Show that one column is a linear combination of the other two

linear algebramatrices

I have the following matrix B. I would like to show that column may be expressed as a linear combination of the other two.

The first step I did was row reduce, but I still do not see one column as a linear combination. Can anyone provide hints? Or is there is a systematic way to solve this kind of problem?

$$ \left[
\begin{array}{ccc}
−1−3i&−8−10i&0−3i\\
−7−3i&−4−9i&−3−2i\\
11-3i& -16-12i&6-5i
\end{array}
\right] $$

In its RREF:

$$ \left[
\begin{array}{ccc}
11−3i&16−12i&6-5i\\
0&\frac{64}{13}-i\frac{190}{13}&\frac{96}{65}-i\frac{92}{65}\\
0& 0&0
\end{array}
\right] $$

Best Answer

Your reduced matrix is not in RREF yet. If we really want the coefficients that give the third vector as a linear combination of the first two, we could continue your reduction to find that the RREF is $$ \pmatrix{ 1 & 0 & \frac{4001}{7730} - \frac{597}{7730}i\\ 0 & 1 & \frac{218}{3865} + \frac{464}{3865}i\\ 0&0&0 }. $$ With that: if $x_1,x_2,x_3$ denote the columns of the original matrix, then we have $$ x_3 = \left( \frac{4001}{7730} - \frac{597}{7730}i\right)x_1 + \left( \frac{218}{3865} + \frac{464}{3865}i \right)x_2. $$

Related Solutions

[Math] Different results for row reduction in Matlab

This is already documented in MATLAB.

Roundoff errors may cause this algorithm to compute a different value for the rank than rank, orth and null.

You need to type format long e. After that you can see the difference if you execute N-eye(2) resulting with

N-eye(2)

ans =

   -5.000000000000004e-002    4.500000000000000e-001
    5.000000000000000e-002   -4.500000000000000e-001

Here, the trouble is already visible in the (1,1) element of the matrix. But also

 [1 1]*(N-eye(2))

ans =

   -4.163336342344337e-017    5.551115123125783e-017

gives you the error between seemingly identical elements.

The reason why you get correct results with an additional zero column is (my personal view) due to the term in the tolerance computation for rref given by (max(size(A))*eps *norm(A,inf)). Here, the first term is 3 instead of 2 and that should make the small difference between selecting a rank 1 and rank 2 matrix.

[Math] Finding the linear combinations of two vectors

What you did is actually incorrect, because you don't want to row-reduce
$$\left(\begin{array}{ccc|c}1 & 2 & 2 & 2\\ 2 & 3 & 4 & 4 \\ 3 & 1 & 5 & 5 \end{array}\right)$$ The problem with this is you're asking if there exist constants $C_1$, $C_2$, $C_3$ such that $C_1 a + C_2 b + C_3 c = c$, which is obvious (pick $C_3 = 1$, $C_1 = C_2 = 0$).

Instead, you want to row reduce the matrix $$\left(\begin{array}{cc|c}1 & 2 & 2\\ 2 & 3 & 4 \\ 3 & 1 & 5 \end{array}\right)$$ So in other words, you know that the vectors $a$ and $b$ are not linear multiples of each other, so you just want to check to see if you can write $c$ as a linear combination of $a$ and $b$, or you want to know if there exist $\alpha, \beta$ such that $\alpha a + \beta b = c$. (This would be the case of the first matrix)

Alternatively you could row reduce the matrix (this is most similar to what you have already done) $$\left(\begin{array}{ccc|c}1 & 2 & 2 & 0\\ 2 & 3 & 4 & 0 \\ 3 & 1 & 5 & 0 \end{array}\right)$$ This would be asking if there exist $C_1, C_2, C_3$ such that $C_1 a + C_2 b + C_3 c = 0$

It should be clear that these are equivalent methods. Once you have done the row-reduction correctly (just chop off the last column and replace it by a column of 0's and see what you find), then the answer should be fairly easy to see. I'll leave the rest to you.

By the way, if you find that the only possibility is $C_1 = C_2 = C_3 = 0$, this would mean that $c$ is not a linear combination of $a$ and $b$.

Best Answer

Related Solutions

[Math] Different results for row reduction in Matlab

[Math] Finding the linear combinations of two vectors

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