9 \\ -6 & -1 & 6 \\ \end{bmatrix} $$ so $(2,1,-2,3)$ are the components of this matrix in the basis $\mathcal{U}$.

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Well, a basis is a set of vectors that spans the vector space and is linear independent. It's clear that these vectors span the subspace U, so your job is to determine whether these matrices are linear independent or not. If they are linear dependent, you can write one matrix as linear combination of the other matrices and leave it out of the set such that the remaining set still spans U. Then, you have to see whether the new set is linear dependent. If so, you have found a basis, if not, you can repeat this procedure.

To show that a matrix is an element of the subspace U, show that it can be written as linear combination of basis vectors of U (or other vectors of U, but you will need to write it as linear combination of basis vectors for the next question anyways.

I'm not to sure what you mean with coordinate vector.

EDIT: To find the coordinate vector, write the matrix as linear combination of the basis you found and the scalars that belong with each basis vector will be the elements (in the same order as the basis vectors) in the coordinate vector.

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  • $\begingroup$ The coordinate vector for a vector $v$ with respect to a basis $\{\,u_1,\dots,u_n\,\}$ is the vector $(c_1,\dots,c_n)$ such that $v=\sum_ic_iu_i$. $\endgroup$ – Gerry Myerson Dec 2 '16 at 22:12
  • $\begingroup$ Ah thanks, I edited my answer so the OP sees this. $\endgroup$ – user370967 Dec 2 '16 at 22:15

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