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The question asks: Find the "coordinates" of

$v=\begin{bmatrix} -2 & -2 \\ -2 & 4 \end{bmatrix}$

relative to the ordered basis, $F=(f_1, f_2, f_3, f_4)$ where

$f_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, f_2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, f_3 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, f_4 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

I am aware I am supposed to solve this problem using elimination, but at the moment, I am stuck.

I tried to transform these vectors into something I am familiar with. So, I transformed each vector to 4x1 making

$v = \begin{bmatrix} -2 \\ -2 \\ -2 \\ 4 \end{bmatrix}$

and I tried to extend it to each of the $f$ vectors, but elimination forces me to use row exchanges with $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0 \end{bmatrix}$, the columns corresponding to $f_1, f_2, f_3, f_4$ respectively.

At this point, I'm fairly certain that this problem cannot be this difficult to solve so I'm wondering where I did something wrong. I'm assuming the transformation of the 2x2 matrices into 4x1 is the primary culprit, but I cannot think of an alternative method to solving this problem. Is there a simpler way?

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I don't see the problem here. Take your $4 \times 4$ matrix, augment it with $v$, and start row reducing. –  wj32 Oct 13 '12 at 1:36
    
@wj32 I was under the false impression that row exchanges would alter my answer in some drastic way. I just tried it out and I got the exact same answer as EuYu. Thank you for your input! –  Goyatuzo Oct 13 '12 at 4:17

2 Answers 2

up vote 2 down vote accepted

Notice that your ordered basis is in fact an orthogonal basis under the standard inner product. In particular, that means we have $$v = \frac{\langle v,\ f_1\rangle}{2} f_1 + \frac{\langle v,\ f_2\rangle}{2} f_2 + \frac{\langle v,\ f_3\rangle}{2} f_3 + \frac{\langle v,\ f_4\rangle}{2} f_4$$ This gives you $$v = f_1 - 3f_2 -2f_3$$ so your coordinates are $$[v]_F = \begin{pmatrix}1\\-3\\-2\\0\end{pmatrix}$$

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Thank you! I was fearful that the answer I had gotten previously was incorrect because I did some row exchanges. I got the exact same answer as you in the general case as Gerry mentioned. It's good to know there are some shortcuts. –  Goyatuzo Oct 13 '12 at 4:18

"Find the co-ordinates of $v$ with respect to $F$" is just a fancy way of saying find the coefficients of $v$ as a linear combination of elements of $F$, so it's just asking for real numbers $a_1,a_2,a_3,a_4$ such that $$v=a_1f_1+a_2f_2+a_3f_3+a_4f_4$$ Two matrices are equal if and only if each entry in one equals the corresponding entry in the other, so this one matrix equation gives rise to four (linear) equations in the four unknowns $a_1,a_2,a_3,a_4$. So write down that system and solve it.

This approach has the advantage of always working, and not requiring you to remember anything about "transforming vectors." EuYu's approach has the advantage of working much faster, but only when your $F$ is an orthogonal set.

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